matematika - fizika - halapaPri gibanju tijela po kružnici brzina v stalno se mijenja. Ona ostaje jednaka veli činom, ali joj se neprestano mijenja smjer. Linearna ili obodna brzina

1

Zadatak 101 (Ivan gimnazija) Koliki moraju biti polumjer kružne staze umjetnog Zemljina satelita i njegova brzina da

njegov period bude jednak periodu okretanja Zemlje tj da se sa Zemlje čini nepomičnim

(masa Zemlje m = 6 middot 1024

kg gravitacijska konstanta G = 667 middot 10-11

N middot m2 kg

2)

Rješenje 101 TS = TZ = T = 24 middot 60 middot 60 = 86400 s m = 6 middot 1024 kg G = 667 middot 10-11 N middot m2 kg2

mS masa satelita r = v =

Opći zakon gravitacije

Ako se bilo koja dva tijela masa m1 i m2 nalaze u međusobnoj udaljenosti r među njima djeluje

privlačna gravitacijska sila čiji je iznos

1 22

m mF G

r

sdot= sdot

gdje je G gravitacijska konstanta koja ima jednaku vrijednost za privlačenje između bilo koja dva

tijela

Da bi se tijelo gibalo po kružnici potrebno je da na nj djeluje centripetalna sila

2v

F mcpr

= sdot

koja ima smjer prema središtu kružnice

Pri gibanju tijela po kružnici brzina v stalno se mijenja Ona ostaje jednaka veličinom ali joj se

neprestano mijenja smjer Linearna ili obodna brzina v računa se po formuli

v 2 r

T

πsdot sdot=

gdje je r polumjer kružnice T perioda tj vrijeme jednog ophoda

Sila gravitacije između satelita mase mS i Zemlje mase m na udaljenosti r mora biti jednaka

centripetalnoj sili na satelit na udaljenosti r od središta vrtnje

2 24 4

2 2

2 2

42 22

m m m mr rS r T

mS

SF F m G m Gcp g S ST r T r

π π

π

sdotsdot sdotsdot sdot sdot sdot= rArr sdot = sdot rArr sdot

sdot

sdot

sdot

sdot = rArr

32 2 23 3 3

2 2 24

4 4

m T m T m Tr G r G r G

π π π

sdot sdot sdotrArr = sdot rArr = sdot rArr = sdot =

sdot sdot sdot

( )2242

6 10 86 40011 73 667 10 423 10 2 2

4

kg sN mm

kg π

sdot sdotsdotminus= sdot sdot = sdot

sdot

Brzina satelita da njegov period bude jednak periodu okretanja Zemlje iznosi

72 2 423 10

307614 86 400

r m mv

T s s

π πsdot sdot sdot sdot sdot= = =

Vježba 101 Koliki mora biti polumjer kružne staze umjetnog Zemljina satelita da njegov period bude

jednak periodu okretanja Zemlje tj da se sa Zemlje čini nepomičnim

(masa Zemlje m = 6 middot 1021

t gravitacijska konstanta G = 667 middot 10-11

N middot m2 kg

2)

Rezultat 423 107 m

Zadatak 102 (Ana gimnazija) Kolikom silom Mars privlači kamen mase 1 kg koji se nalazina na njegovoj površini Masa

Marsa je 65 1023

kg a polumjer 3400 km (gravitacijska konstanta G = 667 middot 10-11

N middot m2 kg

2)

2

Rješenje 102 m = 1 kg mM = 65 middot 10

23 kg r = 3400 km = 34 10

6 m

G = 667 middot 10-11

N middot m2 kg

2 F =




1 22

m mF G

r

sdot= sdot


tijela

Na površini Marsa privlačna sila između tijela mase m i Marsa mase mM iznosi

( )

2 231 65 1011

667 10 375 2 2 2

634 10

m m N m kg kgMF G N

r kgm

sdot sdot sdot sdotminus= sdot = sdot sdot =

sdot

Vježba 102 Kolikom silom Mars privlači kamen mase 100 dag koji se nalazina na njegovoj površini

Masa Marsa je 65 1021

t a polumjer 3400 km (gravitacijska konstanta G = 667 middot 10-11

N middot m2 kg

2)

Rezultat 375 N

Zadatak 103 (Ana gimnazija) Masa Marsa je 65 10

23 kg a polumjer 3400 km Kolika je akceleracija slobodnoga pada na

površini Marsa (gravitacijska konstanta G = 667 middot 10-11

N middot m2 kg

2)

Rješenje 103 mM = 65 1023 kg r = 3400 km = 34 middot 106 m G = 667 middot 10-11 N middot m2 kg2 g =




1 22

m mF G

r

sdot= sdot


tijela

Silu kojom Zemlja privlači sva tijela nazivamo silom težom Pod djelovanjem sile teže sva tijela

padaju na Zemlju ili pritišću na njezinu površinu

Akceleracija kojom tijela padaju na Zemlju naziva se akceleracijom slobodnog pada Prema drugom

Newtonovom poučku

G m g= sdot

gdje je G sila teža m masa tijela i g akceleracija slobodnog pada koja je za sva tijela na istome mjestu

na Zemlji jednaka

Gravitacijska sila privlačenja tijela i Marsa djeluje prema trećem Newtonovu zakonu na tijelo i Mars

jednako samo u suprotnim smjerovima Sila teža na Marsu je naziv za gravitacijsku silu kojom Mars privlači sva tijela u blizini svoje površine Za privlačenje nekog tijela mase m i Marsa mase mM

možemo napisati

2 2 2

m m m m m

M M MG F m g G m g G m g G

r r r

sdot sdot= rArr sdot = sdot rArr sdot = sdot rArr = sdot =

3

( )

2 2365 1011

667 10 375 2 2 2

634 10

N m kg m

kg sm

sdot sdotminus= sdot sdot =

sdot

Vježba 103 Masa Marsa je 65 10

20 t a polumjer 3400 km Kolika je akceleracija slobodnoga pada na

površini Marsa (gravitacijska konstanta G = 667 middot 10-11 N middot m2 kg2)

Rezultat 375 ms2

Zadatak 104 (Ivana gimnazija)

Brzina satelita u orbiti oko Zemlje neovisna je o

A masi satelita

B masi Zemlje

C udaljenosti satelita od površine Zemlje

D gravitacijskoj konstanti

Rješenje 104 m ndash masa satelita M ndash masa Zemlje r ndash udaljenost satelita od površine Zemlje

G ndash gravitacijska konstanta




1 22

m mF G

r

sdot= sdot


tijela


2v

F mcpr

= sdot


Neka je m masa satelita a M masa Zemlje Budući da ulogu centripetalne sile ima sila gravitacije

vrijedi

2 2

2

2 2

v m M v m M MF F m G m G v Gcp g

r r rr

r

mr

sdot sdot= rArr sdot = sdot rArr sdot = sdot rArr = sdotsdot rArr

2

M M

v G v Gr r

rArr = sdot rArr = sdot

Iz formule vidi se da brzina satelita ovisi isključivo o njegovoj udaljenosti od središta Zemlje a ne o

njegovoj masi Odgovor je pod A

mr

Vježba 104 Brzina satelita u orbiti oko Zemlje ovisna je o

A masi satelita

B težini satelita


D veličini satelita

Rezultat C

4

Zadatak 105 (Josipa medicinska škola)

Polumjer Zemljine putanje oko Sunca je 390 puta veći od polumjera Mjesečeve putanje oko

Zemlje Mjesec obiđe Zemlju približno 13 puta u godini dana Koliki je omjer brzine kruženja

Zemlje oko Sunca (vz) i brzine kruženja Mjeseca oko Zemlje (vm)

1 13 13 1 1 30 30 1A v v B v v C v v D v vz m z m z m z m= = = =

Rješenje 105

rzs = 390 middot rmz 1

13

T godm = Tz = 1 god vz vm =

Tijelo rotira kada se njegove čestice gibaju po kružnicama čija središta leže u istoj točki ili na istom

pravcu Kada kruto tijelo rotira oko čvrste osi sve se njegove čestice gibaju po koncentričnim

kružnicama (koncentrične kružnice imaju zajedničko središte) Obodna (linearna) brzina iznosi

v 2 r

T

πsdot sdot=

gdje je r polumjer kružnice T perioda (ophodno vrijeme vrijeme jednog okreta)

Brzina kruženja

bull Zemlje oko Sunca dana je formulom

2v

rzsz

Tz

πsdot sdot=

bull Mjeseca oko Zemlje dana je formulom

2v

rmzm

Tm

πsdot sdot=

Gledamo omjer brzina

2 1390

v v v v v 132v v v2 v v 1

2r r rzs zs zsr godmzr TT T T zs mz z z z z z z z

r r r r T r godmz mz mz zm m m m mz m mzT T Tm m m

π π

ππ

sdot sdot sdotsdot sdot

sdot= rArr = rArr = rArr = rArr = rArr

sdot sdotsdotsdot

sdot

sdot sdot

1390

v v v390 3013 v v 30 1v 1 v 13 v 1

z zr godmz

r god

zz

mm mzm

m

sdot sdot

rArr = rArr = rArr = rArr =sdot

Odgovor je pod D

rmzrzs

Vježba 105 Polumjer Zemljine putanje oko Sunca je 390 puta veći od polumjera Mjesečeve putanje oko


Mjeseca oko Zemlje (vm) i brzine kruženja Zemlje oko Sunca (vz)

1 13 13 1 1 30 30 1A v v B v v C v v D v vm z m z m z m z= = = =

Rezultat C

5

Zadatak 106 (Iva srednja škola)


A masi satelita B masi Zemlje

tanC udaljenosti satelita od površine Zemlje D gravitacijskoj kons ti

Rješenje 106




1 22

m mF G

r

sdot= sdot


tijela


2v

F mcpr

= sdot


Budući da satelit mase m kruži oko Zemlje mase M na udaljenosti r ulogu centripetalne sile Fcp ima

gravitacijska Fg pa vrijedi

2 2

2

2 2


r r rr

r

mr


2

M M

v G v Gr r


Iz formule vidi se da brzina satelita ovisi isključivo o njegovoj udaljenosti od središta Zemlje masi

Zemlje i gravitacijskoj konstanti a ne o njegovoj masi

Odgovor je pod A

Vježba 106

Newtonov zakon gravitacije vrijedi

A za tijela velikih masa B za tijela malih masa

C za dvije mase na bilo kojoj udaljenosti D za dvije mase na velikoj udaljenosti

Rezultat C

Zadatak 107 (Luka tehnička škola)

Kolika je masa Sunca kad znamo da je srednja brzina Zemlje pri kruženju oko Sunca 30 kms

a polumjer njezine staze 15 middot 108 km (gravitacijska konstanta G = 667 middot 10-11 N middot m2 kg2)

Rješenje 107

v = 30 kms = 3 middot 104 ms r = 15 108 km = 15 middot 1011 m G = 667 middot 10-11 N middot m2 kg2

mS =

Da bi se tijelo mase m gibalo po kružnici polumjera r potrebno je da na nj djeluje centripetalna sila

2v

F mcpr

= sdot

gdje je v obodna brzina tijela




6

1 22

m mF G

r

sdot= sdot


tijela Taj zakon zovemo općim zakonom gravitacije

Sila gravitacije je uzrok kružnoga gibanja Zemlje oko Sunca Sila gravitacije Fg između Zemlje mase

m i Sunca mase mS na udaljenosti r mora biti jednaka centripetalnoj sili Fcp na Zemlju na udaljenosti r

od središta vrtnje

2

2

2 2

2

m m m mv vS SF F G m G mg cmrr G

prr

rsdot sdot= rArr sdot = sdot rArr sdot = sdot sdot

sdotrArr

24 11

3 10 15 10230

202 10 2

11667 10

2

mm

v r sm kg

S G N m

kg

sdot sdot sdotsdot

rArr = = = sdot

sdotminussdot

vvvv

rrrrmmmmmmmmSSSS

Vježba 107

Kolika je masa Sunca kad znamo da je srednja brzina Zemlje pri kruženju oko Sunca

1800 kmmin a polumjer njezine staze 15 middot 108 km (gravitacijska konstanta

G = 667 middot 10-11 N middot m2 kg2)

Rezultat 202 middot 1030

kg

Zadatak 108 (Tina gimnazija)

Znajući srednji polumjer Zemlje R = 6371 km odredi visinu na kojoj ubrzanje iznosi 60

ubrzanja g0 uz površinu

Rješenje 108

R = 6371 km = 6371 middot 106 m g0 g = 60 middot g0 = 060 middot g0 h =




1 22

m mF G

r

sdot= sdot



Težina tijela jest sila kojom tijelo zbog Zemljina privlačenja djeluje na horizontalnu podlogu ili ovjes

Za slučaj kad tijelo i podloga odnosno ovjes miruju ili se gibaju jednoliko po pravcu s obzirom na

Zemlju težina tijela je veličinom jednaka sili teže

G m g= sdot

Na tijelo mase m koje se nalazi u blizini Zemljine površine djeluje vertikalno prema dolje sila teža

G = m middot g koja je rezultanta gravitacijske i centrifugalne sile zbog vrtnje Zemlje oko svoje osi U

većini slučajeva može se zanemariti utjecaj centrifugalne sile i uzeti da je sila teža jednaka

gravitacijskoj sili Ako je m masa tijela g0 ubrzanje sile teže na površini Zemlje M masa Zemlje R

srednji polumjer Zemlje g ubrzanje sile teže na visini h od površine Zemlje tada je

7

bull na površini Zemlje

0 2

m Mm g G

R

sdotsdot = sdot

bull na visini h

( )

2

m Mm g G

R h

sdotsdot = sdot

+

Iz ovih dviju relacija dobivamo

( ) ( ) ( )

podijelimo

jednadžbe

0 2 2 20 0

2 2 2

m M m Mm g G Gm g gR R R

m M m Mm g gm g G G

R h R h

m

R h

MG

m

m MmG

sdot sdotsdot = sdot sdot sdotsdot sdot

rArr rArr = rArr = rArrsdot sdotsdot sdot

sdot = sdot sdot sdot

+ + +

sdot

sdot

( )

( )1

2 2 220 0 0 0

21

2

g g g gR h R h R hR

g g g R R gR

R h

+ + +rArr = rArr = rArr = rArr = rArr

+

20 0 0 0

g g g gR h R h R hR hR R

R g R g R g g

+ + +rArr = rArr = rArr = rArr + = sdot rArrsdot

0 0 01 1 1060 0

0

60 0

0

g g gh R R h R h

g gg

gR h R

grArr = sdot minus rArr = sdot minus rArr = sdot minus rArr = sdot minus rArr

sdot sdot

1 1 161 1 6371 10 1 185392563

060 060 060h R h R m mrArr = sdot minus rArr = sdot minus = sdot sdot minus = =

185392563 1854 km km= asymp

mmmmmmmmMMMMMMMM

gggg0000 gggg

hhhh

Vježba 108 Znajući srednji polumjer Zemlje R = 6371 km odredi visinu na kojoj ubrzanje iznosi 80


Rezultat 752 km

Zadatak 109 (Neven tehnička škola)

Gdje se između Mjeseca i Zemlje poništavaju njihova gravitacijska polja Masa Mjeseca je

oko 81 put manja od mase Zemlje a udaljenost Zemlje od Mjeseca iznosi d = 384 000 km

Rješenje 109

M ndash masa Zemlje m ndash masa Mjeseca M = 81 middot m d = 384 000 km = 384 middot 108 m

x =

Prostor oko masivnog tijela mase m u kojem se osjeća gravitacijsko djelovanje tog tijela nazivamo

gravitacijsko polje Jakost gravitacijskog polja γ u nekoj točki polja na udaljenosti r od tijela dana je

izrazom

8

2

mG

r

γ = sdot


tijela




1 22

m mF G

r

sdot= sdot



γγγγMMMMγγγγZZZZ PPPP

dddd

Mjesec - mMjesec - mMjesec - mMjesec - mZemlja - MZemlja - MZemlja - MZemlja - M

d - xd - xd - xd - xxxxx

C

Najprije odredimo gravitacijska polja Zemlje i Mjeseca u točki P

bull gravitacijsko polje Zemlje

2

MG

Zx

γ = sdot

bull gravitacijsko polje Mjeseca

( )

2

mG

Md x

γ = sdot

minus

1inačica

Gravitacijsko polje Zemlje usmjereno je radijalno prema Zemlji a gravitacijsko polje Mjeseca prema

Mjesecu Budući da se u točki P polja moraju poništiti (rezultantno gravitacijsko polje mora biti

jednako nuli) u njoj su ona jednakog iznosa ali suprotnih smjerova pa vrijedi

( ) ( )

( )2 2 2 2

22

m M m M

G G G GM Z

x xd x d

x d x

Gx

γ γ= rArr sdot = sdot rArr sdot =sdot minus

sdotsdot rArrminus minus

( ) ( ) ( )2 2 22 2 21

m

m x M d x m x M d x x xMM

drArr sdot = sdot minus rArr sdot = sdot minus rArr sdot = minussdot rArr

( )radiciramo jednadžbu jer time

izbjegavamo kvadratnu jednadžbu

22mx d x

MrArr rArr sdot = minus rArr

( )22

1m m m m

x d x x d x x x d x dM M M M

rArr sdot = minus rArr sdot = minus rArr sdot + = rArr sdot + = rArr

9

1

1

1 1m m

x d x dM

M

M mrArr sdot + = rArr sdot + = sdot

+

rArr

11 1 1 1

81 81 81

d d d dx x x

m

m

xm m

M m

rArr = rArr = rArr = rArr = rArr

+ + + +sdot sdot

911 1 1 9 1 10 10 10

19 1 9 9 9 9

d

d d d d dx x x x x x

sdotrArr = rArr = rArr = rArr = rArr = rArr = =

++ +

89 384 10 8

3456 10 10

mm

sdot sdot= = sdot

2inačica




( ) ( )

( )2 2 2 2

22

m M m M

G G G GM Z

x xd x d

x d x

Gx



( ) ( ) ( )2 22 2 2 2 22m x M d x M d x m x M d d x x m xrArr sdot = sdot minus rArr sdot minus = sdot rArr sdot minus sdot sdot + = sdot rArr

2 2 2 2 2 22 2 0M d M d x M x m x M d M d x M x m xrArr sdot minus sdot sdot sdot + sdot = sdot rArr sdot minus sdot sdot sdot + sdot minus sdot = rArr

( )2 2 2 2 2

2 0 2 0M x m x M d x M d M m x M d x M drArr sdot minus sdot minus sdot sdot sdot + sdot = rArr minus sdot minus sdot sdot sdot + sdot = rArr

( )2

2 2 22 0

242

2 12 2

a M m b M d c M dM m x M d x M d

b b a ca M m b M d c M d x

a

= minus = minus sdot sdot = sdotminus sdot minus sdot sdot sdot + sdot =

rArr rArr rArrminus plusmn minus sdot sdot

= minus = minus sdot sdot = sdot =sdot

( ) ( ) ( )

( )

2 22 2 4

12 2

M d M d M m M dx

M m

minus minus sdot sdot plusmn minus sdot sdot minus sdot minus sdot sdotrArr = rArr

sdot minus

( )

2 2 2 2 22 4 4 4

12 2

M d M d M d M m dx

M m

sdot sdot plusmn sdot sdot minus sdot sdot + sdot sdot sdotrArr = rArr

sdot minus

( ) ( )

2 2 2 2 22 4 2 4

12 12

4

2 2

24M d M m d M d M mM d

x xM m

d M d

M m

sdot sdot plusmn + sdot sdot sdotsdot sdot minus sdot sdot sdot sdot plusmn sdot sdot sdotrArr = rArr = rArr

sdot minus sdot minus

( )

( )( )

( )( )

22 2

12 12 122 2

2

2

d M M m d M M mM d d M mx x x

M m M m M m

sdot sdot plusmn sdot sdot sdot plusmn sdotsdot sdot plusmn sdot sdot sdotrArr = rArr = rArr = rArr

sdot minus sdot minus sdot minus

10

81 81

1 1 81

1281 81

2 2 81

M M m m m mx d x d

M m m mM M mx d

M m M M m m m mx d x d

M m m m

+ sdot sdot + sdot sdot= sdot = sdot

minus sdot minusplusmn sdotrArr = sdot rArr rArr rArr

minus minus sdot sdot minus sdot sdot= sdot = sdot

minus sdot minus

281 81 81 9 90

1 1 180 80 80

81 9 72281 81

2 280 802 80

m m m m mx d x d x d

m m m

m m mm m x d x d

x d m mm

sdot + sdot sdot + sdot sdot= sdot = sdot = sdot

sdot sdot sdotrArr rArr rArr rArr

sdot minus sdot sdotsdot minus sdot = sdot = sdot

= sdot sdot sdotsdot

9 9 8 8384 10 432 101 1 18 8 1

89 9 8 3456 10384 10 22

90

80

2 210 10

72

80

x d x d xm

mm x m

x mx d x d x mm

m

= sdot = sdot = sdot sdot = sdotrArr rArr rArr rArr rArr

= sdot= sdot = sdot

sdot

sdot

sdotsdot sdot

sdot=

rješenje odbacujemo jer je točka P izvan spojnice Zemlja Mjese1

8

3456 102

cx

x mrArr

= sdot

minusminus

xxxx d - xd - xd - xd - x

Zemlja - MZemlja - MZemlja - MZemlja - MMjesec - mMjesec - mMjesec - mMjesec - m

dddd

mmmm1111FFFFgZgZgZgZ FFFFgMgMgMgM

G

3inačica

Kada se neko tijelo mase m1 nađe u točki P bit će u bestežinskom stanju jer će na nj djelovati

gravitacijske sile Zemlje i Mjeseca koje su jednake po iznosu ali imaju suprotne smjerove

Gravitacijska sila kojom Zemlja mase M djeluje na tijelo mase m1 dana je izrazom

1 2

M mF G

gZx

sdot= sdot

Gravitacijska sila kojom Mjesec mase m djeluje na tijelo mase m1 dana je izrazom

( )

1 2

m mF G

gMd x

sdot= sdot

minus

Budući da tijelo mora biti u stanju mirovanja gravitacijske sile FgZ i FgM jednake su po iznosu ali

suprotnog smjera pa vrijedi

( ) ( )

( )1 1 1 12 2

2

2 2

2

1

m m M m m m M mF F G G G G

gM gZx xd x d x

x d x

G m

sdot sdot sdot sdot= rArr sdot = sdot rArr sdot =

minussdot sdot

minus sdotminus

sdotrArr

( ) ( ) ( )2 2 22 2 21

m





22mx d x


11

( )22m m m

x d x x d x x x dM M M

rArr sdot = minus rArr sdot = minus rArr sdot + = rArr

1

1 1 11

m m m

x d x d x dM M M m

M

rArr sdot + = rArr sdot + = sdot + rArrsdot=

+

rArr

11 1 1 1

81 81 81

d d d dx x x

m

m

xm m

M m


+ + + +sdot sdot

911 1 1 9 1 10 10 10

19 1 9 9 9 9

d



++ +

89 384 10 8

3456 10 10

mm

sdot sdot= = sdot

4inačica




1 2

M mF G

gZx

sdot= sdot


( )

1 2

m mF G

gMd x

sdot= sdot

minus



( ) ( )

( )1 1 1 12 2

2

2 2

2

1


gM gZx xd x d x

x d x

G m


minussdot sdot

minus sdotminus

sdotrArr




( )2 2 2 2 2


( )2

2 2 22 0

242

2 12 2



a




( ) ( ) ( )

( )

2 22 2 4

12 2

M d M d M m M dx

M m


sdot minus

12

( )

2 2 2 2 22 4 4 4

12 2

M d M d M d M m dx

M m


sdot minus

( ) ( )

2 2 2 2 22 4 2 4

12 12

4

2 2


x xM m

d M d

M m



( )

( )( )

( )( )

22 2

12 12 122 2

2

2


M m M m M m



81 81

1 1 81

1281 81

2 2 81

M M m m m mx d x d

M m m mM M mx d


M m m m




minus sdot minus

281 81 81 9 90

1 1 180 80 80

81 9 72281 81

2 280 802 80


m m m

m m mm m x d x d

x d m mm





9 9 8 8384 10 432 101 1 18 8 1

89 9 8 3456 10384 10 22

90

80

2 210 10

72

80

x d x d xm

mm x m

x mx d x d x mm

m


= sdot= sdot = sdot

sdot

sdot

sdotsdot sdot

sdot=


8

3456 102

cx

x mrArr

= sdot

minusminus

Vježba 109 Gdje se između Mjeseca i Zemlje poništavaju njihova gravitacijska polja Masa Zemlje je 81

puta veća od mase Mjeseca a udaljenost Zemlje od Mjeseca iznosi d = 384 000 000 m

Rezultat 3456 middot 108 m


Kolika bi morala biti perioda Zemljine vrtnje oko osi da bi tijelo na ekvatoru bilo u

bestežinskom stanju (gravitacijska konstanta G = 667 middot 10-11 (N middot m2) kg2 akceleracija slobodnog

pada g = 981 ms2 srednji polumjer Zemlje r = 64 middot 106 m masa Zemlje M = 6 middot 1024 kg)

Rješenje 110

G = 667 middot 10-11

(N middot m2) kg

2 g = 981 ms

2 r = 64 middot 10

6 m M = 6 middot 10

24 kg

T =




1 22

m mF G

r

sdot= sdot




13

24

2

rF mcp

T

πsdot sdot= sdot

gdje je T ophodno vrijeme




G m g= sdot




gravitacijskoj sili

1inačica

Iskoristit ćemo gravitaciju kao centripetalnu silu Ona prisiljava da tijelo mase m uz potrebnu kružnu

brzinu kruži oko Zemlje čiji je polumjer r Sila gravitacije između tijela mase m i Zemlje mase M na

površini Zemlje mora biti jednaka centripetalnoj sili na tijelo na udaljenosti r od središta vrtnje

2 24 4

2 2 2 2

2 2

M m r M T rm r

F FG M

G m G mg cpr T r mT

π πsdot sdot sdot sdot sdot sdot= rArr sdot

sdotsdot

sdot= rArr sdot = sdot

sdotsdot rArr

2 3 2 3 2 34 4 42 2

2r r r r

T T T T rG M G M G M G M

π π ππ

sdot sdot sdot sdot sdot sdotrArr = rArr = rArr = rArr = sdot sdot sdot =

sdot sdot sdot sdot

664 106 3

2 64 10 5085 102

11 24667 10 6 10

2

mm s

N mkg

kg

πsdot

= sdot sdot sdot sdot = sdot =

sdotminussdot sdot sdot

35085 10 3600 141 h= sdot =

2inačica

Privlačna sila između Zemlje mase M i tijela mase m daje tijelu centripetalnu akceleraciju te

jednadžba gibanja tijela ima oblik (sila teža je uzrok kružnog gibanja tijela mase m)

2 2 2

4 4

2 2

r rG F m g m m

Tg mcp

T T m g

π πsdot sdot sdot sdot= rArr sdot = sdot rArr sdot sdot=

sdotsdot rArr

2 2 24 4 42 2

2r r r r

T T T Tg g g g

π π ππ

sdot sdot sdot sdot sdot sdotrArr = rArr = rArr = rArr = sdot sdot =

664 10 3 3

5082 55085 10 141

9

10 3600

812

ms h

m

s

πsdot

= sdot sdot = sdot = sdot =

Vježba 110 Kolika bi morala biti perioda Zemljine vrtnje oko osi da bi tijelo na ekvatoru bilo u


pada g = 981 ms2 srednji polumjer Zemlje r = 6400 km masa Zemlje M = 6 middot 10

21 t)

Rezultat 141 h

14

Zadatak 111 (VM gimnazija)

Kolika bi morala biti gustoća materijala da sila kugli u dodiru promjera 1 m bude jednaka

10-4 N (gravitacijska konstanta G = 667 middot 10-11 (N middot m2) kg2)

Rješenje 111

2 middot r1 = 2 middot r2 = 1 m =gt r1 = r2 = r = 05 m R = 2 middot r F = 10-4 N

G = 667 middot 10-11

(N middot m2) kg

2 ρ =




1 22

m mF G

r

sdot= sdot



Gustoću ρ neke tvari možemo naći iz omjera mase tijela i njegova obujma

m

Vρ =

Kugla polumjera r ima obujam (volumen)

3

4

3V r π= sdot sdot

RRRR

rrrr rrrr

Budući da kugle imaju jednake mase a udaljenost između njihovih središta je R vrijedi

2

1 2 2 2 22

1 2 2 22

m m m

m m R FF G F G mm m

GF G R R

R

R

G

= =sdot

rArr = sdot rArr = sdot rArr = rArrsdot= sdot

sdot

2 22

2 R F R F F F

m m m R m rG G G G

sdot sdotrArr = rArr = rArr = sdot rArr = sdot sdot

Iz podataka za masu m kugle i polumjera r dobijemo gustoću materijala

6metoda

3supstitucij

2 6

234 3 4

34 3

3

e 4

m

V F F Fr r

F G G Gm r

G r rr

V r

r

ρ

ρ ρ ρπ ππ

π

=

sdot sdot sdot sdot sdot

= sdot sdot rArr rArr = rArr = rArr = rArrsdot sdot sdot sdotsdot sdot

=

sdot

sdot sdot

15

( )

3 43 3 10

233851 2 2 2 2 3

2 2 2 05 11667 10

2

F

F N kgG

Gr r N m mm

kg

ρ ρπ π π

sdot minus

rArr = rArr = sdot = sdot =

sdot sdot sdot sdot sdotsdot sdot minussdot

Vježba 111

Kolika bi morala biti gustoća materijala da sila kugli u dodiru promjera 100 cm bude jednaka

01 mN (gravitacijska konstanta G = 667 middot 10-11 (N middot m2) kg2)

Rezultat 233851 kgm3

Zadatak 112 (Petra gimnazija)

Perioda kruženja umjetnog satelita oko planeta iznosi T Udaljenost satelita od središta planeta

iznosi r Na kolikoj udaljenosti od središta planeta kruži drugi satelit kojemu je perioda kruženja 8

T

4 88 4

r rA B C r D rsdot sdot

Rješenje 112

r1 = r T1 = T 1

2 8

T T= sdot r2 =

Keplerovi zakoni

bull Svaki planet giba se po elipsi a Sunce je u jednom žarištu te elipse

bull Spojnica Sunce ndash planet u jednakim vremenskim intervalima opisuje jednake površine

bull Kvadrati ophodnih vremena planeta odnose se kao kubovi njihovih srednjih udaljenosti od

Sunca 2 3

1 12 3

2

2

T r

T r

=

2 3 2 32 3 2 3 2 3 2 3 31 1 2 1

1 2 2 1 1 2 2 1 2

1

21

2 3 22 2 1

T r T rT r T r T r T r r

r TT T

sdot= rArr sdot = sdot rArr sdot rArr =sdot= sdot rArr

21 3 1 12 3 3

183 3 3 3 364 642 2 2 22 2 2 6

2

4

T r T r r

r r r r r

T T

T

T

sdot sdot sdot sdot sdot sdot

rArr = rArr = rArr = rArr = sdot rArr

1 1 13 3 3 33 2 2 26 4 4

4 6

r r r r r rrArr = sdot rArr = sdot rArr = sdot

Odgovor je pod B

Vježba 112



T

3 99 81


Rezultat A

16

Zadatak 113 (Tihomir srednja škola)

Na kojoj visini iznad površine Zemlje akceleracija sile teže iznosi 733 m s2 Masa Zemlje je

6 middot 1024 kg a polumjer 64 middot 106 m (gravitacijska konstanta G = 667 middot 10-11 (N middot m2) kg2)

Rješenje 113

g = 733 m s2 M = 6 middot 1024 kg R = 64 middot 106 m G = 667 middot 10-11 (N middot m2) kg2

h =




1 22

m mF G

r

sdot= sdot


tijela Nalazi li se tijelo mase m na visini h iznad površine planeta polumjera R i mase M tada je

gravitacijska sila jednaka

( )2

m M

F G

R h

sdot= sdot

+




Newtonovom poučku

G m g= sdot


na Zemlji jednaka

Budući da je sila teža G na površini Zemlje jednaka privlačnoj sili F tijela mase m i Zemlje mase M

(na visini h iznad površine Zemlje) vrijedi

( ) ( )

( )2

2

2

m M m MG F m g G m g G

R h R h

R h

m g

sdot sdot= rArr sdot = sdot rArr sdot = sdot rArr

+ +

+sdot

sdot

( ) ( ) 2 2G M G M G M G M

R h R h R h h Rg g g g

sdot sdot sdot sdotrArr + = rArr + = rArr + = rArr = minus =

211 24

667 10 6 102

664 10 98901511 989

7332

N mkg

kgh m m km

m

s


rArr = minus sdot = asymp

Vježba 113


6 middot 1021

t a polumjer 64 middot 103 km (gravitacijska konstanta G = 667 middot 10

-11 (N middot m

2) kg

2)

Rezultat 989 km

Zadatak 114 (Ivica srednja škola)

Satelit obilazi Zemlju po kružnoj stazi brzinom 652 km s Koliko je ophodno vrijeme (masa

Zemlje M = 596 middot 1024


(N middot m2) kg

2)

Rješenje 114

v = 652 km s = 6520 m s M = 596 middot 1024

kg G = 667 middot 10-11

(N middot m2) kg

2

T =


17



1 22

m mF G

r

sdot= sdot


tijela


2v

F mcpr

= sdot

gdje je v obodna ili linearna brzina




2

rv

T

πsdot sdot=

gdje je r polumjer kružnice T perioda vrijeme jednog ophoda

Sila gravitacije Fg između satelita mase m i Zemlje mase M na udaljenosti r mora biti jednaka

centripetalnoj sili Fcp na satelit na udaljenosti r od središta vrtnje

2 2

2 2

2

2 2

m M v m M v G MF F G m G m rg cp

r

mr vvrr r

sdotsdot

sdot sdot sdot= rArr sdot = sdot rArr sdot = sdot rArr = rArr

2

G Mr

v

sdotrArr =

Ophodno vrijeme dobit ćemo iz sustava jednadžbi

metod

2 222

2

2 2

a

zamj2

ene

G Mr rrv Tv

T vT vTG M

T

v

G M G M vr r rv v v

π πππ

sdotsdot sdot sdot sdotsdot sdot sdot sdot= ==

rArr rArr rArr rArr = rArr sdot sdot sdot = = =

sdot

[ ]

211 24

2 667 10 596 102

2901175 1502 min

3 3

6520

901175 60

N mkg

G M kgT s

v m

s

ππ

sdotminussdot sdot sdot sdot sdot

sdot sdot sdotrArr = = = = =

Vježba 114


Zemlje M = 596 middot 1024 kg gravitacijska konstanta G = 667 middot 10-11 (N middot m2) kg2)

Rezultat 12016 min

Zadatak 115 (Marija gimnazija)

Na koju visinu iznad površine Zemlje treba podići tijelo da bi se težina umanjila dva puta

(polumjer zakrivljenosti Zemlje R = 64 middot 106 m)

5 3 6 6 3 10 4 10 265 10 7 10A m B m C m D msdot sdot sdot sdot

Rješenje 115

m ndash masa tijela M ndash masa Zemlje 1

1 2

G G= sdot R = 64 middot 106 m h =


18



1 22

m mF G

r

sdot= sdot



Za privlačenje tijela mase m i Zemlje mase M možemo napisati sljedeće jednadžbe

( )

( )( ) ( )

težina tijela mase m na površini Zemlje

težina tijela mase

2 2

1

1 2m na visini h

22

m M m MG G G G

R R

m M m MG G G G

R h R h

sdot sdot= sdot = sdot

rArr rArrsdot sdot

= sdot sdot = sdot

+ +

( ) ( ) ( )

podijelimo

jednadžbe

1

2 2 22

1 1 1

2 2 22 2

m MG

G R R Rm

m MG

G

m MMG G

R h R h R

G

h

G

sdot sdotsdot sdot

rArr rArr = rArr = rArr = rArrsdot sdot

sdot sdot sdot sdot

+ + +

( ) ( ) ( )( )

22 2 2

2 22 2 2 2

2

2 2

R h R h R hR h R

R R

R

R

+ + +rArr = rArr = rArr = rArr + = sdotsdot rArr

( )2 2 2

2 2 2 2R h R R h R R h R h R RrArr + = sdot rArr + = sdot rArr + = sdot rArr = sdot minus rArr

( ) ( )6 62 1 64 10 2 1 265 10 h R m mrArr = sdot minus = sdot sdot minus = sdot

Odgovor je pod C

Vježba 115

Na koju visinu iznad površine Zemlje treba podići tijelo da bi se težina umanjila četiri puta



Rezultat A

Zadatak 116 (Martin tehnička škola)

Odredite brzinu satelita koji obilazi Zemlju po kružnoj stazi s ophodnim vremenom 150 min

(gravitacijska konstanta G = 667 middot 10-11

N middot m2 kg

2 masa Zemlje M = 596 middot 10

24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rješenje 116

T = 150 min = [ 150 middot 60 ] = 9000 s G = 667 middot 10-11 N middot m2 kg2 M = 596 middot 1024 kg

v =




1 22

m mF G

r

sdot= sdot



19


2v

F mcpr

= sdot


Kada kruto tijelo rotira oko čvrste osi sve se njegove čestice gibaju po koncentričnim kružnicama

(koncentrične kružnice imaju zajedničko središte) Obodna (linearna) brzina iznosi

v 2 r

T

πsdot sdot=

gdje je r polumjer kružnice T perioda (vrijeme jednog okreta)

Sila gravitacije između satelita mase m i Zemlje mase M na udaljenosti r mora biti jednaka


2 22

2 2

v m M v m M M

F F m G m G v Gcpr r

r

m rr r

sdot sdot= rArr sdot = sdot rArr sdot = sdot rArr = sdotsdot

Iz sustava jednadžbi dobije se brzina satelita

pomnožimo

jednadžbe

2

2 22 3

2

rv

r M MTv v G v G

M T r Tv

r

rG

r

π

π π

sdot sdot=

sdot sdot sdot sdotrArr rArr sdot = sdot sdot rArr = sdot sdot rArr

= sdot

3

2 2 23 3 3G M G M G Mv v v

T T T

π π πsdot sdot sdot sdot sdot sdot sdot sdot sdotrArr = rArr = rArr = =

211 24

2 667 10 596 1023

652284 652 9 000

N mkg

m kmkg

s s s

πsdotminus


= = asymp

Odgovor je pod B

Vježba 116

Odredite brzinu satelita koji obilazi Zemlju po kružnoj stazi s ophodnim vremenom

2 h 30 min (gravitacijska konstanta G = 667 middot 10-11

N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rezultat B

Zadatak 117 (Stela maturantica)

Označimo sa Z gravitacijsku silu kojom Zemlja djeluje na satelit kad je na njezinoj površini

Gravitacijska sila Zemlje na satelit kada je na visini R 50 (R ndash polumjer Zemlje) jest

104 102 098 096A Z B Z C Z D Zsdot sdot sdot sdot

Rješenje 117

F1 = Z r1 = R 50 51

2 50 1 50 50 50

R R R R Rr R R

sdot += + = + = = sdot F2 =


20



1 22

m mF G

r

sdot= sdot



Gravitacijska sila kojom Zemlja mase M djeluje na satelit mase m kad je na njezinoj površini je

1

1

1 2 2

1

F Z

r R

m M m MF G Z G

r R

sdot sdot= sdot rArr rArr = sdot

=

=

Gravitacijska sila Zemlje mase M na satelit mase m kada je na visini r2 iznosi

51

22 2 22 2 2601 250 512 250050

m M m M m MF G F G F G

r R

r

R

Rsdot sdot sdot

= sdot rArr rArr = sdot rArr = sdot rArr sdotsdot

= sdot

2500

2 22601

m MF G

R

sdot sdotrArr = sdot

sdot

Iz sustava jednadžbi dobijemo gravitacijsku silu F2

2podijelimo

jednadžbe

2500 2500

222601 26012 2

25

2

002 22

2601

m M m MZ G G

F FRR

m

m MG

R

M m MZ Z m MGF G

RG

RR

sdot sdot sdot sdot sdot= sdot sdot sdot

sdot sdotrArr rArr = rArr = rArr sdot sdot sdot sdot sdot sdot= sdot

sdot

2500 2500 25002 2 096 2 22601 2601 2 1

60

F FZ F Z F Z

Z ZsdotrArr = rArr = rArr = sdot rArr = sdot

Odgovor je pod D

Vježba 117




Rezultat C

Zadatak 118 (Petra medicinska škola)

Satelit se giba oko Zemlje na udaljenosti 230 km od morske površine Kolika je brzina

satelita (srednji polumjer Zemlje R = 6400 km masa Zemlje 6 middot 1024 kg)

Rješenje 118

h = 230 km = 23 middot 105 m R = 6400 km = 64 middot 10

6 m M = 6 middot 10

24 kg v =




1 22

m mF G

r

sdot= sdot


tijela

21

211

6 7 6 102

N mG

kg

sdotminus= sdot

Taj zakon zovemo općim zakonom gravitacije

Da bi se tijelo mase m gibalo po kružnici polumjera r brzinom v potrebno je da na nj djeluje

centripetalna sila

2v

F mcpr

= sdot


Sila gravitacije Fg između satelita mase m i Zemlje mase M na udaljenosti R + h mora biti jednaka

centripetalnoj sili Fcp na satelit na udaljenosti R + h od središta vrtnje

( ) ( )

2

2

2 2

v m M v m MF F F F m G m Gg cp cp g

R h R

R h

mhR h R h

sdot sdot= rArr = rArr sdot = sdot rArr sdot = sdot rArr

+ ++

+sdot

+

2 2G M G M G M

v v vR h R h R h

sdot sdot sdotrArr = rArr = rArr = =

+ + +

211 24

667 10 6 102

77693 6 5

64 10 23 10

N mkg

mkg

sm m


= =sdot + sdot

Vježba 118


satelita (srednji polumjer Zemlje R = 6400 km masa Zemlje 6 middot 1024

kg)

Rezultat 7000 m

sasymp

Zadatak 119 (Lux gimnazija)

Znajući masu Zemlje M = 6 middot 1024

kg i srednji polumjer R = 6400 km odredi ubrzanje na visini

h = 1000 km iznad površine Zemlje (ubrzanje slobodnog pada na površini Zemlje g = 981 m s2)

Rješenje 119

M = 6 middot 1024

kg R = 6400 km = 64 middot 106 m h = 1000 km = 1 middot 10

6 m

g = 981 m s2 g1 =




1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot


22




Newtonovom poučku

G m g= sdot


na Zemlji jednaka

1inačica

Za privlačenje tijela mase m koje je na visini h i Zemlje mase M možemo napisati

( ) ( )1 2

1

2

mm M m M

G F m g G m g G

R h R h


+ +

( ) ( )

211 24

667 10 6 102

731 1 2 2 2

6 664 10 1 10

N mkg

G M mkgg

sR h m m


sdotrArr = = =

+ sdot + sdot

2inačica

Na površini Zemlje je h = 0 pa za privlačenje tijela mase m i Zemlje mase M možemo napisati

2 2

2

m M m M G MG F m g G m g G g

R

m

R R

sdot sdot sdot= rArr sdot = sdot rArr sdot = sdot rArr =


( ) ( ) ( )

1 1 12 2

2

m M m M G MG F m g G m g G m g

R h R h R h


+ + +

Iz sustava jednadžbi izračunamo g1

( )

( ) ( ) ( )

1

2 2 221 1 1

1

podijelimo

jednadž2

e2 2 2

b1

G MG Mg

g g gR h R h R hR

G M G Mg g g

G M

G Mg

R R RR h

sdot sdotsdot=

+ + +rArr rArr = rArr = rArr = rArr

sdot sdot sdot=

+

( )

2 221 1 1

2

g g gR R R

g g R h g R hR

g

h

rArr = rArr = rArr = rArr+

sdot++

22 664 10

981 734 1 2 6 6 2

64 10 1 10

R m m mg g

R h s m m s

sdotrArr = sdot = sdot =

+ sdot + sdot

(Odstupanje u rezultatu je zbog zaokruživanja)

Vježba 119


t i srednji polumjer R = 6400 km odredi ubrzanje na visini


Rezultat 734 2

m

s

23


Odredi prosječnu gustoću planeta Merkura ako je poznato da je perioda obilaska satelita

neposredno uz površinu (prva svemirska brzina) 292 km s Koliko je gravitacijsko ubrzanje na tom

planetu ako mu je polumjer 2439 km

Rješenje 120

v = 292 km s = 2920 m s R = 2439 km = 2439 middot 106m ρ = g =




1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot





Newtonovom poučku

G m g= sdot


na Zemlji jednaka


2v

F mcpr

= sdot



m

m VV

ρ ρ= rArr = sdot


3

4

3V r π= sdot sdot

Sila gravitacije Fg između satelita mase m i Merkura mase M na udaljenosti R mora biti jednaka

centripetalnoj sili Fcp na satelit na udaljenosti R od središta vrtnje

2 2

2

2

2

m M v m M vF F G m G mg cp

R R G mR R

Rsdot sdot= rArr sdot = sdot rArr sdot = sdot sdot

sdotrArr

4 3

3

4 3

2 24 3

3

3

v R v RM R

G G

M V V R

M R

ρ π

ρ

ρ

π

π

= sdot = sdot sdot

= sdot

sdot sdot

rArr = rArr rArr sdot sdot sdot = rArr

sdot

sdot

2 24 33

23 4

3

34

v R vR

G GR R

ρ π ρππ

sdotsdot

sdot sdotrArr sdot sdot sdot = =

sdot sdot sdotsdotrArr =

24

( )

2

3 2920

5130 2 32

11 64 667 10 2439 10

2

m

kgs

N m mm

kgπ

sdot

= =sdotminus


U tom je slučaju sila teža uzrok kružnoga gibanja satelita Zato mora biti Fcp jednako sili teži G

2 2 21

v v v

F G G F m g m m g m gcp cpR R Rm

= rArr = rArr sdot = sdot rArr rArrsdotsdot = sdot = =

2

2920

350 6 2

2439 10

m

ms

m s

= =sdot

Vježba 120



planetu ako mu je promjer 4870 km

Rezultat 5130 350 3 2

kg m

m s

2

Rješenje 102 m = 1 kg mM = 65 middot 10

23 kg r = 3400 km = 34 10

6 m

G = 667 middot 10-11

N middot m2 kg

2 F =




1 22

m mF G

r

sdot= sdot


tijela

Na površini Marsa privlačna sila između tijela mase m i Marsa mase mM iznosi

( )

2 231 65 1011

667 10 375 2 2 2

634 10

m m N m kg kgMF G N

r kgm

sdot sdot sdot sdotminus= sdot = sdot sdot =

sdot

Vježba 102 Kolikom silom Mars privlači kamen mase 100 dag koji se nalazina na njegovoj površini

Masa Marsa je 65 1021

t a polumjer 3400 km (gravitacijska konstanta G = 667 middot 10-11

N middot m2 kg

2)

Rezultat 375 N

Zadatak 103 (Ana gimnazija) Masa Marsa je 65 10

23 kg a polumjer 3400 km Kolika je akceleracija slobodnoga pada na

površini Marsa (gravitacijska konstanta G = 667 middot 10-11

N middot m2 kg

2)

Rješenje 103 mM = 65 1023 kg r = 3400 km = 34 middot 106 m G = 667 middot 10-11 N middot m2 kg2 g =




1 22

m mF G

r

sdot= sdot


tijela




Newtonovom poučku

G m g= sdot


na Zemlji jednaka

Gravitacijska sila privlačenja tijela i Marsa djeluje prema trećem Newtonovu zakonu na tijelo i Mars

jednako samo u suprotnim smjerovima Sila teža na Marsu je naziv za gravitacijsku silu kojom Mars privlači sva tijela u blizini svoje površine Za privlačenje nekog tijela mase m i Marsa mase mM

možemo napisati

2 2 2

m m m m m

M M MG F m g G m g G m g G

r r r

sdot sdot= rArr sdot = sdot rArr sdot = sdot rArr = sdot =

3

( )

2 2365 1011

667 10 375 2 2 2

634 10

N m kg m

kg sm


sdot




Rezultat 375 ms2



A masi satelita

B masi Zemlje








1 22

m mF G

r

sdot= sdot


tijela


2v

F mcpr

= sdot



vrijedi

2 2

2

2 2


r r rr

r

mr


2

M M

v G v Gr r




mr


A masi satelita

B težini satelita



Rezultat C

4






Rješenje 105


13





v 2 r

T

πsdot sdot=


Brzina kruženja


2v

rzsz

Tz

πsdot sdot=


2v

rmzm

Tm

πsdot sdot=


2 1390




π π

ππ



sdot sdotsdotsdot

sdot

sdot sdot

1390

v v v390 3013 v v 30 1v 1 v 13 v 1

z zr godmz

r god

zz

mm mzm

m

sdot sdot


Odgovor je pod D

rmzrzs





Rezultat C

5





Rješenje 106




1 22

m mF G

r

sdot= sdot


tijela


2v

F mcpr

= sdot




2 2

2

2 2


r r rr

r

mr


2

M M

v G v Gr r




Odgovor je pod A

Vježba 106




Rezultat C




Rješenje 107


mS =


2v

F mcpr

= sdot





6

1 22

m mF G

r

sdot= sdot





od središta vrtnje

2

2

2 2

2


prr


sdotrArr

24 11

3 10 15 10230

202 10 2

11667 10

2

mm

v r sm kg

S G N m

kg

sdot sdot sdotsdot

rArr = = = sdot

sdotminussdot

vvvv

rrrrmmmmmmmmSSSS

Vježba 107





kg




Rješenje 108





1 22

m mF G

r

sdot= sdot






G m g= sdot






7


0 2

m Mm g G

R

sdotsdot = sdot

bull na visini h

( )

2

m Mm g G

R h

sdotsdot = sdot

+


( ) ( ) ( )

podijelimo

jednadžbe

0 2 2 20 0

2 2 2


m M m Mm g gm g G G

R h R h

m

R h

MG

m

m MmG




+ + +

sdot

sdot

( )

( )1

2 2 220 0 0 0

21

2

g g g gR h R h R hR

g g g R R gR

R h


+

20 0 0 0


R g R g R g g


0 0 01 1 1060 0

0

60 0

0

g g gh R R h R h

g gg

gR h R


sdot sdot

1 1 161 1 6371 10 1 185392563


185392563 1854 km km= asymp

mmmmmmmmMMMMMMMM

gggg0000 gggg

hhhh



Rezultat 752 km




Rješenje 109


x =



izrazom

8

2

mG

r

γ = sdot


tijela




1 22

m mF G

r

sdot= sdot




dddd



C



2

MG

Zx

γ = sdot


( )

2

mG

Md x

γ = sdot

minus

1inačica




( ) ( )

( )2 2 2 2

22

m M m M

G G G GM Z

x xd x d

x d x

Gx



( ) ( ) ( )2 2 22 2 21

m





22mx d x


( )22

1m m m m



9

1

1

1 1m m

x d x dM

M


+

rArr

11 1 1 1

81 81 81

d d d dx x x

m

m

xm m

M m


+ + + +sdot sdot

911 1 1 9 1 10 10 10

19 1 9 9 9 9

d



++ +

89 384 10 8

3456 10 10

mm

sdot sdot= = sdot

2inačica




( ) ( )

( )2 2 2 2

22

m M m M

G G G GM Z

x xd x d

x d x

Gx





( )2 2 2 2 2


( )2

2 2 22 0

242

2 12 2



a




( ) ( ) ( )

( )

2 22 2 4

12 2

M d M d M m M dx

M m


sdot minus

( )

2 2 2 2 22 4 4 4

12 2

M d M d M d M m dx

M m


sdot minus

( ) ( )

2 2 2 2 22 4 2 4

12 12

4

2 2


x xM m

d M d

M m



( )

( )( )

( )( )

22 2

12 12 122 2

2

2


M m M m M m



10

81 81

1 1 81

1281 81

2 2 81

M M m m m mx d x d

M m m mM M mx d


M m m m




minus sdot minus

281 81 81 9 90

1 1 180 80 80

81 9 72281 81

2 280 802 80


m m m

m m mm m x d x d

x d m mm





9 9 8 8384 10 432 101 1 18 8 1

89 9 8 3456 10384 10 22

90

80

2 210 10

72

80

x d x d xm

mm x m

x mx d x d x mm

m


= sdot= sdot = sdot

sdot

sdot

sdotsdot sdot

sdot=


8

3456 102

cx

x mrArr

= sdot

minusminus



dddd


G

3inačica




1 2

M mF G

gZx

sdot= sdot


( )

1 2

m mF G

gMd x

sdot= sdot

minus



( ) ( )

( )1 1 1 12 2

2

2 2

2

1


gM gZx xd x d x

x d x

G m


minussdot sdot

minus sdotminus

sdotrArr

( ) ( ) ( )2 2 22 2 21

m





22mx d x


11

( )22m m m



1

1 1 11

m m m

x d x d x dM M M m

M


+

rArr

11 1 1 1

81 81 81

d d d dx x x

m

m

xm m

M m


+ + + +sdot sdot

911 1 1 9 1 10 10 10

19 1 9 9 9 9

d



++ +

89 384 10 8

3456 10 10

mm

sdot sdot= = sdot

4inačica




1 2

M mF G

gZx

sdot= sdot


( )

1 2

m mF G

gMd x

sdot= sdot

minus



( ) ( )

( )1 1 1 12 2

2

2 2

2

1


gM gZx xd x d x

x d x

G m


minussdot sdot

minus sdotminus

sdotrArr




( )2 2 2 2 2


( )2

2 2 22 0

242

2 12 2



a




( ) ( ) ( )

( )

2 22 2 4

12 2

M d M d M m M dx

M m


sdot minus

12

( )

2 2 2 2 22 4 4 4

12 2

M d M d M d M m dx

M m


sdot minus

( ) ( )

2 2 2 2 22 4 2 4

12 12

4

2 2


x xM m

d M d

M m



( )

( )( )

( )( )

22 2

12 12 122 2

2

2


M m M m M m



81 81

1 1 81

1281 81

2 2 81

M M m m m mx d x d

M m m mM M mx d


M m m m




minus sdot minus

281 81 81 9 90

1 1 180 80 80

81 9 72281 81

2 280 802 80


m m m

m m mm m x d x d

x d m mm





9 9 8 8384 10 432 101 1 18 8 1

89 9 8 3456 10384 10 22

90

80

2 210 10

72

80

x d x d xm

mm x m

x mx d x d x mm

m


= sdot= sdot = sdot

sdot

sdot

sdotsdot sdot

sdot=


8

3456 102

cx

x mrArr

= sdot

minusminus








Rješenje 110

G = 667 middot 10-11

(N middot m2) kg

2 g = 981 ms

2 r = 64 middot 10

6 m M = 6 middot 10

24 kg

T =




1 22

m mF G

r

sdot= sdot




13

24

2

rF mcp

T

πsdot sdot= sdot





G m g= sdot




gravitacijskoj sili

1inačica




2 24 4

2 2 2 2

2 2

M m r M T rm r

F FG M

G m G mg cpr T r mT


sdotsdot


sdotsdot rArr

2 3 2 3 2 34 4 42 2

2r r r r


π π ππ


sdot sdot sdot sdot

664 106 3

2 64 10 5085 102

11 24667 10 6 10

2

mm s

N mkg

kg

πsdot



35085 10 3600 141 h= sdot =

2inačica



2 2 2

4 4

2 2

r rG F m g m m

Tg mcp

T T m g


sdotsdot rArr

2 2 24 4 42 2

2r r r r

T T T Tg g g g

π π ππ


664 10 3 3

5082 55085 10 141

9

10 3600

812

ms h

m

s

πsdot





21 t)

Rezultat 141 h

14




Rješenje 111


G = 667 middot 10-11

(N middot m2) kg

2 ρ =




1 22

m mF G

r

sdot= sdot




m

Vρ =


3

4

3V r π= sdot sdot

RRRR

rrrr rrrr


2

1 2 2 2 22

1 2 2 22

m m m

m m R FF G F G mm m

GF G R R

R

R

G

= =sdot


sdot

2 22

2 R F R F F F

m m m R m rG G G G



6metoda

3supstitucij

2 6

234 3 4

34 3

3

e 4

m

V F F Fr r

F G G Gm r

G r rr

V r

r

ρ

ρ ρ ρπ ππ

π

=



=

sdot

sdot sdot

15

( )

3 43 3 10

233851 2 2 2 2 3

2 2 2 05 11667 10

2

F

F N kgG

Gr r N m mm

kg

ρ ρπ π π

sdot minus



Vježba 111







T

4 88 4


Rješenje 112

r1 = r T1 = T 1

2 8

T T= sdot r2 =

Keplerovi zakoni




Sunca 2 3

1 12 3

2

2

T r

T r

=

2 3 2 32 3 2 3 2 3 2 3 31 1 2 1

1 2 2 1 1 2 2 1 2

1

21

2 3 22 2 1


r TT T


21 3 1 12 3 3

183 3 3 3 364 642 2 2 22 2 2 6

2

4

T r T r r

r r r r r

T T

T

T



1 1 13 3 3 33 2 2 26 4 4

4 6


Odgovor je pod B

Vježba 112



T

3 99 81


Rezultat A

16




Rješenje 113


h =




1 22

m mF G

r

sdot= sdot




( )2

m M

F G

R h

sdot= sdot

+




Newtonovom poučku

G m g= sdot


na Zemlji jednaka



( ) ( )

( )2

2

2


R h R h

R h

m g


+ +

+sdot

sdot

( ) ( ) 2 2G M G M G M G M



211 24

667 10 6 102

664 10 98901511 989

7332

N mkg

kgh m m km

m

s



Vježba 113


6 middot 1021


-11 (N middot m

2) kg

2)

Rezultat 989 km





(N middot m2) kg

2)

Rješenje 114

v = 652 km s = 6520 m s M = 596 middot 1024

kg G = 667 middot 10-11

(N middot m2) kg

2

T =


17



1 22

m mF G

r

sdot= sdot


tijela


2v

F mcpr

= sdot





2

rv

T

πsdot sdot=




2 2

2 2

2

2 2


r

mr vvrr r

sdotsdot


2

G Mr

v

sdotrArr =


metod

2 222

2

2 2

a

zamj2

ene

G Mr rrv Tv

T vT vTG M

T

v

G M G M vr r rv v v

π πππ



sdot

[ ]

211 24

2 667 10 596 102

2901175 1502 min

3 3

6520

901175 60

N mkg

G M kgT s

v m

s

ππ



Vježba 114



Rezultat 12016 min





Rješenje 115


1 2



18



1 22

m mF G

r

sdot= sdot




( )

( )( ) ( )


težina tijela mase

2 2

1

1 2m na visini h

22

m M m MG G G G

R R

m M m MG G G G

R h R h


rArr rArrsdot sdot

= sdot sdot = sdot

+ +

( ) ( ) ( )

podijelimo

jednadžbe

1

2 2 22

1 1 1

2 2 22 2

m MG

G R R Rm

m MG

G

m MMG G

R h R h R

G

h

G

sdot sdotsdot sdot


sdot sdot sdot sdot

+ + +

( ) ( ) ( )( )

22 2 2

2 22 2 2 2

2

2 2

R h R h R hR h R

R R

R

R


( )2 2 2



Odgovor je pod C

Vježba 115




Rezultat A




N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rješenje 116


v =




1 22

m mF G

r

sdot= sdot



19


2v

F mcpr

= sdot




v 2 r

T

πsdot sdot=




2 22

2 2

v m M v m M M


r

m rr r



pomnožimo

jednadžbe

2

2 22 3

2

rv

r M MTv v G v G

M T r Tv

r

rG

r

π

π π

sdot sdot=


= sdot

3

2 2 23 3 3G M G M G Mv v v

T T T


211 24

2 667 10 596 1023

652284 652 9 000

N mkg

m kmkg

s s s

πsdotminus


= = asymp

Odgovor je pod B

Vježba 116



N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rezultat B





Rješenje 117

F1 = Z r1 = R 50 51

2 50 1 50 50 50

R R R R Rr R R

sdot += + = + = = sdot F2 =


20



1 22

m mF G

r

sdot= sdot




1

1

1 2 2

1

F Z

r R

m M m MF G Z G

r R


=

=


51

22 2 22 2 2601 250 512 250050


r R

r

R

Rsdot sdot sdot


= sdot

2500

2 22601

m MF G

R


sdot


2podijelimo

jednadžbe

2500 2500

222601 26012 2

25

2

002 22

2601

m M m MZ G G

F FRR

m

m MG

R

M m MZ Z m MGF G

RG

RR



sdot

2500 2500 25002 2 096 2 22601 2601 2 1

60

F FZ F Z F Z


Odgovor je pod D

Vježba 117




Rezultat C




Rješenje 118


6 m M = 6 middot 10

24 kg v =




1 22

m mF G

r

sdot= sdot


tijela

21

211

6 7 6 102

N mG

kg

sdotminus= sdot



centripetalna sila

2v

F mcpr

= sdot




( ) ( )

2

2

2 2


R h R

R h

mhR h R h


+ ++

+sdot

+

2 2G M G M G M

v v vR h R h R h


+ + +

211 24

667 10 6 102

77693 6 5

64 10 23 10

N mkg

mkg

sm m


= =sdot + sdot

Vježba 118



kg)

Rezultat 7000 m

sasymp





Rješenje 119

M = 6 middot 1024


6 m

g = 981 m s2 g1 =




1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot


22




Newtonovom poučku

G m g= sdot


na Zemlji jednaka

1inačica


( ) ( )1 2

1

2

mm M m M

G F m g G m g G

R h R h


+ +

( ) ( )

211 24

667 10 6 102

731 1 2 2 2

6 664 10 1 10

N mkg

G M mkgg

sR h m m


sdotrArr = = =

+ sdot + sdot

2inačica


2 2

2


R

m

R R



( ) ( ) ( )

1 1 12 2

2


R h R h R h


+ + +


( )

( ) ( ) ( )

1

2 2 221 1 1

1

podijelimo

jednadž2

e2 2 2

b1

G MG Mg

g g gR h R h R hR

G M G Mg g g

G M

G Mg

R R RR h

sdot sdotsdot=


sdot sdot sdot=

+

( )

2 221 1 1

2

g g gR R R

g g R h g R hR

g

h


sdot++

22 664 10

981 734 1 2 6 6 2

64 10 1 10

R m m mg g

R h s m m s


+ sdot + sdot


Vježba 119




Rezultat 734 2

m

s

23





Rješenje 120





1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot





Newtonovom poučku

G m g= sdot


na Zemlji jednaka


2v

F mcpr

= sdot



m

m VV

ρ ρ= rArr = sdot


3

4

3V r π= sdot sdot



2 2

2

2

2


R R G mR R


sdotrArr

4 3

3

4 3

2 24 3

3

3

v R v RM R

G G

M V V R

M R

ρ π

ρ

ρ

π

π

= sdot = sdot sdot

= sdot

sdot sdot


sdot

sdot

2 24 33

23 4

3

34

v R vR

G GR R

ρ π ρππ

sdotsdot



24

( )

2

3 2920

5130 2 32

11 64 667 10 2439 10

2

m

kgs

N m mm

kgπ

sdot

= =sdotminus



2 2 21

v v v



2

2920

350 6 2

2439 10

m

ms

m s

= =sdot

Vježba 120




Rezultat 5130 350 3 2

kg m

m s

3

( )

2 2365 1011

667 10 375 2 2 2

634 10

N m kg m

kg sm


sdot




Rezultat 375 ms2



A masi satelita

B masi Zemlje








1 22

m mF G

r

sdot= sdot


tijela


2v

F mcpr

= sdot



vrijedi

2 2

2

2 2


r r rr

r

mr


2

M M

v G v Gr r




mr


A masi satelita

B težini satelita



Rezultat C

4






Rješenje 105


13





v 2 r

T

πsdot sdot=


Brzina kruženja


2v

rzsz

Tz

πsdot sdot=


2v

rmzm

Tm

πsdot sdot=


2 1390




π π

ππ



sdot sdotsdotsdot

sdot

sdot sdot

1390

v v v390 3013 v v 30 1v 1 v 13 v 1

z zr godmz

r god

zz

mm mzm

m

sdot sdot


Odgovor je pod D

rmzrzs





Rezultat C

5





Rješenje 106




1 22

m mF G

r

sdot= sdot


tijela


2v

F mcpr

= sdot




2 2

2

2 2


r r rr

r

mr


2

M M

v G v Gr r




Odgovor je pod A

Vježba 106




Rezultat C




Rješenje 107


mS =


2v

F mcpr

= sdot





6

1 22

m mF G

r

sdot= sdot





od središta vrtnje

2

2

2 2

2


prr


sdotrArr

24 11

3 10 15 10230

202 10 2

11667 10

2

mm

v r sm kg

S G N m

kg

sdot sdot sdotsdot

rArr = = = sdot

sdotminussdot

vvvv

rrrrmmmmmmmmSSSS

Vježba 107





kg




Rješenje 108





1 22

m mF G

r

sdot= sdot






G m g= sdot






7


0 2

m Mm g G

R

sdotsdot = sdot

bull na visini h

( )

2

m Mm g G

R h

sdotsdot = sdot

+


( ) ( ) ( )

podijelimo

jednadžbe

0 2 2 20 0

2 2 2


m M m Mm g gm g G G

R h R h

m

R h

MG

m

m MmG




+ + +

sdot

sdot

( )

( )1

2 2 220 0 0 0

21

2

g g g gR h R h R hR

g g g R R gR

R h


+

20 0 0 0


R g R g R g g


0 0 01 1 1060 0

0

60 0

0

g g gh R R h R h

g gg

gR h R


sdot sdot

1 1 161 1 6371 10 1 185392563


185392563 1854 km km= asymp

mmmmmmmmMMMMMMMM

gggg0000 gggg

hhhh



Rezultat 752 km




Rješenje 109


x =



izrazom

8

2

mG

r

γ = sdot


tijela




1 22

m mF G

r

sdot= sdot




dddd



C



2

MG

Zx

γ = sdot


( )

2

mG

Md x

γ = sdot

minus

1inačica




( ) ( )

( )2 2 2 2

22

m M m M

G G G GM Z

x xd x d

x d x

Gx



( ) ( ) ( )2 2 22 2 21

m





22mx d x


( )22

1m m m m



9

1

1

1 1m m

x d x dM

M


+

rArr

11 1 1 1

81 81 81

d d d dx x x

m

m

xm m

M m


+ + + +sdot sdot

911 1 1 9 1 10 10 10

19 1 9 9 9 9

d



++ +

89 384 10 8

3456 10 10

mm

sdot sdot= = sdot

2inačica




( ) ( )

( )2 2 2 2

22

m M m M

G G G GM Z

x xd x d

x d x

Gx





( )2 2 2 2 2


( )2

2 2 22 0

242

2 12 2



a




( ) ( ) ( )

( )

2 22 2 4

12 2

M d M d M m M dx

M m


sdot minus

( )

2 2 2 2 22 4 4 4

12 2

M d M d M d M m dx

M m


sdot minus

( ) ( )

2 2 2 2 22 4 2 4

12 12

4

2 2


x xM m

d M d

M m



( )

( )( )

( )( )

22 2

12 12 122 2

2

2


M m M m M m



10

81 81

1 1 81

1281 81

2 2 81

M M m m m mx d x d

M m m mM M mx d


M m m m




minus sdot minus

281 81 81 9 90

1 1 180 80 80

81 9 72281 81

2 280 802 80


m m m

m m mm m x d x d

x d m mm





9 9 8 8384 10 432 101 1 18 8 1

89 9 8 3456 10384 10 22

90

80

2 210 10

72

80

x d x d xm

mm x m

x mx d x d x mm

m


= sdot= sdot = sdot

sdot

sdot

sdotsdot sdot

sdot=


8

3456 102

cx

x mrArr

= sdot

minusminus



dddd


G

3inačica




1 2

M mF G

gZx

sdot= sdot


( )

1 2

m mF G

gMd x

sdot= sdot

minus



( ) ( )

( )1 1 1 12 2

2

2 2

2

1


gM gZx xd x d x

x d x

G m


minussdot sdot

minus sdotminus

sdotrArr

( ) ( ) ( )2 2 22 2 21

m





22mx d x


11

( )22m m m



1

1 1 11

m m m

x d x d x dM M M m

M


+

rArr

11 1 1 1

81 81 81

d d d dx x x

m

m

xm m

M m


+ + + +sdot sdot

911 1 1 9 1 10 10 10

19 1 9 9 9 9

d



++ +

89 384 10 8

3456 10 10

mm

sdot sdot= = sdot

4inačica




1 2

M mF G

gZx

sdot= sdot


( )

1 2

m mF G

gMd x

sdot= sdot

minus



( ) ( )

( )1 1 1 12 2

2

2 2

2

1


gM gZx xd x d x

x d x

G m


minussdot sdot

minus sdotminus

sdotrArr




( )2 2 2 2 2


( )2

2 2 22 0

242

2 12 2



a




( ) ( ) ( )

( )

2 22 2 4

12 2

M d M d M m M dx

M m


sdot minus

12

( )

2 2 2 2 22 4 4 4

12 2

M d M d M d M m dx

M m


sdot minus

( ) ( )

2 2 2 2 22 4 2 4

12 12

4

2 2


x xM m

d M d

M m



( )

( )( )

( )( )

22 2

12 12 122 2

2

2


M m M m M m



81 81

1 1 81

1281 81

2 2 81

M M m m m mx d x d

M m m mM M mx d


M m m m




minus sdot minus

281 81 81 9 90

1 1 180 80 80

81 9 72281 81

2 280 802 80


m m m

m m mm m x d x d

x d m mm





9 9 8 8384 10 432 101 1 18 8 1

89 9 8 3456 10384 10 22

90

80

2 210 10

72

80

x d x d xm

mm x m

x mx d x d x mm

m


= sdot= sdot = sdot

sdot

sdot

sdotsdot sdot

sdot=


8

3456 102

cx

x mrArr

= sdot

minusminus








Rješenje 110

G = 667 middot 10-11

(N middot m2) kg

2 g = 981 ms

2 r = 64 middot 10

6 m M = 6 middot 10

24 kg

T =




1 22

m mF G

r

sdot= sdot




13

24

2

rF mcp

T

πsdot sdot= sdot





G m g= sdot




gravitacijskoj sili

1inačica




2 24 4

2 2 2 2

2 2

M m r M T rm r

F FG M

G m G mg cpr T r mT


sdotsdot


sdotsdot rArr

2 3 2 3 2 34 4 42 2

2r r r r


π π ππ


sdot sdot sdot sdot

664 106 3

2 64 10 5085 102

11 24667 10 6 10

2

mm s

N mkg

kg

πsdot



35085 10 3600 141 h= sdot =

2inačica



2 2 2

4 4

2 2

r rG F m g m m

Tg mcp

T T m g


sdotsdot rArr

2 2 24 4 42 2

2r r r r

T T T Tg g g g

π π ππ


664 10 3 3

5082 55085 10 141

9

10 3600

812

ms h

m

s

πsdot





21 t)

Rezultat 141 h

14




Rješenje 111


G = 667 middot 10-11

(N middot m2) kg

2 ρ =




1 22

m mF G

r

sdot= sdot




m

Vρ =


3

4

3V r π= sdot sdot

RRRR

rrrr rrrr


2

1 2 2 2 22

1 2 2 22

m m m

m m R FF G F G mm m

GF G R R

R

R

G

= =sdot


sdot

2 22

2 R F R F F F

m m m R m rG G G G



6metoda

3supstitucij

2 6

234 3 4

34 3

3

e 4

m

V F F Fr r

F G G Gm r

G r rr

V r

r

ρ

ρ ρ ρπ ππ

π

=



=

sdot

sdot sdot

15

( )

3 43 3 10

233851 2 2 2 2 3

2 2 2 05 11667 10

2

F

F N kgG

Gr r N m mm

kg

ρ ρπ π π

sdot minus



Vježba 111







T

4 88 4


Rješenje 112

r1 = r T1 = T 1

2 8

T T= sdot r2 =

Keplerovi zakoni




Sunca 2 3

1 12 3

2

2

T r

T r

=

2 3 2 32 3 2 3 2 3 2 3 31 1 2 1

1 2 2 1 1 2 2 1 2

1

21

2 3 22 2 1


r TT T


21 3 1 12 3 3

183 3 3 3 364 642 2 2 22 2 2 6

2

4

T r T r r

r r r r r

T T

T

T



1 1 13 3 3 33 2 2 26 4 4

4 6


Odgovor je pod B

Vježba 112



T

3 99 81


Rezultat A

16




Rješenje 113


h =




1 22

m mF G

r

sdot= sdot




( )2

m M

F G

R h

sdot= sdot

+




Newtonovom poučku

G m g= sdot


na Zemlji jednaka



( ) ( )

( )2

2

2


R h R h

R h

m g


+ +

+sdot

sdot

( ) ( ) 2 2G M G M G M G M



211 24

667 10 6 102

664 10 98901511 989

7332

N mkg

kgh m m km

m

s



Vježba 113


6 middot 1021


-11 (N middot m

2) kg

2)

Rezultat 989 km





(N middot m2) kg

2)

Rješenje 114

v = 652 km s = 6520 m s M = 596 middot 1024

kg G = 667 middot 10-11

(N middot m2) kg

2

T =


17



1 22

m mF G

r

sdot= sdot


tijela


2v

F mcpr

= sdot





2

rv

T

πsdot sdot=




2 2

2 2

2

2 2


r

mr vvrr r

sdotsdot


2

G Mr

v

sdotrArr =


metod

2 222

2

2 2

a

zamj2

ene

G Mr rrv Tv

T vT vTG M

T

v

G M G M vr r rv v v

π πππ



sdot

[ ]

211 24

2 667 10 596 102

2901175 1502 min

3 3

6520

901175 60

N mkg

G M kgT s

v m

s

ππ



Vježba 114



Rezultat 12016 min





Rješenje 115


1 2



18



1 22

m mF G

r

sdot= sdot




( )

( )( ) ( )


težina tijela mase

2 2

1

1 2m na visini h

22

m M m MG G G G

R R

m M m MG G G G

R h R h


rArr rArrsdot sdot

= sdot sdot = sdot

+ +

( ) ( ) ( )

podijelimo

jednadžbe

1

2 2 22

1 1 1

2 2 22 2

m MG

G R R Rm

m MG

G

m MMG G

R h R h R

G

h

G

sdot sdotsdot sdot


sdot sdot sdot sdot

+ + +

( ) ( ) ( )( )

22 2 2

2 22 2 2 2

2

2 2

R h R h R hR h R

R R

R

R


( )2 2 2



Odgovor je pod C

Vježba 115




Rezultat A




N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rješenje 116


v =




1 22

m mF G

r

sdot= sdot



19


2v

F mcpr

= sdot




v 2 r

T

πsdot sdot=




2 22

2 2

v m M v m M M


r

m rr r



pomnožimo

jednadžbe

2

2 22 3

2

rv

r M MTv v G v G

M T r Tv

r

rG

r

π

π π

sdot sdot=


= sdot

3

2 2 23 3 3G M G M G Mv v v

T T T


211 24

2 667 10 596 1023

652284 652 9 000

N mkg

m kmkg

s s s

πsdotminus


= = asymp

Odgovor je pod B

Vježba 116



N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rezultat B





Rješenje 117

F1 = Z r1 = R 50 51

2 50 1 50 50 50

R R R R Rr R R

sdot += + = + = = sdot F2 =


20



1 22

m mF G

r

sdot= sdot




1

1

1 2 2

1

F Z

r R

m M m MF G Z G

r R


=

=


51

22 2 22 2 2601 250 512 250050


r R

r

R

Rsdot sdot sdot


= sdot

2500

2 22601

m MF G

R


sdot


2podijelimo

jednadžbe

2500 2500

222601 26012 2

25

2

002 22

2601

m M m MZ G G

F FRR

m

m MG

R

M m MZ Z m MGF G

RG

RR



sdot

2500 2500 25002 2 096 2 22601 2601 2 1

60

F FZ F Z F Z


Odgovor je pod D

Vježba 117




Rezultat C




Rješenje 118


6 m M = 6 middot 10

24 kg v =




1 22

m mF G

r

sdot= sdot


tijela

21

211

6 7 6 102

N mG

kg

sdotminus= sdot



centripetalna sila

2v

F mcpr

= sdot




( ) ( )

2

2

2 2


R h R

R h

mhR h R h


+ ++

+sdot

+

2 2G M G M G M

v v vR h R h R h


+ + +

211 24

667 10 6 102

77693 6 5

64 10 23 10

N mkg

mkg

sm m


= =sdot + sdot

Vježba 118



kg)

Rezultat 7000 m

sasymp





Rješenje 119

M = 6 middot 1024


6 m

g = 981 m s2 g1 =




1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot


22




Newtonovom poučku

G m g= sdot


na Zemlji jednaka

1inačica


( ) ( )1 2

1

2

mm M m M

G F m g G m g G

R h R h


+ +

( ) ( )

211 24

667 10 6 102

731 1 2 2 2

6 664 10 1 10

N mkg

G M mkgg

sR h m m


sdotrArr = = =

+ sdot + sdot

2inačica


2 2

2


R

m

R R



( ) ( ) ( )

1 1 12 2

2


R h R h R h


+ + +


( )

( ) ( ) ( )

1

2 2 221 1 1

1

podijelimo

jednadž2

e2 2 2

b1

G MG Mg

g g gR h R h R hR

G M G Mg g g

G M

G Mg

R R RR h

sdot sdotsdot=


sdot sdot sdot=

+

( )

2 221 1 1

2

g g gR R R

g g R h g R hR

g

h


sdot++

22 664 10

981 734 1 2 6 6 2

64 10 1 10

R m m mg g

R h s m m s


+ sdot + sdot


Vježba 119




Rezultat 734 2

m

s

23





Rješenje 120





1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot





Newtonovom poučku

G m g= sdot


na Zemlji jednaka


2v

F mcpr

= sdot



m

m VV

ρ ρ= rArr = sdot


3

4

3V r π= sdot sdot



2 2

2

2

2


R R G mR R


sdotrArr

4 3

3

4 3

2 24 3

3

3

v R v RM R

G G

M V V R

M R

ρ π

ρ

ρ

π

π

= sdot = sdot sdot

= sdot

sdot sdot


sdot

sdot

2 24 33

23 4

3

34

v R vR

G GR R

ρ π ρππ

sdotsdot



24

( )

2

3 2920

5130 2 32

11 64 667 10 2439 10

2

m

kgs

N m mm

kgπ

sdot

= =sdotminus



2 2 21

v v v



2

2920

350 6 2

2439 10

m

ms

m s

= =sdot

Vježba 120




Rezultat 5130 350 3 2

kg m

m s

4






Rješenje 105


13





v 2 r

T

πsdot sdot=


Brzina kruženja


2v

rzsz

Tz

πsdot sdot=


2v

rmzm

Tm

πsdot sdot=


2 1390




π π

ππ



sdot sdotsdotsdot

sdot

sdot sdot

1390

v v v390 3013 v v 30 1v 1 v 13 v 1

z zr godmz

r god

zz

mm mzm

m

sdot sdot


Odgovor je pod D

rmzrzs





Rezultat C

5





Rješenje 106




1 22

m mF G

r

sdot= sdot


tijela


2v

F mcpr

= sdot




2 2

2

2 2


r r rr

r

mr


2

M M

v G v Gr r




Odgovor je pod A

Vježba 106




Rezultat C




Rješenje 107


mS =


2v

F mcpr

= sdot





6

1 22

m mF G

r

sdot= sdot





od središta vrtnje

2

2

2 2

2


prr


sdotrArr

24 11

3 10 15 10230

202 10 2

11667 10

2

mm

v r sm kg

S G N m

kg

sdot sdot sdotsdot

rArr = = = sdot

sdotminussdot

vvvv

rrrrmmmmmmmmSSSS

Vježba 107





kg




Rješenje 108





1 22

m mF G

r

sdot= sdot






G m g= sdot






7


0 2

m Mm g G

R

sdotsdot = sdot

bull na visini h

( )

2

m Mm g G

R h

sdotsdot = sdot

+


( ) ( ) ( )

podijelimo

jednadžbe

0 2 2 20 0

2 2 2


m M m Mm g gm g G G

R h R h

m

R h

MG

m

m MmG




+ + +

sdot

sdot

( )

( )1

2 2 220 0 0 0

21

2

g g g gR h R h R hR

g g g R R gR

R h


+

20 0 0 0


R g R g R g g


0 0 01 1 1060 0

0

60 0

0

g g gh R R h R h

g gg

gR h R


sdot sdot

1 1 161 1 6371 10 1 185392563


185392563 1854 km km= asymp

mmmmmmmmMMMMMMMM

gggg0000 gggg

hhhh



Rezultat 752 km




Rješenje 109


x =



izrazom

8

2

mG

r

γ = sdot


tijela




1 22

m mF G

r

sdot= sdot




dddd



C



2

MG

Zx

γ = sdot


( )

2

mG

Md x

γ = sdot

minus

1inačica




( ) ( )

( )2 2 2 2

22

m M m M

G G G GM Z

x xd x d

x d x

Gx



( ) ( ) ( )2 2 22 2 21

m





22mx d x


( )22

1m m m m



9

1

1

1 1m m

x d x dM

M


+

rArr

11 1 1 1

81 81 81

d d d dx x x

m

m

xm m

M m


+ + + +sdot sdot

911 1 1 9 1 10 10 10

19 1 9 9 9 9

d



++ +

89 384 10 8

3456 10 10

mm

sdot sdot= = sdot

2inačica




( ) ( )

( )2 2 2 2

22

m M m M

G G G GM Z

x xd x d

x d x

Gx





( )2 2 2 2 2


( )2

2 2 22 0

242

2 12 2



a




( ) ( ) ( )

( )

2 22 2 4

12 2

M d M d M m M dx

M m


sdot minus

( )

2 2 2 2 22 4 4 4

12 2

M d M d M d M m dx

M m


sdot minus

( ) ( )

2 2 2 2 22 4 2 4

12 12

4

2 2


x xM m

d M d

M m



( )

( )( )

( )( )

22 2

12 12 122 2

2

2


M m M m M m



10

81 81

1 1 81

1281 81

2 2 81

M M m m m mx d x d

M m m mM M mx d


M m m m




minus sdot minus

281 81 81 9 90

1 1 180 80 80

81 9 72281 81

2 280 802 80


m m m

m m mm m x d x d

x d m mm





9 9 8 8384 10 432 101 1 18 8 1

89 9 8 3456 10384 10 22

90

80

2 210 10

72

80

x d x d xm

mm x m

x mx d x d x mm

m


= sdot= sdot = sdot

sdot

sdot

sdotsdot sdot

sdot=


8

3456 102

cx

x mrArr

= sdot

minusminus



dddd


G

3inačica




1 2

M mF G

gZx

sdot= sdot


( )

1 2

m mF G

gMd x

sdot= sdot

minus



( ) ( )

( )1 1 1 12 2

2

2 2

2

1


gM gZx xd x d x

x d x

G m


minussdot sdot

minus sdotminus

sdotrArr

( ) ( ) ( )2 2 22 2 21

m





22mx d x


11

( )22m m m



1

1 1 11

m m m

x d x d x dM M M m

M


+

rArr

11 1 1 1

81 81 81

d d d dx x x

m

m

xm m

M m


+ + + +sdot sdot

911 1 1 9 1 10 10 10

19 1 9 9 9 9

d



++ +

89 384 10 8

3456 10 10

mm

sdot sdot= = sdot

4inačica




1 2

M mF G

gZx

sdot= sdot


( )

1 2

m mF G

gMd x

sdot= sdot

minus



( ) ( )

( )1 1 1 12 2

2

2 2

2

1


gM gZx xd x d x

x d x

G m


minussdot sdot

minus sdotminus

sdotrArr




( )2 2 2 2 2


( )2

2 2 22 0

242

2 12 2



a




( ) ( ) ( )

( )

2 22 2 4

12 2

M d M d M m M dx

M m


sdot minus

12

( )

2 2 2 2 22 4 4 4

12 2

M d M d M d M m dx

M m


sdot minus

( ) ( )

2 2 2 2 22 4 2 4

12 12

4

2 2


x xM m

d M d

M m



( )

( )( )

( )( )

22 2

12 12 122 2

2

2


M m M m M m



81 81

1 1 81

1281 81

2 2 81

M M m m m mx d x d

M m m mM M mx d


M m m m




minus sdot minus

281 81 81 9 90

1 1 180 80 80

81 9 72281 81

2 280 802 80


m m m

m m mm m x d x d

x d m mm





9 9 8 8384 10 432 101 1 18 8 1

89 9 8 3456 10384 10 22

90

80

2 210 10

72

80

x d x d xm

mm x m

x mx d x d x mm

m


= sdot= sdot = sdot

sdot

sdot

sdotsdot sdot

sdot=


8

3456 102

cx

x mrArr

= sdot

minusminus








Rješenje 110

G = 667 middot 10-11

(N middot m2) kg

2 g = 981 ms

2 r = 64 middot 10

6 m M = 6 middot 10

24 kg

T =




1 22

m mF G

r

sdot= sdot




13

24

2

rF mcp

T

πsdot sdot= sdot





G m g= sdot




gravitacijskoj sili

1inačica




2 24 4

2 2 2 2

2 2

M m r M T rm r

F FG M

G m G mg cpr T r mT


sdotsdot


sdotsdot rArr

2 3 2 3 2 34 4 42 2

2r r r r


π π ππ


sdot sdot sdot sdot

664 106 3

2 64 10 5085 102

11 24667 10 6 10

2

mm s

N mkg

kg

πsdot



35085 10 3600 141 h= sdot =

2inačica



2 2 2

4 4

2 2

r rG F m g m m

Tg mcp

T T m g


sdotsdot rArr

2 2 24 4 42 2

2r r r r

T T T Tg g g g

π π ππ


664 10 3 3

5082 55085 10 141

9

10 3600

812

ms h

m

s

πsdot





21 t)

Rezultat 141 h

14




Rješenje 111


G = 667 middot 10-11

(N middot m2) kg

2 ρ =




1 22

m mF G

r

sdot= sdot




m

Vρ =


3

4

3V r π= sdot sdot

RRRR

rrrr rrrr


2

1 2 2 2 22

1 2 2 22

m m m

m m R FF G F G mm m

GF G R R

R

R

G

= =sdot


sdot

2 22

2 R F R F F F

m m m R m rG G G G



6metoda

3supstitucij

2 6

234 3 4

34 3

3

e 4

m

V F F Fr r

F G G Gm r

G r rr

V r

r

ρ

ρ ρ ρπ ππ

π

=



=

sdot

sdot sdot

15

( )

3 43 3 10

233851 2 2 2 2 3

2 2 2 05 11667 10

2

F

F N kgG

Gr r N m mm

kg

ρ ρπ π π

sdot minus



Vježba 111







T

4 88 4


Rješenje 112

r1 = r T1 = T 1

2 8

T T= sdot r2 =

Keplerovi zakoni




Sunca 2 3

1 12 3

2

2

T r

T r

=

2 3 2 32 3 2 3 2 3 2 3 31 1 2 1

1 2 2 1 1 2 2 1 2

1

21

2 3 22 2 1


r TT T


21 3 1 12 3 3

183 3 3 3 364 642 2 2 22 2 2 6

2

4

T r T r r

r r r r r

T T

T

T



1 1 13 3 3 33 2 2 26 4 4

4 6


Odgovor je pod B

Vježba 112



T

3 99 81


Rezultat A

16




Rješenje 113


h =




1 22

m mF G

r

sdot= sdot




( )2

m M

F G

R h

sdot= sdot

+




Newtonovom poučku

G m g= sdot


na Zemlji jednaka



( ) ( )

( )2

2

2


R h R h

R h

m g


+ +

+sdot

sdot

( ) ( ) 2 2G M G M G M G M



211 24

667 10 6 102

664 10 98901511 989

7332

N mkg

kgh m m km

m

s



Vježba 113


6 middot 1021


-11 (N middot m

2) kg

2)

Rezultat 989 km





(N middot m2) kg

2)

Rješenje 114

v = 652 km s = 6520 m s M = 596 middot 1024

kg G = 667 middot 10-11

(N middot m2) kg

2

T =


17



1 22

m mF G

r

sdot= sdot


tijela


2v

F mcpr

= sdot





2

rv

T

πsdot sdot=




2 2

2 2

2

2 2


r

mr vvrr r

sdotsdot


2

G Mr

v

sdotrArr =


metod

2 222

2

2 2

a

zamj2

ene

G Mr rrv Tv

T vT vTG M

T

v

G M G M vr r rv v v

π πππ



sdot

[ ]

211 24

2 667 10 596 102

2901175 1502 min

3 3

6520

901175 60

N mkg

G M kgT s

v m

s

ππ



Vježba 114



Rezultat 12016 min





Rješenje 115


1 2



18



1 22

m mF G

r

sdot= sdot




( )

( )( ) ( )


težina tijela mase

2 2

1

1 2m na visini h

22

m M m MG G G G

R R

m M m MG G G G

R h R h


rArr rArrsdot sdot

= sdot sdot = sdot

+ +

( ) ( ) ( )

podijelimo

jednadžbe

1

2 2 22

1 1 1

2 2 22 2

m MG

G R R Rm

m MG

G

m MMG G

R h R h R

G

h

G

sdot sdotsdot sdot


sdot sdot sdot sdot

+ + +

( ) ( ) ( )( )

22 2 2

2 22 2 2 2

2

2 2

R h R h R hR h R

R R

R

R


( )2 2 2



Odgovor je pod C

Vježba 115




Rezultat A




N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rješenje 116


v =




1 22

m mF G

r

sdot= sdot



19


2v

F mcpr

= sdot




v 2 r

T

πsdot sdot=




2 22

2 2

v m M v m M M


r

m rr r



pomnožimo

jednadžbe

2

2 22 3

2

rv

r M MTv v G v G

M T r Tv

r

rG

r

π

π π

sdot sdot=


= sdot

3

2 2 23 3 3G M G M G Mv v v

T T T


211 24

2 667 10 596 1023

652284 652 9 000

N mkg

m kmkg

s s s

πsdotminus


= = asymp

Odgovor je pod B

Vježba 116



N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rezultat B





Rješenje 117

F1 = Z r1 = R 50 51

2 50 1 50 50 50

R R R R Rr R R

sdot += + = + = = sdot F2 =


20



1 22

m mF G

r

sdot= sdot




1

1

1 2 2

1

F Z

r R

m M m MF G Z G

r R


=

=


51

22 2 22 2 2601 250 512 250050


r R

r

R

Rsdot sdot sdot


= sdot

2500

2 22601

m MF G

R


sdot


2podijelimo

jednadžbe

2500 2500

222601 26012 2

25

2

002 22

2601

m M m MZ G G

F FRR

m

m MG

R

M m MZ Z m MGF G

RG

RR



sdot

2500 2500 25002 2 096 2 22601 2601 2 1

60

F FZ F Z F Z


Odgovor je pod D

Vježba 117




Rezultat C




Rješenje 118


6 m M = 6 middot 10

24 kg v =




1 22

m mF G

r

sdot= sdot


tijela

21

211

6 7 6 102

N mG

kg

sdotminus= sdot



centripetalna sila

2v

F mcpr

= sdot




( ) ( )

2

2

2 2


R h R

R h

mhR h R h


+ ++

+sdot

+

2 2G M G M G M

v v vR h R h R h


+ + +

211 24

667 10 6 102

77693 6 5

64 10 23 10

N mkg

mkg

sm m


= =sdot + sdot

Vježba 118



kg)

Rezultat 7000 m

sasymp





Rješenje 119

M = 6 middot 1024


6 m

g = 981 m s2 g1 =




1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot


22




Newtonovom poučku

G m g= sdot


na Zemlji jednaka

1inačica


( ) ( )1 2

1

2

mm M m M

G F m g G m g G

R h R h


+ +

( ) ( )

211 24

667 10 6 102

731 1 2 2 2

6 664 10 1 10

N mkg

G M mkgg

sR h m m


sdotrArr = = =

+ sdot + sdot

2inačica


2 2

2


R

m

R R



( ) ( ) ( )

1 1 12 2

2


R h R h R h


+ + +


( )

( ) ( ) ( )

1

2 2 221 1 1

1

podijelimo

jednadž2

e2 2 2

b1

G MG Mg

g g gR h R h R hR

G M G Mg g g

G M

G Mg

R R RR h

sdot sdotsdot=


sdot sdot sdot=

+

( )

2 221 1 1

2

g g gR R R

g g R h g R hR

g

h


sdot++

22 664 10

981 734 1 2 6 6 2

64 10 1 10

R m m mg g

R h s m m s


+ sdot + sdot


Vježba 119




Rezultat 734 2

m

s

23





Rješenje 120





1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot





Newtonovom poučku

G m g= sdot


na Zemlji jednaka


2v

F mcpr

= sdot



m

m VV

ρ ρ= rArr = sdot


3

4

3V r π= sdot sdot



2 2

2

2

2


R R G mR R


sdotrArr

4 3

3

4 3

2 24 3

3

3

v R v RM R

G G

M V V R

M R

ρ π

ρ

ρ

π

π

= sdot = sdot sdot

= sdot

sdot sdot


sdot

sdot

2 24 33

23 4

3

34

v R vR

G GR R

ρ π ρππ

sdotsdot



24

( )

2

3 2920

5130 2 32

11 64 667 10 2439 10

2

m

kgs

N m mm

kgπ

sdot

= =sdotminus



2 2 21

v v v



2

2920

350 6 2

2439 10

m

ms

m s

= =sdot

Vježba 120




Rezultat 5130 350 3 2

kg m

m s

5





Rješenje 106




1 22

m mF G

r

sdot= sdot


tijela


2v

F mcpr

= sdot




2 2

2

2 2


r r rr

r

mr


2

M M

v G v Gr r




Odgovor je pod A

Vježba 106




Rezultat C




Rješenje 107


mS =


2v

F mcpr

= sdot





6

1 22

m mF G

r

sdot= sdot





od središta vrtnje

2

2

2 2

2


prr


sdotrArr

24 11

3 10 15 10230

202 10 2

11667 10

2

mm

v r sm kg

S G N m

kg

sdot sdot sdotsdot

rArr = = = sdot

sdotminussdot

vvvv

rrrrmmmmmmmmSSSS

Vježba 107





kg




Rješenje 108





1 22

m mF G

r

sdot= sdot






G m g= sdot






7


0 2

m Mm g G

R

sdotsdot = sdot

bull na visini h

( )

2

m Mm g G

R h

sdotsdot = sdot

+


( ) ( ) ( )

podijelimo

jednadžbe

0 2 2 20 0

2 2 2


m M m Mm g gm g G G

R h R h

m

R h

MG

m

m MmG




+ + +

sdot

sdot

( )

( )1

2 2 220 0 0 0

21

2

g g g gR h R h R hR

g g g R R gR

R h


+

20 0 0 0


R g R g R g g


0 0 01 1 1060 0

0

60 0

0

g g gh R R h R h

g gg

gR h R


sdot sdot

1 1 161 1 6371 10 1 185392563


185392563 1854 km km= asymp

mmmmmmmmMMMMMMMM

gggg0000 gggg

hhhh



Rezultat 752 km




Rješenje 109


x =



izrazom

8

2

mG

r

γ = sdot


tijela




1 22

m mF G

r

sdot= sdot




dddd



C



2

MG

Zx

γ = sdot


( )

2

mG

Md x

γ = sdot

minus

1inačica




( ) ( )

( )2 2 2 2

22

m M m M

G G G GM Z

x xd x d

x d x

Gx



( ) ( ) ( )2 2 22 2 21

m





22mx d x


( )22

1m m m m



9

1

1

1 1m m

x d x dM

M


+

rArr

11 1 1 1

81 81 81

d d d dx x x

m

m

xm m

M m


+ + + +sdot sdot

911 1 1 9 1 10 10 10

19 1 9 9 9 9

d



++ +

89 384 10 8

3456 10 10

mm

sdot sdot= = sdot

2inačica




( ) ( )

( )2 2 2 2

22

m M m M

G G G GM Z

x xd x d

x d x

Gx





( )2 2 2 2 2


( )2

2 2 22 0

242

2 12 2



a




( ) ( ) ( )

( )

2 22 2 4

12 2

M d M d M m M dx

M m


sdot minus

( )

2 2 2 2 22 4 4 4

12 2

M d M d M d M m dx

M m


sdot minus

( ) ( )

2 2 2 2 22 4 2 4

12 12

4

2 2


x xM m

d M d

M m



( )

( )( )

( )( )

22 2

12 12 122 2

2

2


M m M m M m



10

81 81

1 1 81

1281 81

2 2 81

M M m m m mx d x d

M m m mM M mx d


M m m m




minus sdot minus

281 81 81 9 90

1 1 180 80 80

81 9 72281 81

2 280 802 80


m m m

m m mm m x d x d

x d m mm





9 9 8 8384 10 432 101 1 18 8 1

89 9 8 3456 10384 10 22

90

80

2 210 10

72

80

x d x d xm

mm x m

x mx d x d x mm

m


= sdot= sdot = sdot

sdot

sdot

sdotsdot sdot

sdot=


8

3456 102

cx

x mrArr

= sdot

minusminus



dddd


G

3inačica




1 2

M mF G

gZx

sdot= sdot


( )

1 2

m mF G

gMd x

sdot= sdot

minus



( ) ( )

( )1 1 1 12 2

2

2 2

2

1


gM gZx xd x d x

x d x

G m


minussdot sdot

minus sdotminus

sdotrArr

( ) ( ) ( )2 2 22 2 21

m





22mx d x


11

( )22m m m



1

1 1 11

m m m

x d x d x dM M M m

M


+

rArr

11 1 1 1

81 81 81

d d d dx x x

m

m

xm m

M m


+ + + +sdot sdot

911 1 1 9 1 10 10 10

19 1 9 9 9 9

d



++ +

89 384 10 8

3456 10 10

mm

sdot sdot= = sdot

4inačica




1 2

M mF G

gZx

sdot= sdot


( )

1 2

m mF G

gMd x

sdot= sdot

minus



( ) ( )

( )1 1 1 12 2

2

2 2

2

1


gM gZx xd x d x

x d x

G m


minussdot sdot

minus sdotminus

sdotrArr




( )2 2 2 2 2


( )2

2 2 22 0

242

2 12 2



a




( ) ( ) ( )

( )

2 22 2 4

12 2

M d M d M m M dx

M m


sdot minus

12

( )

2 2 2 2 22 4 4 4

12 2

M d M d M d M m dx

M m


sdot minus

( ) ( )

2 2 2 2 22 4 2 4

12 12

4

2 2


x xM m

d M d

M m



( )

( )( )

( )( )

22 2

12 12 122 2

2

2


M m M m M m



81 81

1 1 81

1281 81

2 2 81

M M m m m mx d x d

M m m mM M mx d


M m m m




minus sdot minus

281 81 81 9 90

1 1 180 80 80

81 9 72281 81

2 280 802 80


m m m

m m mm m x d x d

x d m mm





9 9 8 8384 10 432 101 1 18 8 1

89 9 8 3456 10384 10 22

90

80

2 210 10

72

80

x d x d xm

mm x m

x mx d x d x mm

m


= sdot= sdot = sdot

sdot

sdot

sdotsdot sdot

sdot=


8

3456 102

cx

x mrArr

= sdot

minusminus








Rješenje 110

G = 667 middot 10-11

(N middot m2) kg

2 g = 981 ms

2 r = 64 middot 10

6 m M = 6 middot 10

24 kg

T =




1 22

m mF G

r

sdot= sdot




13

24

2

rF mcp

T

πsdot sdot= sdot





G m g= sdot




gravitacijskoj sili

1inačica




2 24 4

2 2 2 2

2 2

M m r M T rm r

F FG M

G m G mg cpr T r mT


sdotsdot


sdotsdot rArr

2 3 2 3 2 34 4 42 2

2r r r r


π π ππ


sdot sdot sdot sdot

664 106 3

2 64 10 5085 102

11 24667 10 6 10

2

mm s

N mkg

kg

πsdot



35085 10 3600 141 h= sdot =

2inačica



2 2 2

4 4

2 2

r rG F m g m m

Tg mcp

T T m g


sdotsdot rArr

2 2 24 4 42 2

2r r r r

T T T Tg g g g

π π ππ


664 10 3 3

5082 55085 10 141

9

10 3600

812

ms h

m

s

πsdot





21 t)

Rezultat 141 h

14




Rješenje 111


G = 667 middot 10-11

(N middot m2) kg

2 ρ =




1 22

m mF G

r

sdot= sdot




m

Vρ =


3

4

3V r π= sdot sdot

RRRR

rrrr rrrr


2

1 2 2 2 22

1 2 2 22

m m m

m m R FF G F G mm m

GF G R R

R

R

G

= =sdot


sdot

2 22

2 R F R F F F

m m m R m rG G G G



6metoda

3supstitucij

2 6

234 3 4

34 3

3

e 4

m

V F F Fr r

F G G Gm r

G r rr

V r

r

ρ

ρ ρ ρπ ππ

π

=



=

sdot

sdot sdot

15

( )

3 43 3 10

233851 2 2 2 2 3

2 2 2 05 11667 10

2

F

F N kgG

Gr r N m mm

kg

ρ ρπ π π

sdot minus



Vježba 111







T

4 88 4


Rješenje 112

r1 = r T1 = T 1

2 8

T T= sdot r2 =

Keplerovi zakoni




Sunca 2 3

1 12 3

2

2

T r

T r

=

2 3 2 32 3 2 3 2 3 2 3 31 1 2 1

1 2 2 1 1 2 2 1 2

1

21

2 3 22 2 1


r TT T


21 3 1 12 3 3

183 3 3 3 364 642 2 2 22 2 2 6

2

4

T r T r r

r r r r r

T T

T

T



1 1 13 3 3 33 2 2 26 4 4

4 6


Odgovor je pod B

Vježba 112



T

3 99 81


Rezultat A

16




Rješenje 113


h =




1 22

m mF G

r

sdot= sdot




( )2

m M

F G

R h

sdot= sdot

+




Newtonovom poučku

G m g= sdot


na Zemlji jednaka



( ) ( )

( )2

2

2


R h R h

R h

m g


+ +

+sdot

sdot

( ) ( ) 2 2G M G M G M G M



211 24

667 10 6 102

664 10 98901511 989

7332

N mkg

kgh m m km

m

s



Vježba 113


6 middot 1021


-11 (N middot m

2) kg

2)

Rezultat 989 km





(N middot m2) kg

2)

Rješenje 114

v = 652 km s = 6520 m s M = 596 middot 1024

kg G = 667 middot 10-11

(N middot m2) kg

2

T =


17



1 22

m mF G

r

sdot= sdot


tijela


2v

F mcpr

= sdot





2

rv

T

πsdot sdot=




2 2

2 2

2

2 2


r

mr vvrr r

sdotsdot


2

G Mr

v

sdotrArr =


metod

2 222

2

2 2

a

zamj2

ene

G Mr rrv Tv

T vT vTG M

T

v

G M G M vr r rv v v

π πππ



sdot

[ ]

211 24

2 667 10 596 102

2901175 1502 min

3 3

6520

901175 60

N mkg

G M kgT s

v m

s

ππ



Vježba 114



Rezultat 12016 min





Rješenje 115


1 2



18



1 22

m mF G

r

sdot= sdot




( )

( )( ) ( )


težina tijela mase

2 2

1

1 2m na visini h

22

m M m MG G G G

R R

m M m MG G G G

R h R h


rArr rArrsdot sdot

= sdot sdot = sdot

+ +

( ) ( ) ( )

podijelimo

jednadžbe

1

2 2 22

1 1 1

2 2 22 2

m MG

G R R Rm

m MG

G

m MMG G

R h R h R

G

h

G

sdot sdotsdot sdot


sdot sdot sdot sdot

+ + +

( ) ( ) ( )( )

22 2 2

2 22 2 2 2

2

2 2

R h R h R hR h R

R R

R

R


( )2 2 2



Odgovor je pod C

Vježba 115




Rezultat A




N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rješenje 116


v =




1 22

m mF G

r

sdot= sdot



19


2v

F mcpr

= sdot




v 2 r

T

πsdot sdot=




2 22

2 2

v m M v m M M


r

m rr r



pomnožimo

jednadžbe

2

2 22 3

2

rv

r M MTv v G v G

M T r Tv

r

rG

r

π

π π

sdot sdot=


= sdot

3

2 2 23 3 3G M G M G Mv v v

T T T


211 24

2 667 10 596 1023

652284 652 9 000

N mkg

m kmkg

s s s

πsdotminus


= = asymp

Odgovor je pod B

Vježba 116



N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rezultat B





Rješenje 117

F1 = Z r1 = R 50 51

2 50 1 50 50 50

R R R R Rr R R

sdot += + = + = = sdot F2 =


20



1 22

m mF G

r

sdot= sdot




1

1

1 2 2

1

F Z

r R

m M m MF G Z G

r R


=

=


51

22 2 22 2 2601 250 512 250050


r R

r

R

Rsdot sdot sdot


= sdot

2500

2 22601

m MF G

R


sdot


2podijelimo

jednadžbe

2500 2500

222601 26012 2

25

2

002 22

2601

m M m MZ G G

F FRR

m

m MG

R

M m MZ Z m MGF G

RG

RR



sdot

2500 2500 25002 2 096 2 22601 2601 2 1

60

F FZ F Z F Z


Odgovor je pod D

Vježba 117




Rezultat C




Rješenje 118


6 m M = 6 middot 10

24 kg v =




1 22

m mF G

r

sdot= sdot


tijela

21

211

6 7 6 102

N mG

kg

sdotminus= sdot



centripetalna sila

2v

F mcpr

= sdot




( ) ( )

2

2

2 2


R h R

R h

mhR h R h


+ ++

+sdot

+

2 2G M G M G M

v v vR h R h R h


+ + +

211 24

667 10 6 102

77693 6 5

64 10 23 10

N mkg

mkg

sm m


= =sdot + sdot

Vježba 118



kg)

Rezultat 7000 m

sasymp





Rješenje 119

M = 6 middot 1024


6 m

g = 981 m s2 g1 =




1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot


22




Newtonovom poučku

G m g= sdot


na Zemlji jednaka

1inačica


( ) ( )1 2

1

2

mm M m M

G F m g G m g G

R h R h


+ +

( ) ( )

211 24

667 10 6 102

731 1 2 2 2

6 664 10 1 10

N mkg

G M mkgg

sR h m m


sdotrArr = = =

+ sdot + sdot

2inačica


2 2

2


R

m

R R



( ) ( ) ( )

1 1 12 2

2


R h R h R h


+ + +


( )

( ) ( ) ( )

1

2 2 221 1 1

1

podijelimo

jednadž2

e2 2 2

b1

G MG Mg

g g gR h R h R hR

G M G Mg g g

G M

G Mg

R R RR h

sdot sdotsdot=


sdot sdot sdot=

+

( )

2 221 1 1

2

g g gR R R

g g R h g R hR

g

h


sdot++

22 664 10

981 734 1 2 6 6 2

64 10 1 10

R m m mg g

R h s m m s


+ sdot + sdot


Vježba 119




Rezultat 734 2

m

s

23





Rješenje 120





1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot





Newtonovom poučku

G m g= sdot


na Zemlji jednaka


2v

F mcpr

= sdot



m

m VV

ρ ρ= rArr = sdot


3

4

3V r π= sdot sdot



2 2

2

2

2


R R G mR R


sdotrArr

4 3

3

4 3

2 24 3

3

3

v R v RM R

G G

M V V R

M R

ρ π

ρ

ρ

π

π

= sdot = sdot sdot

= sdot

sdot sdot


sdot

sdot

2 24 33

23 4

3

34

v R vR

G GR R

ρ π ρππ

sdotsdot



24

( )

2

3 2920

5130 2 32

11 64 667 10 2439 10

2

m

kgs

N m mm

kgπ

sdot

= =sdotminus



2 2 21

v v v



2

2920

350 6 2

2439 10

m

ms

m s

= =sdot

Vježba 120




Rezultat 5130 350 3 2

kg m

m s

6

1 22

m mF G

r

sdot= sdot





od središta vrtnje

2

2

2 2

2


prr


sdotrArr

24 11

3 10 15 10230

202 10 2

11667 10

2

mm

v r sm kg

S G N m

kg

sdot sdot sdotsdot

rArr = = = sdot

sdotminussdot

vvvv

rrrrmmmmmmmmSSSS

Vježba 107





kg




Rješenje 108





1 22

m mF G

r

sdot= sdot






G m g= sdot






7


0 2

m Mm g G

R

sdotsdot = sdot

bull na visini h

( )

2

m Mm g G

R h

sdotsdot = sdot

+


( ) ( ) ( )

podijelimo

jednadžbe

0 2 2 20 0

2 2 2


m M m Mm g gm g G G

R h R h

m

R h

MG

m

m MmG




+ + +

sdot

sdot

( )

( )1

2 2 220 0 0 0

21

2

g g g gR h R h R hR

g g g R R gR

R h


+

20 0 0 0


R g R g R g g


0 0 01 1 1060 0

0

60 0

0

g g gh R R h R h

g gg

gR h R


sdot sdot

1 1 161 1 6371 10 1 185392563


185392563 1854 km km= asymp

mmmmmmmmMMMMMMMM

gggg0000 gggg

hhhh



Rezultat 752 km




Rješenje 109


x =



izrazom

8

2

mG

r

γ = sdot


tijela




1 22

m mF G

r

sdot= sdot




dddd



C



2

MG

Zx

γ = sdot


( )

2

mG

Md x

γ = sdot

minus

1inačica




( ) ( )

( )2 2 2 2

22

m M m M

G G G GM Z

x xd x d

x d x

Gx



( ) ( ) ( )2 2 22 2 21

m





22mx d x


( )22

1m m m m



9

1

1

1 1m m

x d x dM

M


+

rArr

11 1 1 1

81 81 81

d d d dx x x

m

m

xm m

M m


+ + + +sdot sdot

911 1 1 9 1 10 10 10

19 1 9 9 9 9

d



++ +

89 384 10 8

3456 10 10

mm

sdot sdot= = sdot

2inačica




( ) ( )

( )2 2 2 2

22

m M m M

G G G GM Z

x xd x d

x d x

Gx





( )2 2 2 2 2


( )2

2 2 22 0

242

2 12 2



a




( ) ( ) ( )

( )

2 22 2 4

12 2

M d M d M m M dx

M m


sdot minus

( )

2 2 2 2 22 4 4 4

12 2

M d M d M d M m dx

M m


sdot minus

( ) ( )

2 2 2 2 22 4 2 4

12 12

4

2 2


x xM m

d M d

M m



( )

( )( )

( )( )

22 2

12 12 122 2

2

2


M m M m M m



10

81 81

1 1 81

1281 81

2 2 81

M M m m m mx d x d

M m m mM M mx d


M m m m




minus sdot minus

281 81 81 9 90

1 1 180 80 80

81 9 72281 81

2 280 802 80


m m m

m m mm m x d x d

x d m mm





9 9 8 8384 10 432 101 1 18 8 1

89 9 8 3456 10384 10 22

90

80

2 210 10

72

80

x d x d xm

mm x m

x mx d x d x mm

m


= sdot= sdot = sdot

sdot

sdot

sdotsdot sdot

sdot=


8

3456 102

cx

x mrArr

= sdot

minusminus



dddd


G

3inačica




1 2

M mF G

gZx

sdot= sdot


( )

1 2

m mF G

gMd x

sdot= sdot

minus



( ) ( )

( )1 1 1 12 2

2

2 2

2

1


gM gZx xd x d x

x d x

G m


minussdot sdot

minus sdotminus

sdotrArr

( ) ( ) ( )2 2 22 2 21

m





22mx d x


11

( )22m m m



1

1 1 11

m m m

x d x d x dM M M m

M


+

rArr

11 1 1 1

81 81 81

d d d dx x x

m

m

xm m

M m


+ + + +sdot sdot

911 1 1 9 1 10 10 10

19 1 9 9 9 9

d



++ +

89 384 10 8

3456 10 10

mm

sdot sdot= = sdot

4inačica




1 2

M mF G

gZx

sdot= sdot


( )

1 2

m mF G

gMd x

sdot= sdot

minus



( ) ( )

( )1 1 1 12 2

2

2 2

2

1


gM gZx xd x d x

x d x

G m


minussdot sdot

minus sdotminus

sdotrArr




( )2 2 2 2 2


( )2

2 2 22 0

242

2 12 2



a




( ) ( ) ( )

( )

2 22 2 4

12 2

M d M d M m M dx

M m


sdot minus

12

( )

2 2 2 2 22 4 4 4

12 2

M d M d M d M m dx

M m


sdot minus

( ) ( )

2 2 2 2 22 4 2 4

12 12

4

2 2


x xM m

d M d

M m



( )

( )( )

( )( )

22 2

12 12 122 2

2

2


M m M m M m



81 81

1 1 81

1281 81

2 2 81

M M m m m mx d x d

M m m mM M mx d


M m m m




minus sdot minus

281 81 81 9 90

1 1 180 80 80

81 9 72281 81

2 280 802 80


m m m

m m mm m x d x d

x d m mm





9 9 8 8384 10 432 101 1 18 8 1

89 9 8 3456 10384 10 22

90

80

2 210 10

72

80

x d x d xm

mm x m

x mx d x d x mm

m


= sdot= sdot = sdot

sdot

sdot

sdotsdot sdot

sdot=


8

3456 102

cx

x mrArr

= sdot

minusminus








Rješenje 110

G = 667 middot 10-11

(N middot m2) kg

2 g = 981 ms

2 r = 64 middot 10

6 m M = 6 middot 10

24 kg

T =




1 22

m mF G

r

sdot= sdot




13

24

2

rF mcp

T

πsdot sdot= sdot





G m g= sdot




gravitacijskoj sili

1inačica




2 24 4

2 2 2 2

2 2

M m r M T rm r

F FG M

G m G mg cpr T r mT


sdotsdot


sdotsdot rArr

2 3 2 3 2 34 4 42 2

2r r r r


π π ππ


sdot sdot sdot sdot

664 106 3

2 64 10 5085 102

11 24667 10 6 10

2

mm s

N mkg

kg

πsdot



35085 10 3600 141 h= sdot =

2inačica



2 2 2

4 4

2 2

r rG F m g m m

Tg mcp

T T m g


sdotsdot rArr

2 2 24 4 42 2

2r r r r

T T T Tg g g g

π π ππ


664 10 3 3

5082 55085 10 141

9

10 3600

812

ms h

m

s

πsdot





21 t)

Rezultat 141 h

14




Rješenje 111


G = 667 middot 10-11

(N middot m2) kg

2 ρ =




1 22

m mF G

r

sdot= sdot




m

Vρ =


3

4

3V r π= sdot sdot

RRRR

rrrr rrrr


2

1 2 2 2 22

1 2 2 22

m m m

m m R FF G F G mm m

GF G R R

R

R

G

= =sdot


sdot

2 22

2 R F R F F F

m m m R m rG G G G



6metoda

3supstitucij

2 6

234 3 4

34 3

3

e 4

m

V F F Fr r

F G G Gm r

G r rr

V r

r

ρ

ρ ρ ρπ ππ

π

=



=

sdot

sdot sdot

15

( )

3 43 3 10

233851 2 2 2 2 3

2 2 2 05 11667 10

2

F

F N kgG

Gr r N m mm

kg

ρ ρπ π π

sdot minus



Vježba 111







T

4 88 4


Rješenje 112

r1 = r T1 = T 1

2 8

T T= sdot r2 =

Keplerovi zakoni




Sunca 2 3

1 12 3

2

2

T r

T r

=

2 3 2 32 3 2 3 2 3 2 3 31 1 2 1

1 2 2 1 1 2 2 1 2

1

21

2 3 22 2 1


r TT T


21 3 1 12 3 3

183 3 3 3 364 642 2 2 22 2 2 6

2

4

T r T r r

r r r r r

T T

T

T



1 1 13 3 3 33 2 2 26 4 4

4 6


Odgovor je pod B

Vježba 112



T

3 99 81


Rezultat A

16




Rješenje 113


h =




1 22

m mF G

r

sdot= sdot




( )2

m M

F G

R h

sdot= sdot

+




Newtonovom poučku

G m g= sdot


na Zemlji jednaka



( ) ( )

( )2

2

2


R h R h

R h

m g


+ +

+sdot

sdot

( ) ( ) 2 2G M G M G M G M



211 24

667 10 6 102

664 10 98901511 989

7332

N mkg

kgh m m km

m

s



Vježba 113


6 middot 1021


-11 (N middot m

2) kg

2)

Rezultat 989 km





(N middot m2) kg

2)

Rješenje 114

v = 652 km s = 6520 m s M = 596 middot 1024

kg G = 667 middot 10-11

(N middot m2) kg

2

T =


17



1 22

m mF G

r

sdot= sdot


tijela


2v

F mcpr

= sdot





2

rv

T

πsdot sdot=




2 2

2 2

2

2 2


r

mr vvrr r

sdotsdot


2

G Mr

v

sdotrArr =


metod

2 222

2

2 2

a

zamj2

ene

G Mr rrv Tv

T vT vTG M

T

v

G M G M vr r rv v v

π πππ



sdot

[ ]

211 24

2 667 10 596 102

2901175 1502 min

3 3

6520

901175 60

N mkg

G M kgT s

v m

s

ππ



Vježba 114



Rezultat 12016 min





Rješenje 115


1 2



18



1 22

m mF G

r

sdot= sdot




( )

( )( ) ( )


težina tijela mase

2 2

1

1 2m na visini h

22

m M m MG G G G

R R

m M m MG G G G

R h R h


rArr rArrsdot sdot

= sdot sdot = sdot

+ +

( ) ( ) ( )

podijelimo

jednadžbe

1

2 2 22

1 1 1

2 2 22 2

m MG

G R R Rm

m MG

G

m MMG G

R h R h R

G

h

G

sdot sdotsdot sdot


sdot sdot sdot sdot

+ + +

( ) ( ) ( )( )

22 2 2

2 22 2 2 2

2

2 2

R h R h R hR h R

R R

R

R


( )2 2 2



Odgovor je pod C

Vježba 115




Rezultat A




N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rješenje 116


v =




1 22

m mF G

r

sdot= sdot



19


2v

F mcpr

= sdot




v 2 r

T

πsdot sdot=




2 22

2 2

v m M v m M M


r

m rr r



pomnožimo

jednadžbe

2

2 22 3

2

rv

r M MTv v G v G

M T r Tv

r

rG

r

π

π π

sdot sdot=


= sdot

3

2 2 23 3 3G M G M G Mv v v

T T T


211 24

2 667 10 596 1023

652284 652 9 000

N mkg

m kmkg

s s s

πsdotminus


= = asymp

Odgovor je pod B

Vježba 116



N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rezultat B





Rješenje 117

F1 = Z r1 = R 50 51

2 50 1 50 50 50

R R R R Rr R R

sdot += + = + = = sdot F2 =


20



1 22

m mF G

r

sdot= sdot




1

1

1 2 2

1

F Z

r R

m M m MF G Z G

r R


=

=


51

22 2 22 2 2601 250 512 250050


r R

r

R

Rsdot sdot sdot


= sdot

2500

2 22601

m MF G

R


sdot


2podijelimo

jednadžbe

2500 2500

222601 26012 2

25

2

002 22

2601

m M m MZ G G

F FRR

m

m MG

R

M m MZ Z m MGF G

RG

RR



sdot

2500 2500 25002 2 096 2 22601 2601 2 1

60

F FZ F Z F Z


Odgovor je pod D

Vježba 117




Rezultat C




Rješenje 118


6 m M = 6 middot 10

24 kg v =




1 22

m mF G

r

sdot= sdot


tijela

21

211

6 7 6 102

N mG

kg

sdotminus= sdot



centripetalna sila

2v

F mcpr

= sdot




( ) ( )

2

2

2 2


R h R

R h

mhR h R h


+ ++

+sdot

+

2 2G M G M G M

v v vR h R h R h


+ + +

211 24

667 10 6 102

77693 6 5

64 10 23 10

N mkg

mkg

sm m


= =sdot + sdot

Vježba 118



kg)

Rezultat 7000 m

sasymp





Rješenje 119

M = 6 middot 1024


6 m

g = 981 m s2 g1 =




1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot


22




Newtonovom poučku

G m g= sdot


na Zemlji jednaka

1inačica


( ) ( )1 2

1

2

mm M m M

G F m g G m g G

R h R h


+ +

( ) ( )

211 24

667 10 6 102

731 1 2 2 2

6 664 10 1 10

N mkg

G M mkgg

sR h m m


sdotrArr = = =

+ sdot + sdot

2inačica


2 2

2


R

m

R R



( ) ( ) ( )

1 1 12 2

2


R h R h R h


+ + +


( )

( ) ( ) ( )

1

2 2 221 1 1

1

podijelimo

jednadž2

e2 2 2

b1

G MG Mg

g g gR h R h R hR

G M G Mg g g

G M

G Mg

R R RR h

sdot sdotsdot=


sdot sdot sdot=

+

( )

2 221 1 1

2

g g gR R R

g g R h g R hR

g

h


sdot++

22 664 10

981 734 1 2 6 6 2

64 10 1 10

R m m mg g

R h s m m s


+ sdot + sdot


Vježba 119




Rezultat 734 2

m

s

23





Rješenje 120





1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot





Newtonovom poučku

G m g= sdot


na Zemlji jednaka


2v

F mcpr

= sdot



m

m VV

ρ ρ= rArr = sdot


3

4

3V r π= sdot sdot



2 2

2

2

2


R R G mR R


sdotrArr

4 3

3

4 3

2 24 3

3

3

v R v RM R

G G

M V V R

M R

ρ π

ρ

ρ

π

π

= sdot = sdot sdot

= sdot

sdot sdot


sdot

sdot

2 24 33

23 4

3

34

v R vR

G GR R

ρ π ρππ

sdotsdot



24

( )

2

3 2920

5130 2 32

11 64 667 10 2439 10

2

m

kgs

N m mm

kgπ

sdot

= =sdotminus



2 2 21

v v v



2

2920

350 6 2

2439 10

m

ms

m s

= =sdot

Vježba 120




Rezultat 5130 350 3 2

kg m

m s

7


0 2

m Mm g G

R

sdotsdot = sdot

bull na visini h

( )

2

m Mm g G

R h

sdotsdot = sdot

+


( ) ( ) ( )

podijelimo

jednadžbe

0 2 2 20 0

2 2 2


m M m Mm g gm g G G

R h R h

m

R h

MG

m

m MmG




+ + +

sdot

sdot

( )

( )1

2 2 220 0 0 0

21

2

g g g gR h R h R hR

g g g R R gR

R h


+

20 0 0 0


R g R g R g g


0 0 01 1 1060 0

0

60 0

0

g g gh R R h R h

g gg

gR h R


sdot sdot

1 1 161 1 6371 10 1 185392563


185392563 1854 km km= asymp

mmmmmmmmMMMMMMMM

gggg0000 gggg

hhhh



Rezultat 752 km




Rješenje 109


x =



izrazom

8

2

mG

r

γ = sdot


tijela




1 22

m mF G

r

sdot= sdot




dddd



C



2

MG

Zx

γ = sdot


( )

2

mG

Md x

γ = sdot

minus

1inačica




( ) ( )

( )2 2 2 2

22

m M m M

G G G GM Z

x xd x d

x d x

Gx



( ) ( ) ( )2 2 22 2 21

m





22mx d x


( )22

1m m m m



9

1

1

1 1m m

x d x dM

M


+

rArr

11 1 1 1

81 81 81

d d d dx x x

m

m

xm m

M m


+ + + +sdot sdot

911 1 1 9 1 10 10 10

19 1 9 9 9 9

d



++ +

89 384 10 8

3456 10 10

mm

sdot sdot= = sdot

2inačica




( ) ( )

( )2 2 2 2

22

m M m M

G G G GM Z

x xd x d

x d x

Gx





( )2 2 2 2 2


( )2

2 2 22 0

242

2 12 2



a




( ) ( ) ( )

( )

2 22 2 4

12 2

M d M d M m M dx

M m


sdot minus

( )

2 2 2 2 22 4 4 4

12 2

M d M d M d M m dx

M m


sdot minus

( ) ( )

2 2 2 2 22 4 2 4

12 12

4

2 2


x xM m

d M d

M m



( )

( )( )

( )( )

22 2

12 12 122 2

2

2


M m M m M m



10

81 81

1 1 81

1281 81

2 2 81

M M m m m mx d x d

M m m mM M mx d


M m m m




minus sdot minus

281 81 81 9 90

1 1 180 80 80

81 9 72281 81

2 280 802 80


m m m

m m mm m x d x d

x d m mm





9 9 8 8384 10 432 101 1 18 8 1

89 9 8 3456 10384 10 22

90

80

2 210 10

72

80

x d x d xm

mm x m

x mx d x d x mm

m


= sdot= sdot = sdot

sdot

sdot

sdotsdot sdot

sdot=


8

3456 102

cx

x mrArr

= sdot

minusminus



dddd


G

3inačica




1 2

M mF G

gZx

sdot= sdot


( )

1 2

m mF G

gMd x

sdot= sdot

minus



( ) ( )

( )1 1 1 12 2

2

2 2

2

1


gM gZx xd x d x

x d x

G m


minussdot sdot

minus sdotminus

sdotrArr

( ) ( ) ( )2 2 22 2 21

m





22mx d x


11

( )22m m m



1

1 1 11

m m m

x d x d x dM M M m

M


+

rArr

11 1 1 1

81 81 81

d d d dx x x

m

m

xm m

M m


+ + + +sdot sdot

911 1 1 9 1 10 10 10

19 1 9 9 9 9

d



++ +

89 384 10 8

3456 10 10

mm

sdot sdot= = sdot

4inačica




1 2

M mF G

gZx

sdot= sdot


( )

1 2

m mF G

gMd x

sdot= sdot

minus



( ) ( )

( )1 1 1 12 2

2

2 2

2

1


gM gZx xd x d x

x d x

G m


minussdot sdot

minus sdotminus

sdotrArr




( )2 2 2 2 2


( )2

2 2 22 0

242

2 12 2



a




( ) ( ) ( )

( )

2 22 2 4

12 2

M d M d M m M dx

M m


sdot minus

12

( )

2 2 2 2 22 4 4 4

12 2

M d M d M d M m dx

M m


sdot minus

( ) ( )

2 2 2 2 22 4 2 4

12 12

4

2 2


x xM m

d M d

M m



( )

( )( )

( )( )

22 2

12 12 122 2

2

2


M m M m M m



81 81

1 1 81

1281 81

2 2 81

M M m m m mx d x d

M m m mM M mx d


M m m m




minus sdot minus

281 81 81 9 90

1 1 180 80 80

81 9 72281 81

2 280 802 80


m m m

m m mm m x d x d

x d m mm





9 9 8 8384 10 432 101 1 18 8 1

89 9 8 3456 10384 10 22

90

80

2 210 10

72

80

x d x d xm

mm x m

x mx d x d x mm

m


= sdot= sdot = sdot

sdot

sdot

sdotsdot sdot

sdot=


8

3456 102

cx

x mrArr

= sdot

minusminus








Rješenje 110

G = 667 middot 10-11

(N middot m2) kg

2 g = 981 ms

2 r = 64 middot 10

6 m M = 6 middot 10

24 kg

T =




1 22

m mF G

r

sdot= sdot




13

24

2

rF mcp

T

πsdot sdot= sdot





G m g= sdot




gravitacijskoj sili

1inačica




2 24 4

2 2 2 2

2 2

M m r M T rm r

F FG M

G m G mg cpr T r mT


sdotsdot


sdotsdot rArr

2 3 2 3 2 34 4 42 2

2r r r r


π π ππ


sdot sdot sdot sdot

664 106 3

2 64 10 5085 102

11 24667 10 6 10

2

mm s

N mkg

kg

πsdot



35085 10 3600 141 h= sdot =

2inačica



2 2 2

4 4

2 2

r rG F m g m m

Tg mcp

T T m g


sdotsdot rArr

2 2 24 4 42 2

2r r r r

T T T Tg g g g

π π ππ


664 10 3 3

5082 55085 10 141

9

10 3600

812

ms h

m

s

πsdot





21 t)

Rezultat 141 h

14




Rješenje 111


G = 667 middot 10-11

(N middot m2) kg

2 ρ =




1 22

m mF G

r

sdot= sdot




m

Vρ =


3

4

3V r π= sdot sdot

RRRR

rrrr rrrr


2

1 2 2 2 22

1 2 2 22

m m m

m m R FF G F G mm m

GF G R R

R

R

G

= =sdot


sdot

2 22

2 R F R F F F

m m m R m rG G G G



6metoda

3supstitucij

2 6

234 3 4

34 3

3

e 4

m

V F F Fr r

F G G Gm r

G r rr

V r

r

ρ

ρ ρ ρπ ππ

π

=



=

sdot

sdot sdot

15

( )

3 43 3 10

233851 2 2 2 2 3

2 2 2 05 11667 10

2

F

F N kgG

Gr r N m mm

kg

ρ ρπ π π

sdot minus



Vježba 111







T

4 88 4


Rješenje 112

r1 = r T1 = T 1

2 8

T T= sdot r2 =

Keplerovi zakoni




Sunca 2 3

1 12 3

2

2

T r

T r

=

2 3 2 32 3 2 3 2 3 2 3 31 1 2 1

1 2 2 1 1 2 2 1 2

1

21

2 3 22 2 1


r TT T


21 3 1 12 3 3

183 3 3 3 364 642 2 2 22 2 2 6

2

4

T r T r r

r r r r r

T T

T

T



1 1 13 3 3 33 2 2 26 4 4

4 6


Odgovor je pod B

Vježba 112



T

3 99 81


Rezultat A

16




Rješenje 113


h =




1 22

m mF G

r

sdot= sdot




( )2

m M

F G

R h

sdot= sdot

+




Newtonovom poučku

G m g= sdot


na Zemlji jednaka



( ) ( )

( )2

2

2


R h R h

R h

m g


+ +

+sdot

sdot

( ) ( ) 2 2G M G M G M G M



211 24

667 10 6 102

664 10 98901511 989

7332

N mkg

kgh m m km

m

s



Vježba 113


6 middot 1021


-11 (N middot m

2) kg

2)

Rezultat 989 km





(N middot m2) kg

2)

Rješenje 114

v = 652 km s = 6520 m s M = 596 middot 1024

kg G = 667 middot 10-11

(N middot m2) kg

2

T =


17



1 22

m mF G

r

sdot= sdot


tijela


2v

F mcpr

= sdot





2

rv

T

πsdot sdot=




2 2

2 2

2

2 2


r

mr vvrr r

sdotsdot


2

G Mr

v

sdotrArr =


metod

2 222

2

2 2

a

zamj2

ene

G Mr rrv Tv

T vT vTG M

T

v

G M G M vr r rv v v

π πππ



sdot

[ ]

211 24

2 667 10 596 102

2901175 1502 min

3 3

6520

901175 60

N mkg

G M kgT s

v m

s

ππ



Vježba 114



Rezultat 12016 min





Rješenje 115


1 2



18



1 22

m mF G

r

sdot= sdot




( )

( )( ) ( )


težina tijela mase

2 2

1

1 2m na visini h

22

m M m MG G G G

R R

m M m MG G G G

R h R h


rArr rArrsdot sdot

= sdot sdot = sdot

+ +

( ) ( ) ( )

podijelimo

jednadžbe

1

2 2 22

1 1 1

2 2 22 2

m MG

G R R Rm

m MG

G

m MMG G

R h R h R

G

h

G

sdot sdotsdot sdot


sdot sdot sdot sdot

+ + +

( ) ( ) ( )( )

22 2 2

2 22 2 2 2

2

2 2

R h R h R hR h R

R R

R

R


( )2 2 2



Odgovor je pod C

Vježba 115




Rezultat A




N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rješenje 116


v =




1 22

m mF G

r

sdot= sdot



19


2v

F mcpr

= sdot




v 2 r

T

πsdot sdot=




2 22

2 2

v m M v m M M


r

m rr r



pomnožimo

jednadžbe

2

2 22 3

2

rv

r M MTv v G v G

M T r Tv

r

rG

r

π

π π

sdot sdot=


= sdot

3

2 2 23 3 3G M G M G Mv v v

T T T


211 24

2 667 10 596 1023

652284 652 9 000

N mkg

m kmkg

s s s

πsdotminus


= = asymp

Odgovor je pod B

Vježba 116



N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rezultat B





Rješenje 117

F1 = Z r1 = R 50 51

2 50 1 50 50 50

R R R R Rr R R

sdot += + = + = = sdot F2 =


20



1 22

m mF G

r

sdot= sdot




1

1

1 2 2

1

F Z

r R

m M m MF G Z G

r R


=

=


51

22 2 22 2 2601 250 512 250050


r R

r

R

Rsdot sdot sdot


= sdot

2500

2 22601

m MF G

R


sdot


2podijelimo

jednadžbe

2500 2500

222601 26012 2

25

2

002 22

2601

m M m MZ G G

F FRR

m

m MG

R

M m MZ Z m MGF G

RG

RR



sdot

2500 2500 25002 2 096 2 22601 2601 2 1

60

F FZ F Z F Z


Odgovor je pod D

Vježba 117




Rezultat C




Rješenje 118


6 m M = 6 middot 10

24 kg v =




1 22

m mF G

r

sdot= sdot


tijela

21

211

6 7 6 102

N mG

kg

sdotminus= sdot



centripetalna sila

2v

F mcpr

= sdot




( ) ( )

2

2

2 2


R h R

R h

mhR h R h


+ ++

+sdot

+

2 2G M G M G M

v v vR h R h R h


+ + +

211 24

667 10 6 102

77693 6 5

64 10 23 10

N mkg

mkg

sm m


= =sdot + sdot

Vježba 118



kg)

Rezultat 7000 m

sasymp





Rješenje 119

M = 6 middot 1024


6 m

g = 981 m s2 g1 =




1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot


22




Newtonovom poučku

G m g= sdot


na Zemlji jednaka

1inačica


( ) ( )1 2

1

2

mm M m M

G F m g G m g G

R h R h


+ +

( ) ( )

211 24

667 10 6 102

731 1 2 2 2

6 664 10 1 10

N mkg

G M mkgg

sR h m m


sdotrArr = = =

+ sdot + sdot

2inačica


2 2

2


R

m

R R



( ) ( ) ( )

1 1 12 2

2


R h R h R h


+ + +


( )

( ) ( ) ( )

1

2 2 221 1 1

1

podijelimo

jednadž2

e2 2 2

b1

G MG Mg

g g gR h R h R hR

G M G Mg g g

G M

G Mg

R R RR h

sdot sdotsdot=


sdot sdot sdot=

+

( )

2 221 1 1

2

g g gR R R

g g R h g R hR

g

h


sdot++

22 664 10

981 734 1 2 6 6 2

64 10 1 10

R m m mg g

R h s m m s


+ sdot + sdot


Vježba 119




Rezultat 734 2

m

s

23





Rješenje 120





1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot





Newtonovom poučku

G m g= sdot


na Zemlji jednaka


2v

F mcpr

= sdot



m

m VV

ρ ρ= rArr = sdot


3

4

3V r π= sdot sdot



2 2

2

2

2


R R G mR R


sdotrArr

4 3

3

4 3

2 24 3

3

3

v R v RM R

G G

M V V R

M R

ρ π

ρ

ρ

π

π

= sdot = sdot sdot

= sdot

sdot sdot


sdot

sdot

2 24 33

23 4

3

34

v R vR

G GR R

ρ π ρππ

sdotsdot



24

( )

2

3 2920

5130 2 32

11 64 667 10 2439 10

2

m

kgs

N m mm

kgπ

sdot

= =sdotminus



2 2 21

v v v



2

2920

350 6 2

2439 10

m

ms

m s

= =sdot

Vježba 120




Rezultat 5130 350 3 2

kg m

m s

8

2

mG

r

γ = sdot


tijela




1 22

m mF G

r

sdot= sdot




dddd



C



2

MG

Zx

γ = sdot


( )

2

mG

Md x

γ = sdot

minus

1inačica




( ) ( )

( )2 2 2 2

22

m M m M

G G G GM Z

x xd x d

x d x

Gx



( ) ( ) ( )2 2 22 2 21

m





22mx d x


( )22

1m m m m



9

1

1

1 1m m

x d x dM

M


+

rArr

11 1 1 1

81 81 81

d d d dx x x

m

m

xm m

M m


+ + + +sdot sdot

911 1 1 9 1 10 10 10

19 1 9 9 9 9

d



++ +

89 384 10 8

3456 10 10

mm

sdot sdot= = sdot

2inačica




( ) ( )

( )2 2 2 2

22

m M m M

G G G GM Z

x xd x d

x d x

Gx





( )2 2 2 2 2


( )2

2 2 22 0

242

2 12 2



a




( ) ( ) ( )

( )

2 22 2 4

12 2

M d M d M m M dx

M m


sdot minus

( )

2 2 2 2 22 4 4 4

12 2

M d M d M d M m dx

M m


sdot minus

( ) ( )

2 2 2 2 22 4 2 4

12 12

4

2 2


x xM m

d M d

M m



( )

( )( )

( )( )

22 2

12 12 122 2

2

2


M m M m M m



10

81 81

1 1 81

1281 81

2 2 81

M M m m m mx d x d

M m m mM M mx d


M m m m




minus sdot minus

281 81 81 9 90

1 1 180 80 80

81 9 72281 81

2 280 802 80


m m m

m m mm m x d x d

x d m mm





9 9 8 8384 10 432 101 1 18 8 1

89 9 8 3456 10384 10 22

90

80

2 210 10

72

80

x d x d xm

mm x m

x mx d x d x mm

m


= sdot= sdot = sdot

sdot

sdot

sdotsdot sdot

sdot=


8

3456 102

cx

x mrArr

= sdot

minusminus



dddd


G

3inačica




1 2

M mF G

gZx

sdot= sdot


( )

1 2

m mF G

gMd x

sdot= sdot

minus



( ) ( )

( )1 1 1 12 2

2

2 2

2

1


gM gZx xd x d x

x d x

G m


minussdot sdot

minus sdotminus

sdotrArr

( ) ( ) ( )2 2 22 2 21

m





22mx d x


11

( )22m m m



1

1 1 11

m m m

x d x d x dM M M m

M


+

rArr

11 1 1 1

81 81 81

d d d dx x x

m

m

xm m

M m


+ + + +sdot sdot

911 1 1 9 1 10 10 10

19 1 9 9 9 9

d



++ +

89 384 10 8

3456 10 10

mm

sdot sdot= = sdot

4inačica




1 2

M mF G

gZx

sdot= sdot


( )

1 2

m mF G

gMd x

sdot= sdot

minus



( ) ( )

( )1 1 1 12 2

2

2 2

2

1


gM gZx xd x d x

x d x

G m


minussdot sdot

minus sdotminus

sdotrArr




( )2 2 2 2 2


( )2

2 2 22 0

242

2 12 2



a




( ) ( ) ( )

( )

2 22 2 4

12 2

M d M d M m M dx

M m


sdot minus

12

( )

2 2 2 2 22 4 4 4

12 2

M d M d M d M m dx

M m


sdot minus

( ) ( )

2 2 2 2 22 4 2 4

12 12

4

2 2


x xM m

d M d

M m



( )

( )( )

( )( )

22 2

12 12 122 2

2

2


M m M m M m



81 81

1 1 81

1281 81

2 2 81

M M m m m mx d x d

M m m mM M mx d


M m m m




minus sdot minus

281 81 81 9 90

1 1 180 80 80

81 9 72281 81

2 280 802 80


m m m

m m mm m x d x d

x d m mm





9 9 8 8384 10 432 101 1 18 8 1

89 9 8 3456 10384 10 22

90

80

2 210 10

72

80

x d x d xm

mm x m

x mx d x d x mm

m


= sdot= sdot = sdot

sdot

sdot

sdotsdot sdot

sdot=


8

3456 102

cx

x mrArr

= sdot

minusminus








Rješenje 110

G = 667 middot 10-11

(N middot m2) kg

2 g = 981 ms

2 r = 64 middot 10

6 m M = 6 middot 10

24 kg

T =




1 22

m mF G

r

sdot= sdot




13

24

2

rF mcp

T

πsdot sdot= sdot





G m g= sdot




gravitacijskoj sili

1inačica




2 24 4

2 2 2 2

2 2

M m r M T rm r

F FG M

G m G mg cpr T r mT


sdotsdot


sdotsdot rArr

2 3 2 3 2 34 4 42 2

2r r r r


π π ππ


sdot sdot sdot sdot

664 106 3

2 64 10 5085 102

11 24667 10 6 10

2

mm s

N mkg

kg

πsdot



35085 10 3600 141 h= sdot =

2inačica



2 2 2

4 4

2 2

r rG F m g m m

Tg mcp

T T m g


sdotsdot rArr

2 2 24 4 42 2

2r r r r

T T T Tg g g g

π π ππ


664 10 3 3

5082 55085 10 141

9

10 3600

812

ms h

m

s

πsdot





21 t)

Rezultat 141 h

14




Rješenje 111


G = 667 middot 10-11

(N middot m2) kg

2 ρ =




1 22

m mF G

r

sdot= sdot




m

Vρ =


3

4

3V r π= sdot sdot

RRRR

rrrr rrrr


2

1 2 2 2 22

1 2 2 22

m m m

m m R FF G F G mm m

GF G R R

R

R

G

= =sdot


sdot

2 22

2 R F R F F F

m m m R m rG G G G



6metoda

3supstitucij

2 6

234 3 4

34 3

3

e 4

m

V F F Fr r

F G G Gm r

G r rr

V r

r

ρ

ρ ρ ρπ ππ

π

=



=

sdot

sdot sdot

15

( )

3 43 3 10

233851 2 2 2 2 3

2 2 2 05 11667 10

2

F

F N kgG

Gr r N m mm

kg

ρ ρπ π π

sdot minus



Vježba 111







T

4 88 4


Rješenje 112

r1 = r T1 = T 1

2 8

T T= sdot r2 =

Keplerovi zakoni




Sunca 2 3

1 12 3

2

2

T r

T r

=

2 3 2 32 3 2 3 2 3 2 3 31 1 2 1

1 2 2 1 1 2 2 1 2

1

21

2 3 22 2 1


r TT T


21 3 1 12 3 3

183 3 3 3 364 642 2 2 22 2 2 6

2

4

T r T r r

r r r r r

T T

T

T



1 1 13 3 3 33 2 2 26 4 4

4 6


Odgovor je pod B

Vježba 112



T

3 99 81


Rezultat A

16




Rješenje 113


h =




1 22

m mF G

r

sdot= sdot




( )2

m M

F G

R h

sdot= sdot

+




Newtonovom poučku

G m g= sdot


na Zemlji jednaka



( ) ( )

( )2

2

2


R h R h

R h

m g


+ +

+sdot

sdot

( ) ( ) 2 2G M G M G M G M



211 24

667 10 6 102

664 10 98901511 989

7332

N mkg

kgh m m km

m

s



Vježba 113


6 middot 1021


-11 (N middot m

2) kg

2)

Rezultat 989 km





(N middot m2) kg

2)

Rješenje 114

v = 652 km s = 6520 m s M = 596 middot 1024

kg G = 667 middot 10-11

(N middot m2) kg

2

T =


17



1 22

m mF G

r

sdot= sdot


tijela


2v

F mcpr

= sdot





2

rv

T

πsdot sdot=




2 2

2 2

2

2 2


r

mr vvrr r

sdotsdot


2

G Mr

v

sdotrArr =


metod

2 222

2

2 2

a

zamj2

ene

G Mr rrv Tv

T vT vTG M

T

v

G M G M vr r rv v v

π πππ



sdot

[ ]

211 24

2 667 10 596 102

2901175 1502 min

3 3

6520

901175 60

N mkg

G M kgT s

v m

s

ππ



Vježba 114



Rezultat 12016 min





Rješenje 115


1 2



18



1 22

m mF G

r

sdot= sdot




( )

( )( ) ( )


težina tijela mase

2 2

1

1 2m na visini h

22

m M m MG G G G

R R

m M m MG G G G

R h R h


rArr rArrsdot sdot

= sdot sdot = sdot

+ +

( ) ( ) ( )

podijelimo

jednadžbe

1

2 2 22

1 1 1

2 2 22 2

m MG

G R R Rm

m MG

G

m MMG G

R h R h R

G

h

G

sdot sdotsdot sdot


sdot sdot sdot sdot

+ + +

( ) ( ) ( )( )

22 2 2

2 22 2 2 2

2

2 2

R h R h R hR h R

R R

R

R


( )2 2 2



Odgovor je pod C

Vježba 115




Rezultat A




N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rješenje 116


v =




1 22

m mF G

r

sdot= sdot



19


2v

F mcpr

= sdot




v 2 r

T

πsdot sdot=




2 22

2 2

v m M v m M M


r

m rr r



pomnožimo

jednadžbe

2

2 22 3

2

rv

r M MTv v G v G

M T r Tv

r

rG

r

π

π π

sdot sdot=


= sdot

3

2 2 23 3 3G M G M G Mv v v

T T T


211 24

2 667 10 596 1023

652284 652 9 000

N mkg

m kmkg

s s s

πsdotminus


= = asymp

Odgovor je pod B

Vježba 116



N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rezultat B





Rješenje 117

F1 = Z r1 = R 50 51

2 50 1 50 50 50

R R R R Rr R R

sdot += + = + = = sdot F2 =


20



1 22

m mF G

r

sdot= sdot




1

1

1 2 2

1

F Z

r R

m M m MF G Z G

r R


=

=


51

22 2 22 2 2601 250 512 250050


r R

r

R

Rsdot sdot sdot


= sdot

2500

2 22601

m MF G

R


sdot


2podijelimo

jednadžbe

2500 2500

222601 26012 2

25

2

002 22

2601

m M m MZ G G

F FRR

m

m MG

R

M m MZ Z m MGF G

RG

RR



sdot

2500 2500 25002 2 096 2 22601 2601 2 1

60

F FZ F Z F Z


Odgovor je pod D

Vježba 117




Rezultat C




Rješenje 118


6 m M = 6 middot 10

24 kg v =




1 22

m mF G

r

sdot= sdot


tijela

21

211

6 7 6 102

N mG

kg

sdotminus= sdot



centripetalna sila

2v

F mcpr

= sdot




( ) ( )

2

2

2 2


R h R

R h

mhR h R h


+ ++

+sdot

+

2 2G M G M G M

v v vR h R h R h


+ + +

211 24

667 10 6 102

77693 6 5

64 10 23 10

N mkg

mkg

sm m


= =sdot + sdot

Vježba 118



kg)

Rezultat 7000 m

sasymp





Rješenje 119

M = 6 middot 1024


6 m

g = 981 m s2 g1 =




1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot


22




Newtonovom poučku

G m g= sdot


na Zemlji jednaka

1inačica


( ) ( )1 2

1

2

mm M m M

G F m g G m g G

R h R h


+ +

( ) ( )

211 24

667 10 6 102

731 1 2 2 2

6 664 10 1 10

N mkg

G M mkgg

sR h m m


sdotrArr = = =

+ sdot + sdot

2inačica


2 2

2


R

m

R R



( ) ( ) ( )

1 1 12 2

2


R h R h R h


+ + +


( )

( ) ( ) ( )

1

2 2 221 1 1

1

podijelimo

jednadž2

e2 2 2

b1

G MG Mg

g g gR h R h R hR

G M G Mg g g

G M

G Mg

R R RR h

sdot sdotsdot=


sdot sdot sdot=

+

( )

2 221 1 1

2

g g gR R R

g g R h g R hR

g

h


sdot++

22 664 10

981 734 1 2 6 6 2

64 10 1 10

R m m mg g

R h s m m s


+ sdot + sdot


Vježba 119




Rezultat 734 2

m

s

23





Rješenje 120





1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot





Newtonovom poučku

G m g= sdot


na Zemlji jednaka


2v

F mcpr

= sdot



m

m VV

ρ ρ= rArr = sdot


3

4

3V r π= sdot sdot



2 2

2

2

2


R R G mR R


sdotrArr

4 3

3

4 3

2 24 3

3

3

v R v RM R

G G

M V V R

M R

ρ π

ρ

ρ

π

π

= sdot = sdot sdot

= sdot

sdot sdot


sdot

sdot

2 24 33

23 4

3

34

v R vR

G GR R

ρ π ρππ

sdotsdot



24

( )

2

3 2920

5130 2 32

11 64 667 10 2439 10

2

m

kgs

N m mm

kgπ

sdot

= =sdotminus



2 2 21

v v v



2

2920

350 6 2

2439 10

m

ms

m s

= =sdot

Vježba 120




Rezultat 5130 350 3 2

kg m

m s

9

1

1

1 1m m

x d x dM

M


+

rArr

11 1 1 1

81 81 81

d d d dx x x

m

m

xm m

M m


+ + + +sdot sdot

911 1 1 9 1 10 10 10

19 1 9 9 9 9

d



++ +

89 384 10 8

3456 10 10

mm

sdot sdot= = sdot

2inačica




( ) ( )

( )2 2 2 2

22

m M m M

G G G GM Z

x xd x d

x d x

Gx





( )2 2 2 2 2


( )2

2 2 22 0

242

2 12 2



a




( ) ( ) ( )

( )

2 22 2 4

12 2

M d M d M m M dx

M m


sdot minus

( )

2 2 2 2 22 4 4 4

12 2

M d M d M d M m dx

M m


sdot minus

( ) ( )

2 2 2 2 22 4 2 4

12 12

4

2 2


x xM m

d M d

M m



( )

( )( )

( )( )

22 2

12 12 122 2

2

2


M m M m M m



10

81 81

1 1 81

1281 81

2 2 81

M M m m m mx d x d

M m m mM M mx d


M m m m




minus sdot minus

281 81 81 9 90

1 1 180 80 80

81 9 72281 81

2 280 802 80


m m m

m m mm m x d x d

x d m mm





9 9 8 8384 10 432 101 1 18 8 1

89 9 8 3456 10384 10 22

90

80

2 210 10

72

80

x d x d xm

mm x m

x mx d x d x mm

m


= sdot= sdot = sdot

sdot

sdot

sdotsdot sdot

sdot=


8

3456 102

cx

x mrArr

= sdot

minusminus



dddd


G

3inačica




1 2

M mF G

gZx

sdot= sdot


( )

1 2

m mF G

gMd x

sdot= sdot

minus



( ) ( )

( )1 1 1 12 2

2

2 2

2

1


gM gZx xd x d x

x d x

G m


minussdot sdot

minus sdotminus

sdotrArr

( ) ( ) ( )2 2 22 2 21

m





22mx d x


11

( )22m m m



1

1 1 11

m m m

x d x d x dM M M m

M


+

rArr

11 1 1 1

81 81 81

d d d dx x x

m

m

xm m

M m


+ + + +sdot sdot

911 1 1 9 1 10 10 10

19 1 9 9 9 9

d



++ +

89 384 10 8

3456 10 10

mm

sdot sdot= = sdot

4inačica




1 2

M mF G

gZx

sdot= sdot


( )

1 2

m mF G

gMd x

sdot= sdot

minus



( ) ( )

( )1 1 1 12 2

2

2 2

2

1


gM gZx xd x d x

x d x

G m


minussdot sdot

minus sdotminus

sdotrArr




( )2 2 2 2 2


( )2

2 2 22 0

242

2 12 2



a




( ) ( ) ( )

( )

2 22 2 4

12 2

M d M d M m M dx

M m


sdot minus

12

( )

2 2 2 2 22 4 4 4

12 2

M d M d M d M m dx

M m


sdot minus

( ) ( )

2 2 2 2 22 4 2 4

12 12

4

2 2


x xM m

d M d

M m



( )

( )( )

( )( )

22 2

12 12 122 2

2

2


M m M m M m



81 81

1 1 81

1281 81

2 2 81

M M m m m mx d x d

M m m mM M mx d


M m m m




minus sdot minus

281 81 81 9 90

1 1 180 80 80

81 9 72281 81

2 280 802 80


m m m

m m mm m x d x d

x d m mm





9 9 8 8384 10 432 101 1 18 8 1

89 9 8 3456 10384 10 22

90

80

2 210 10

72

80

x d x d xm

mm x m

x mx d x d x mm

m


= sdot= sdot = sdot

sdot

sdot

sdotsdot sdot

sdot=


8

3456 102

cx

x mrArr

= sdot

minusminus








Rješenje 110

G = 667 middot 10-11

(N middot m2) kg

2 g = 981 ms

2 r = 64 middot 10

6 m M = 6 middot 10

24 kg

T =




1 22

m mF G

r

sdot= sdot




13

24

2

rF mcp

T

πsdot sdot= sdot





G m g= sdot




gravitacijskoj sili

1inačica




2 24 4

2 2 2 2

2 2

M m r M T rm r

F FG M

G m G mg cpr T r mT


sdotsdot


sdotsdot rArr

2 3 2 3 2 34 4 42 2

2r r r r


π π ππ


sdot sdot sdot sdot

664 106 3

2 64 10 5085 102

11 24667 10 6 10

2

mm s

N mkg

kg

πsdot



35085 10 3600 141 h= sdot =

2inačica



2 2 2

4 4

2 2

r rG F m g m m

Tg mcp

T T m g


sdotsdot rArr

2 2 24 4 42 2

2r r r r

T T T Tg g g g

π π ππ


664 10 3 3

5082 55085 10 141

9

10 3600

812

ms h

m

s

πsdot





21 t)

Rezultat 141 h

14




Rješenje 111


G = 667 middot 10-11

(N middot m2) kg

2 ρ =




1 22

m mF G

r

sdot= sdot




m

Vρ =


3

4

3V r π= sdot sdot

RRRR

rrrr rrrr


2

1 2 2 2 22

1 2 2 22

m m m

m m R FF G F G mm m

GF G R R

R

R

G

= =sdot


sdot

2 22

2 R F R F F F

m m m R m rG G G G



6metoda

3supstitucij

2 6

234 3 4

34 3

3

e 4

m

V F F Fr r

F G G Gm r

G r rr

V r

r

ρ

ρ ρ ρπ ππ

π

=



=

sdot

sdot sdot

15

( )

3 43 3 10

233851 2 2 2 2 3

2 2 2 05 11667 10

2

F

F N kgG

Gr r N m mm

kg

ρ ρπ π π

sdot minus



Vježba 111







T

4 88 4


Rješenje 112

r1 = r T1 = T 1

2 8

T T= sdot r2 =

Keplerovi zakoni




Sunca 2 3

1 12 3

2

2

T r

T r

=

2 3 2 32 3 2 3 2 3 2 3 31 1 2 1

1 2 2 1 1 2 2 1 2

1

21

2 3 22 2 1


r TT T


21 3 1 12 3 3

183 3 3 3 364 642 2 2 22 2 2 6

2

4

T r T r r

r r r r r

T T

T

T



1 1 13 3 3 33 2 2 26 4 4

4 6


Odgovor je pod B

Vježba 112



T

3 99 81


Rezultat A

16




Rješenje 113


h =




1 22

m mF G

r

sdot= sdot




( )2

m M

F G

R h

sdot= sdot

+




Newtonovom poučku

G m g= sdot


na Zemlji jednaka



( ) ( )

( )2

2

2


R h R h

R h

m g


+ +

+sdot

sdot

( ) ( ) 2 2G M G M G M G M



211 24

667 10 6 102

664 10 98901511 989

7332

N mkg

kgh m m km

m

s



Vježba 113


6 middot 1021


-11 (N middot m

2) kg

2)

Rezultat 989 km





(N middot m2) kg

2)

Rješenje 114

v = 652 km s = 6520 m s M = 596 middot 1024

kg G = 667 middot 10-11

(N middot m2) kg

2

T =


17



1 22

m mF G

r

sdot= sdot


tijela


2v

F mcpr

= sdot





2

rv

T

πsdot sdot=




2 2

2 2

2

2 2


r

mr vvrr r

sdotsdot


2

G Mr

v

sdotrArr =


metod

2 222

2

2 2

a

zamj2

ene

G Mr rrv Tv

T vT vTG M

T

v

G M G M vr r rv v v

π πππ



sdot

[ ]

211 24

2 667 10 596 102

2901175 1502 min

3 3

6520

901175 60

N mkg

G M kgT s

v m

s

ππ



Vježba 114



Rezultat 12016 min





Rješenje 115


1 2



18



1 22

m mF G

r

sdot= sdot




( )

( )( ) ( )


težina tijela mase

2 2

1

1 2m na visini h

22

m M m MG G G G

R R

m M m MG G G G

R h R h


rArr rArrsdot sdot

= sdot sdot = sdot

+ +

( ) ( ) ( )

podijelimo

jednadžbe

1

2 2 22

1 1 1

2 2 22 2

m MG

G R R Rm

m MG

G

m MMG G

R h R h R

G

h

G

sdot sdotsdot sdot


sdot sdot sdot sdot

+ + +

( ) ( ) ( )( )

22 2 2

2 22 2 2 2

2

2 2

R h R h R hR h R

R R

R

R


( )2 2 2



Odgovor je pod C

Vježba 115




Rezultat A




N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rješenje 116


v =




1 22

m mF G

r

sdot= sdot



19


2v

F mcpr

= sdot




v 2 r

T

πsdot sdot=




2 22

2 2

v m M v m M M


r

m rr r



pomnožimo

jednadžbe

2

2 22 3

2

rv

r M MTv v G v G

M T r Tv

r

rG

r

π

π π

sdot sdot=


= sdot

3

2 2 23 3 3G M G M G Mv v v

T T T


211 24

2 667 10 596 1023

652284 652 9 000

N mkg

m kmkg

s s s

πsdotminus


= = asymp

Odgovor je pod B

Vježba 116



N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rezultat B





Rješenje 117

F1 = Z r1 = R 50 51

2 50 1 50 50 50

R R R R Rr R R

sdot += + = + = = sdot F2 =


20



1 22

m mF G

r

sdot= sdot




1

1

1 2 2

1

F Z

r R

m M m MF G Z G

r R


=

=


51

22 2 22 2 2601 250 512 250050


r R

r

R

Rsdot sdot sdot


= sdot

2500

2 22601

m MF G

R


sdot


2podijelimo

jednadžbe

2500 2500

222601 26012 2

25

2

002 22

2601

m M m MZ G G

F FRR

m

m MG

R

M m MZ Z m MGF G

RG

RR



sdot

2500 2500 25002 2 096 2 22601 2601 2 1

60

F FZ F Z F Z


Odgovor je pod D

Vježba 117




Rezultat C




Rješenje 118


6 m M = 6 middot 10

24 kg v =




1 22

m mF G

r

sdot= sdot


tijela

21

211

6 7 6 102

N mG

kg

sdotminus= sdot



centripetalna sila

2v

F mcpr

= sdot




( ) ( )

2

2

2 2


R h R

R h

mhR h R h


+ ++

+sdot

+

2 2G M G M G M

v v vR h R h R h


+ + +

211 24

667 10 6 102

77693 6 5

64 10 23 10

N mkg

mkg

sm m


= =sdot + sdot

Vježba 118



kg)

Rezultat 7000 m

sasymp





Rješenje 119

M = 6 middot 1024


6 m

g = 981 m s2 g1 =




1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot


22




Newtonovom poučku

G m g= sdot


na Zemlji jednaka

1inačica


( ) ( )1 2

1

2

mm M m M

G F m g G m g G

R h R h


+ +

( ) ( )

211 24

667 10 6 102

731 1 2 2 2

6 664 10 1 10

N mkg

G M mkgg

sR h m m


sdotrArr = = =

+ sdot + sdot

2inačica


2 2

2


R

m

R R



( ) ( ) ( )

1 1 12 2

2


R h R h R h


+ + +


( )

( ) ( ) ( )

1

2 2 221 1 1

1

podijelimo

jednadž2

e2 2 2

b1

G MG Mg

g g gR h R h R hR

G M G Mg g g

G M

G Mg

R R RR h

sdot sdotsdot=


sdot sdot sdot=

+

( )

2 221 1 1

2

g g gR R R

g g R h g R hR

g

h


sdot++

22 664 10

981 734 1 2 6 6 2

64 10 1 10

R m m mg g

R h s m m s


+ sdot + sdot


Vježba 119




Rezultat 734 2

m

s

23





Rješenje 120





1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot





Newtonovom poučku

G m g= sdot


na Zemlji jednaka


2v

F mcpr

= sdot



m

m VV

ρ ρ= rArr = sdot


3

4

3V r π= sdot sdot



2 2

2

2

2


R R G mR R


sdotrArr

4 3

3

4 3

2 24 3

3

3

v R v RM R

G G

M V V R

M R

ρ π

ρ

ρ

π

π

= sdot = sdot sdot

= sdot

sdot sdot


sdot

sdot

2 24 33

23 4

3

34

v R vR

G GR R

ρ π ρππ

sdotsdot



24

( )

2

3 2920

5130 2 32

11 64 667 10 2439 10

2

m

kgs

N m mm

kgπ

sdot

= =sdotminus



2 2 21

v v v



2

2920

350 6 2

2439 10

m

ms

m s

= =sdot

Vježba 120




Rezultat 5130 350 3 2

kg m

m s

10

81 81

1 1 81

1281 81

2 2 81

M M m m m mx d x d

M m m mM M mx d


M m m m




minus sdot minus

281 81 81 9 90

1 1 180 80 80

81 9 72281 81

2 280 802 80


m m m

m m mm m x d x d

x d m mm





9 9 8 8384 10 432 101 1 18 8 1

89 9 8 3456 10384 10 22

90

80

2 210 10

72

80

x d x d xm

mm x m

x mx d x d x mm

m


= sdot= sdot = sdot

sdot

sdot

sdotsdot sdot

sdot=


8

3456 102

cx

x mrArr

= sdot

minusminus



dddd


G

3inačica




1 2

M mF G

gZx

sdot= sdot


( )

1 2

m mF G

gMd x

sdot= sdot

minus



( ) ( )

( )1 1 1 12 2

2

2 2

2

1


gM gZx xd x d x

x d x

G m


minussdot sdot

minus sdotminus

sdotrArr

( ) ( ) ( )2 2 22 2 21

m





22mx d x


11

( )22m m m



1

1 1 11

m m m

x d x d x dM M M m

M


+

rArr

11 1 1 1

81 81 81

d d d dx x x

m

m

xm m

M m


+ + + +sdot sdot

911 1 1 9 1 10 10 10

19 1 9 9 9 9

d



++ +

89 384 10 8

3456 10 10

mm

sdot sdot= = sdot

4inačica




1 2

M mF G

gZx

sdot= sdot


( )

1 2

m mF G

gMd x

sdot= sdot

minus



( ) ( )

( )1 1 1 12 2

2

2 2

2

1


gM gZx xd x d x

x d x

G m


minussdot sdot

minus sdotminus

sdotrArr




( )2 2 2 2 2


( )2

2 2 22 0

242

2 12 2



a




( ) ( ) ( )

( )

2 22 2 4

12 2

M d M d M m M dx

M m


sdot minus

12

( )

2 2 2 2 22 4 4 4

12 2

M d M d M d M m dx

M m


sdot minus

( ) ( )

2 2 2 2 22 4 2 4

12 12

4

2 2


x xM m

d M d

M m



( )

( )( )

( )( )

22 2

12 12 122 2

2

2


M m M m M m



81 81

1 1 81

1281 81

2 2 81

M M m m m mx d x d

M m m mM M mx d


M m m m




minus sdot minus

281 81 81 9 90

1 1 180 80 80

81 9 72281 81

2 280 802 80


m m m

m m mm m x d x d

x d m mm





9 9 8 8384 10 432 101 1 18 8 1

89 9 8 3456 10384 10 22

90

80

2 210 10

72

80

x d x d xm

mm x m

x mx d x d x mm

m


= sdot= sdot = sdot

sdot

sdot

sdotsdot sdot

sdot=


8

3456 102

cx

x mrArr

= sdot

minusminus








Rješenje 110

G = 667 middot 10-11

(N middot m2) kg

2 g = 981 ms

2 r = 64 middot 10

6 m M = 6 middot 10

24 kg

T =




1 22

m mF G

r

sdot= sdot




13

24

2

rF mcp

T

πsdot sdot= sdot





G m g= sdot




gravitacijskoj sili

1inačica




2 24 4

2 2 2 2

2 2

M m r M T rm r

F FG M

G m G mg cpr T r mT


sdotsdot


sdotsdot rArr

2 3 2 3 2 34 4 42 2

2r r r r


π π ππ


sdot sdot sdot sdot

664 106 3

2 64 10 5085 102

11 24667 10 6 10

2

mm s

N mkg

kg

πsdot



35085 10 3600 141 h= sdot =

2inačica



2 2 2

4 4

2 2

r rG F m g m m

Tg mcp

T T m g


sdotsdot rArr

2 2 24 4 42 2

2r r r r

T T T Tg g g g

π π ππ


664 10 3 3

5082 55085 10 141

9

10 3600

812

ms h

m

s

πsdot





21 t)

Rezultat 141 h

14




Rješenje 111


G = 667 middot 10-11

(N middot m2) kg

2 ρ =




1 22

m mF G

r

sdot= sdot




m

Vρ =


3

4

3V r π= sdot sdot

RRRR

rrrr rrrr


2

1 2 2 2 22

1 2 2 22

m m m

m m R FF G F G mm m

GF G R R

R

R

G

= =sdot


sdot

2 22

2 R F R F F F

m m m R m rG G G G



6metoda

3supstitucij

2 6

234 3 4

34 3

3

e 4

m

V F F Fr r

F G G Gm r

G r rr

V r

r

ρ

ρ ρ ρπ ππ

π

=



=

sdot

sdot sdot

15

( )

3 43 3 10

233851 2 2 2 2 3

2 2 2 05 11667 10

2

F

F N kgG

Gr r N m mm

kg

ρ ρπ π π

sdot minus



Vježba 111







T

4 88 4


Rješenje 112

r1 = r T1 = T 1

2 8

T T= sdot r2 =

Keplerovi zakoni




Sunca 2 3

1 12 3

2

2

T r

T r

=

2 3 2 32 3 2 3 2 3 2 3 31 1 2 1

1 2 2 1 1 2 2 1 2

1

21

2 3 22 2 1


r TT T


21 3 1 12 3 3

183 3 3 3 364 642 2 2 22 2 2 6

2

4

T r T r r

r r r r r

T T

T

T



1 1 13 3 3 33 2 2 26 4 4

4 6


Odgovor je pod B

Vježba 112



T

3 99 81


Rezultat A

16




Rješenje 113


h =




1 22

m mF G

r

sdot= sdot




( )2

m M

F G

R h

sdot= sdot

+




Newtonovom poučku

G m g= sdot


na Zemlji jednaka



( ) ( )

( )2

2

2


R h R h

R h

m g


+ +

+sdot

sdot

( ) ( ) 2 2G M G M G M G M



211 24

667 10 6 102

664 10 98901511 989

7332

N mkg

kgh m m km

m

s



Vježba 113


6 middot 1021


-11 (N middot m

2) kg

2)

Rezultat 989 km





(N middot m2) kg

2)

Rješenje 114

v = 652 km s = 6520 m s M = 596 middot 1024

kg G = 667 middot 10-11

(N middot m2) kg

2

T =


17



1 22

m mF G

r

sdot= sdot


tijela


2v

F mcpr

= sdot





2

rv

T

πsdot sdot=




2 2

2 2

2

2 2


r

mr vvrr r

sdotsdot


2

G Mr

v

sdotrArr =


metod

2 222

2

2 2

a

zamj2

ene

G Mr rrv Tv

T vT vTG M

T

v

G M G M vr r rv v v

π πππ



sdot

[ ]

211 24

2 667 10 596 102

2901175 1502 min

3 3

6520

901175 60

N mkg

G M kgT s

v m

s

ππ



Vježba 114



Rezultat 12016 min





Rješenje 115


1 2



18



1 22

m mF G

r

sdot= sdot




( )

( )( ) ( )


težina tijela mase

2 2

1

1 2m na visini h

22

m M m MG G G G

R R

m M m MG G G G

R h R h


rArr rArrsdot sdot

= sdot sdot = sdot

+ +

( ) ( ) ( )

podijelimo

jednadžbe

1

2 2 22

1 1 1

2 2 22 2

m MG

G R R Rm

m MG

G

m MMG G

R h R h R

G

h

G

sdot sdotsdot sdot


sdot sdot sdot sdot

+ + +

( ) ( ) ( )( )

22 2 2

2 22 2 2 2

2

2 2

R h R h R hR h R

R R

R

R


( )2 2 2



Odgovor je pod C

Vježba 115




Rezultat A




N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rješenje 116


v =




1 22

m mF G

r

sdot= sdot



19


2v

F mcpr

= sdot




v 2 r

T

πsdot sdot=




2 22

2 2

v m M v m M M


r

m rr r



pomnožimo

jednadžbe

2

2 22 3

2

rv

r M MTv v G v G

M T r Tv

r

rG

r

π

π π

sdot sdot=


= sdot

3

2 2 23 3 3G M G M G Mv v v

T T T


211 24

2 667 10 596 1023

652284 652 9 000

N mkg

m kmkg

s s s

πsdotminus


= = asymp

Odgovor je pod B

Vježba 116



N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rezultat B





Rješenje 117

F1 = Z r1 = R 50 51

2 50 1 50 50 50

R R R R Rr R R

sdot += + = + = = sdot F2 =


20



1 22

m mF G

r

sdot= sdot




1

1

1 2 2

1

F Z

r R

m M m MF G Z G

r R


=

=


51

22 2 22 2 2601 250 512 250050


r R

r

R

Rsdot sdot sdot


= sdot

2500

2 22601

m MF G

R


sdot


2podijelimo

jednadžbe

2500 2500

222601 26012 2

25

2

002 22

2601

m M m MZ G G

F FRR

m

m MG

R

M m MZ Z m MGF G

RG

RR



sdot

2500 2500 25002 2 096 2 22601 2601 2 1

60

F FZ F Z F Z


Odgovor je pod D

Vježba 117




Rezultat C




Rješenje 118


6 m M = 6 middot 10

24 kg v =




1 22

m mF G

r

sdot= sdot


tijela

21

211

6 7 6 102

N mG

kg

sdotminus= sdot



centripetalna sila

2v

F mcpr

= sdot




( ) ( )

2

2

2 2


R h R

R h

mhR h R h


+ ++

+sdot

+

2 2G M G M G M

v v vR h R h R h


+ + +

211 24

667 10 6 102

77693 6 5

64 10 23 10

N mkg

mkg

sm m


= =sdot + sdot

Vježba 118



kg)

Rezultat 7000 m

sasymp





Rješenje 119

M = 6 middot 1024


6 m

g = 981 m s2 g1 =




1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot


22




Newtonovom poučku

G m g= sdot


na Zemlji jednaka

1inačica


( ) ( )1 2

1

2

mm M m M

G F m g G m g G

R h R h


+ +

( ) ( )

211 24

667 10 6 102

731 1 2 2 2

6 664 10 1 10

N mkg

G M mkgg

sR h m m


sdotrArr = = =

+ sdot + sdot

2inačica


2 2

2


R

m

R R



( ) ( ) ( )

1 1 12 2

2


R h R h R h


+ + +


( )

( ) ( ) ( )

1

2 2 221 1 1

1

podijelimo

jednadž2

e2 2 2

b1

G MG Mg

g g gR h R h R hR

G M G Mg g g

G M

G Mg

R R RR h

sdot sdotsdot=


sdot sdot sdot=

+

( )

2 221 1 1

2

g g gR R R

g g R h g R hR

g

h


sdot++

22 664 10

981 734 1 2 6 6 2

64 10 1 10

R m m mg g

R h s m m s


+ sdot + sdot


Vježba 119




Rezultat 734 2

m

s

23





Rješenje 120





1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot





Newtonovom poučku

G m g= sdot


na Zemlji jednaka


2v

F mcpr

= sdot



m

m VV

ρ ρ= rArr = sdot


3

4

3V r π= sdot sdot



2 2

2

2

2


R R G mR R


sdotrArr

4 3

3

4 3

2 24 3

3

3

v R v RM R

G G

M V V R

M R

ρ π

ρ

ρ

π

π

= sdot = sdot sdot

= sdot

sdot sdot


sdot

sdot

2 24 33

23 4

3

34

v R vR

G GR R

ρ π ρππ

sdotsdot



24

( )

2

3 2920

5130 2 32

11 64 667 10 2439 10

2

m

kgs

N m mm

kgπ

sdot

= =sdotminus



2 2 21

v v v



2

2920

350 6 2

2439 10

m

ms

m s

= =sdot

Vježba 120




Rezultat 5130 350 3 2

kg m

m s

11

( )22m m m



1

1 1 11

m m m

x d x d x dM M M m

M


+

rArr

11 1 1 1

81 81 81

d d d dx x x

m

m

xm m

M m


+ + + +sdot sdot

911 1 1 9 1 10 10 10

19 1 9 9 9 9

d



++ +

89 384 10 8

3456 10 10

mm

sdot sdot= = sdot

4inačica




1 2

M mF G

gZx

sdot= sdot


( )

1 2

m mF G

gMd x

sdot= sdot

minus



( ) ( )

( )1 1 1 12 2

2

2 2

2

1


gM gZx xd x d x

x d x

G m


minussdot sdot

minus sdotminus

sdotrArr




( )2 2 2 2 2


( )2

2 2 22 0

242

2 12 2



a




( ) ( ) ( )

( )

2 22 2 4

12 2

M d M d M m M dx

M m


sdot minus

12

( )

2 2 2 2 22 4 4 4

12 2

M d M d M d M m dx

M m


sdot minus

( ) ( )

2 2 2 2 22 4 2 4

12 12

4

2 2


x xM m

d M d

M m



( )

( )( )

( )( )

22 2

12 12 122 2

2

2


M m M m M m



81 81

1 1 81

1281 81

2 2 81

M M m m m mx d x d

M m m mM M mx d


M m m m




minus sdot minus

281 81 81 9 90

1 1 180 80 80

81 9 72281 81

2 280 802 80


m m m

m m mm m x d x d

x d m mm





9 9 8 8384 10 432 101 1 18 8 1

89 9 8 3456 10384 10 22

90

80

2 210 10

72

80

x d x d xm

mm x m

x mx d x d x mm

m


= sdot= sdot = sdot

sdot

sdot

sdotsdot sdot

sdot=


8

3456 102

cx

x mrArr

= sdot

minusminus








Rješenje 110

G = 667 middot 10-11

(N middot m2) kg

2 g = 981 ms

2 r = 64 middot 10

6 m M = 6 middot 10

24 kg

T =




1 22

m mF G

r

sdot= sdot




13

24

2

rF mcp

T

πsdot sdot= sdot





G m g= sdot




gravitacijskoj sili

1inačica




2 24 4

2 2 2 2

2 2

M m r M T rm r

F FG M

G m G mg cpr T r mT


sdotsdot


sdotsdot rArr

2 3 2 3 2 34 4 42 2

2r r r r


π π ππ


sdot sdot sdot sdot

664 106 3

2 64 10 5085 102

11 24667 10 6 10

2

mm s

N mkg

kg

πsdot



35085 10 3600 141 h= sdot =

2inačica



2 2 2

4 4

2 2

r rG F m g m m

Tg mcp

T T m g


sdotsdot rArr

2 2 24 4 42 2

2r r r r

T T T Tg g g g

π π ππ


664 10 3 3

5082 55085 10 141

9

10 3600

812

ms h

m

s

πsdot





21 t)

Rezultat 141 h

14




Rješenje 111


G = 667 middot 10-11

(N middot m2) kg

2 ρ =




1 22

m mF G

r

sdot= sdot




m

Vρ =


3

4

3V r π= sdot sdot

RRRR

rrrr rrrr


2

1 2 2 2 22

1 2 2 22

m m m

m m R FF G F G mm m

GF G R R

R

R

G

= =sdot


sdot

2 22

2 R F R F F F

m m m R m rG G G G



6metoda

3supstitucij

2 6

234 3 4

34 3

3

e 4

m

V F F Fr r

F G G Gm r

G r rr

V r

r

ρ

ρ ρ ρπ ππ

π

=



=

sdot

sdot sdot

15

( )

3 43 3 10

233851 2 2 2 2 3

2 2 2 05 11667 10

2

F

F N kgG

Gr r N m mm

kg

ρ ρπ π π

sdot minus



Vježba 111







T

4 88 4


Rješenje 112

r1 = r T1 = T 1

2 8

T T= sdot r2 =

Keplerovi zakoni




Sunca 2 3

1 12 3

2

2

T r

T r

=

2 3 2 32 3 2 3 2 3 2 3 31 1 2 1

1 2 2 1 1 2 2 1 2

1

21

2 3 22 2 1


r TT T


21 3 1 12 3 3

183 3 3 3 364 642 2 2 22 2 2 6

2

4

T r T r r

r r r r r

T T

T

T



1 1 13 3 3 33 2 2 26 4 4

4 6


Odgovor je pod B

Vježba 112



T

3 99 81


Rezultat A

16




Rješenje 113


h =




1 22

m mF G

r

sdot= sdot




( )2

m M

F G

R h

sdot= sdot

+




Newtonovom poučku

G m g= sdot


na Zemlji jednaka



( ) ( )

( )2

2

2


R h R h

R h

m g


+ +

+sdot

sdot

( ) ( ) 2 2G M G M G M G M



211 24

667 10 6 102

664 10 98901511 989

7332

N mkg

kgh m m km

m

s



Vježba 113


6 middot 1021


-11 (N middot m

2) kg

2)

Rezultat 989 km





(N middot m2) kg

2)

Rješenje 114

v = 652 km s = 6520 m s M = 596 middot 1024

kg G = 667 middot 10-11

(N middot m2) kg

2

T =


17



1 22

m mF G

r

sdot= sdot


tijela


2v

F mcpr

= sdot





2

rv

T

πsdot sdot=




2 2

2 2

2

2 2


r

mr vvrr r

sdotsdot


2

G Mr

v

sdotrArr =


metod

2 222

2

2 2

a

zamj2

ene

G Mr rrv Tv

T vT vTG M

T

v

G M G M vr r rv v v

π πππ



sdot

[ ]

211 24

2 667 10 596 102

2901175 1502 min

3 3

6520

901175 60

N mkg

G M kgT s

v m

s

ππ



Vježba 114



Rezultat 12016 min





Rješenje 115


1 2



18



1 22

m mF G

r

sdot= sdot




( )

( )( ) ( )


težina tijela mase

2 2

1

1 2m na visini h

22

m M m MG G G G

R R

m M m MG G G G

R h R h


rArr rArrsdot sdot

= sdot sdot = sdot

+ +

( ) ( ) ( )

podijelimo

jednadžbe

1

2 2 22

1 1 1

2 2 22 2

m MG

G R R Rm

m MG

G

m MMG G

R h R h R

G

h

G

sdot sdotsdot sdot


sdot sdot sdot sdot

+ + +

( ) ( ) ( )( )

22 2 2

2 22 2 2 2

2

2 2

R h R h R hR h R

R R

R

R


( )2 2 2



Odgovor je pod C

Vježba 115




Rezultat A




N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rješenje 116


v =




1 22

m mF G

r

sdot= sdot



19


2v

F mcpr

= sdot




v 2 r

T

πsdot sdot=




2 22

2 2

v m M v m M M


r

m rr r



pomnožimo

jednadžbe

2

2 22 3

2

rv

r M MTv v G v G

M T r Tv

r

rG

r

π

π π

sdot sdot=


= sdot

3

2 2 23 3 3G M G M G Mv v v

T T T


211 24

2 667 10 596 1023

652284 652 9 000

N mkg

m kmkg

s s s

πsdotminus


= = asymp

Odgovor je pod B

Vježba 116



N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rezultat B





Rješenje 117

F1 = Z r1 = R 50 51

2 50 1 50 50 50

R R R R Rr R R

sdot += + = + = = sdot F2 =


20



1 22

m mF G

r

sdot= sdot




1

1

1 2 2

1

F Z

r R

m M m MF G Z G

r R


=

=


51

22 2 22 2 2601 250 512 250050


r R

r

R

Rsdot sdot sdot


= sdot

2500

2 22601

m MF G

R


sdot


2podijelimo

jednadžbe

2500 2500

222601 26012 2

25

2

002 22

2601

m M m MZ G G

F FRR

m

m MG

R

M m MZ Z m MGF G

RG

RR



sdot

2500 2500 25002 2 096 2 22601 2601 2 1

60

F FZ F Z F Z


Odgovor je pod D

Vježba 117




Rezultat C




Rješenje 118


6 m M = 6 middot 10

24 kg v =




1 22

m mF G

r

sdot= sdot


tijela

21

211

6 7 6 102

N mG

kg

sdotminus= sdot



centripetalna sila

2v

F mcpr

= sdot




( ) ( )

2

2

2 2


R h R

R h

mhR h R h


+ ++

+sdot

+

2 2G M G M G M

v v vR h R h R h


+ + +

211 24

667 10 6 102

77693 6 5

64 10 23 10

N mkg

mkg

sm m


= =sdot + sdot

Vježba 118



kg)

Rezultat 7000 m

sasymp





Rješenje 119

M = 6 middot 1024


6 m

g = 981 m s2 g1 =




1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot


22




Newtonovom poučku

G m g= sdot


na Zemlji jednaka

1inačica


( ) ( )1 2

1

2

mm M m M

G F m g G m g G

R h R h


+ +

( ) ( )

211 24

667 10 6 102

731 1 2 2 2

6 664 10 1 10

N mkg

G M mkgg

sR h m m


sdotrArr = = =

+ sdot + sdot

2inačica


2 2

2


R

m

R R



( ) ( ) ( )

1 1 12 2

2


R h R h R h


+ + +


( )

( ) ( ) ( )

1

2 2 221 1 1

1

podijelimo

jednadž2

e2 2 2

b1

G MG Mg

g g gR h R h R hR

G M G Mg g g

G M

G Mg

R R RR h

sdot sdotsdot=


sdot sdot sdot=

+

( )

2 221 1 1

2

g g gR R R

g g R h g R hR

g

h


sdot++

22 664 10

981 734 1 2 6 6 2

64 10 1 10

R m m mg g

R h s m m s


+ sdot + sdot


Vježba 119




Rezultat 734 2

m

s

23





Rješenje 120





1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot





Newtonovom poučku

G m g= sdot


na Zemlji jednaka


2v

F mcpr

= sdot



m

m VV

ρ ρ= rArr = sdot


3

4

3V r π= sdot sdot



2 2

2

2

2


R R G mR R


sdotrArr

4 3

3

4 3

2 24 3

3

3

v R v RM R

G G

M V V R

M R

ρ π

ρ

ρ

π

π

= sdot = sdot sdot

= sdot

sdot sdot


sdot

sdot

2 24 33

23 4

3

34

v R vR

G GR R

ρ π ρππ

sdotsdot



24

( )

2

3 2920

5130 2 32

11 64 667 10 2439 10

2

m

kgs

N m mm

kgπ

sdot

= =sdotminus



2 2 21

v v v



2

2920

350 6 2

2439 10

m

ms

m s

= =sdot

Vježba 120




Rezultat 5130 350 3 2

kg m

m s

12

( )

2 2 2 2 22 4 4 4

12 2

M d M d M d M m dx

M m


sdot minus

( ) ( )

2 2 2 2 22 4 2 4

12 12

4

2 2


x xM m

d M d

M m



( )

( )( )

( )( )

22 2

12 12 122 2

2

2


M m M m M m



81 81

1 1 81

1281 81

2 2 81

M M m m m mx d x d

M m m mM M mx d


M m m m




minus sdot minus

281 81 81 9 90

1 1 180 80 80

81 9 72281 81

2 280 802 80


m m m

m m mm m x d x d

x d m mm





9 9 8 8384 10 432 101 1 18 8 1

89 9 8 3456 10384 10 22

90

80

2 210 10

72

80

x d x d xm

mm x m

x mx d x d x mm

m


= sdot= sdot = sdot

sdot

sdot

sdotsdot sdot

sdot=


8

3456 102

cx

x mrArr

= sdot

minusminus








Rješenje 110

G = 667 middot 10-11

(N middot m2) kg

2 g = 981 ms

2 r = 64 middot 10

6 m M = 6 middot 10

24 kg

T =




1 22

m mF G

r

sdot= sdot




13

24

2

rF mcp

T

πsdot sdot= sdot





G m g= sdot




gravitacijskoj sili

1inačica




2 24 4

2 2 2 2

2 2

M m r M T rm r

F FG M

G m G mg cpr T r mT


sdotsdot


sdotsdot rArr

2 3 2 3 2 34 4 42 2

2r r r r


π π ππ


sdot sdot sdot sdot

664 106 3

2 64 10 5085 102

11 24667 10 6 10

2

mm s

N mkg

kg

πsdot



35085 10 3600 141 h= sdot =

2inačica



2 2 2

4 4

2 2

r rG F m g m m

Tg mcp

T T m g


sdotsdot rArr

2 2 24 4 42 2

2r r r r

T T T Tg g g g

π π ππ


664 10 3 3

5082 55085 10 141

9

10 3600

812

ms h

m

s

πsdot





21 t)

Rezultat 141 h

14




Rješenje 111


G = 667 middot 10-11

(N middot m2) kg

2 ρ =




1 22

m mF G

r

sdot= sdot




m

Vρ =


3

4

3V r π= sdot sdot

RRRR

rrrr rrrr


2

1 2 2 2 22

1 2 2 22

m m m

m m R FF G F G mm m

GF G R R

R

R

G

= =sdot


sdot

2 22

2 R F R F F F

m m m R m rG G G G



6metoda

3supstitucij

2 6

234 3 4

34 3

3

e 4

m

V F F Fr r

F G G Gm r

G r rr

V r

r

ρ

ρ ρ ρπ ππ

π

=



=

sdot

sdot sdot

15

( )

3 43 3 10

233851 2 2 2 2 3

2 2 2 05 11667 10

2

F

F N kgG

Gr r N m mm

kg

ρ ρπ π π

sdot minus



Vježba 111







T

4 88 4


Rješenje 112

r1 = r T1 = T 1

2 8

T T= sdot r2 =

Keplerovi zakoni




Sunca 2 3

1 12 3

2

2

T r

T r

=

2 3 2 32 3 2 3 2 3 2 3 31 1 2 1

1 2 2 1 1 2 2 1 2

1

21

2 3 22 2 1


r TT T


21 3 1 12 3 3

183 3 3 3 364 642 2 2 22 2 2 6

2

4

T r T r r

r r r r r

T T

T

T



1 1 13 3 3 33 2 2 26 4 4

4 6


Odgovor je pod B

Vježba 112



T

3 99 81


Rezultat A

16




Rješenje 113


h =




1 22

m mF G

r

sdot= sdot




( )2

m M

F G

R h

sdot= sdot

+




Newtonovom poučku

G m g= sdot


na Zemlji jednaka



( ) ( )

( )2

2

2


R h R h

R h

m g


+ +

+sdot

sdot

( ) ( ) 2 2G M G M G M G M



211 24

667 10 6 102

664 10 98901511 989

7332

N mkg

kgh m m km

m

s



Vježba 113


6 middot 1021


-11 (N middot m

2) kg

2)

Rezultat 989 km





(N middot m2) kg

2)

Rješenje 114

v = 652 km s = 6520 m s M = 596 middot 1024

kg G = 667 middot 10-11

(N middot m2) kg

2

T =


17



1 22

m mF G

r

sdot= sdot


tijela


2v

F mcpr

= sdot





2

rv

T

πsdot sdot=




2 2

2 2

2

2 2


r

mr vvrr r

sdotsdot


2

G Mr

v

sdotrArr =


metod

2 222

2

2 2

a

zamj2

ene

G Mr rrv Tv

T vT vTG M

T

v

G M G M vr r rv v v

π πππ



sdot

[ ]

211 24

2 667 10 596 102

2901175 1502 min

3 3

6520

901175 60

N mkg

G M kgT s

v m

s

ππ



Vježba 114



Rezultat 12016 min





Rješenje 115


1 2



18



1 22

m mF G

r

sdot= sdot




( )

( )( ) ( )


težina tijela mase

2 2

1

1 2m na visini h

22

m M m MG G G G

R R

m M m MG G G G

R h R h


rArr rArrsdot sdot

= sdot sdot = sdot

+ +

( ) ( ) ( )

podijelimo

jednadžbe

1

2 2 22

1 1 1

2 2 22 2

m MG

G R R Rm

m MG

G

m MMG G

R h R h R

G

h

G

sdot sdotsdot sdot


sdot sdot sdot sdot

+ + +

( ) ( ) ( )( )

22 2 2

2 22 2 2 2

2

2 2

R h R h R hR h R

R R

R

R


( )2 2 2



Odgovor je pod C

Vježba 115




Rezultat A




N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rješenje 116


v =




1 22

m mF G

r

sdot= sdot



19


2v

F mcpr

= sdot




v 2 r

T

πsdot sdot=




2 22

2 2

v m M v m M M


r

m rr r



pomnožimo

jednadžbe

2

2 22 3

2

rv

r M MTv v G v G

M T r Tv

r

rG

r

π

π π

sdot sdot=


= sdot

3

2 2 23 3 3G M G M G Mv v v

T T T


211 24

2 667 10 596 1023

652284 652 9 000

N mkg

m kmkg

s s s

πsdotminus


= = asymp

Odgovor je pod B

Vježba 116



N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rezultat B





Rješenje 117

F1 = Z r1 = R 50 51

2 50 1 50 50 50

R R R R Rr R R

sdot += + = + = = sdot F2 =


20



1 22

m mF G

r

sdot= sdot




1

1

1 2 2

1

F Z

r R

m M m MF G Z G

r R


=

=


51

22 2 22 2 2601 250 512 250050


r R

r

R

Rsdot sdot sdot


= sdot

2500

2 22601

m MF G

R


sdot


2podijelimo

jednadžbe

2500 2500

222601 26012 2

25

2

002 22

2601

m M m MZ G G

F FRR

m

m MG

R

M m MZ Z m MGF G

RG

RR



sdot

2500 2500 25002 2 096 2 22601 2601 2 1

60

F FZ F Z F Z


Odgovor je pod D

Vježba 117




Rezultat C




Rješenje 118


6 m M = 6 middot 10

24 kg v =




1 22

m mF G

r

sdot= sdot


tijela

21

211

6 7 6 102

N mG

kg

sdotminus= sdot



centripetalna sila

2v

F mcpr

= sdot




( ) ( )

2

2

2 2


R h R

R h

mhR h R h


+ ++

+sdot

+

2 2G M G M G M

v v vR h R h R h


+ + +

211 24

667 10 6 102

77693 6 5

64 10 23 10

N mkg

mkg

sm m


= =sdot + sdot

Vježba 118



kg)

Rezultat 7000 m

sasymp





Rješenje 119

M = 6 middot 1024


6 m

g = 981 m s2 g1 =




1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot


22




Newtonovom poučku

G m g= sdot


na Zemlji jednaka

1inačica


( ) ( )1 2

1

2

mm M m M

G F m g G m g G

R h R h


+ +

( ) ( )

211 24

667 10 6 102

731 1 2 2 2

6 664 10 1 10

N mkg

G M mkgg

sR h m m


sdotrArr = = =

+ sdot + sdot

2inačica


2 2

2


R

m

R R



( ) ( ) ( )

1 1 12 2

2


R h R h R h


+ + +


( )

( ) ( ) ( )

1

2 2 221 1 1

1

podijelimo

jednadž2

e2 2 2

b1

G MG Mg

g g gR h R h R hR

G M G Mg g g

G M

G Mg

R R RR h

sdot sdotsdot=


sdot sdot sdot=

+

( )

2 221 1 1

2

g g gR R R

g g R h g R hR

g

h


sdot++

22 664 10

981 734 1 2 6 6 2

64 10 1 10

R m m mg g

R h s m m s


+ sdot + sdot


Vježba 119




Rezultat 734 2

m

s

23





Rješenje 120





1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot





Newtonovom poučku

G m g= sdot


na Zemlji jednaka


2v

F mcpr

= sdot



m

m VV

ρ ρ= rArr = sdot


3

4

3V r π= sdot sdot



2 2

2

2

2


R R G mR R


sdotrArr

4 3

3

4 3

2 24 3

3

3

v R v RM R

G G

M V V R

M R

ρ π

ρ

ρ

π

π

= sdot = sdot sdot

= sdot

sdot sdot


sdot

sdot

2 24 33

23 4

3

34

v R vR

G GR R

ρ π ρππ

sdotsdot



24

( )

2

3 2920

5130 2 32

11 64 667 10 2439 10

2

m

kgs

N m mm

kgπ

sdot

= =sdotminus



2 2 21

v v v



2

2920

350 6 2

2439 10

m

ms

m s

= =sdot

Vježba 120




Rezultat 5130 350 3 2

kg m

m s

13

24

2

rF mcp

T

πsdot sdot= sdot





G m g= sdot




gravitacijskoj sili

1inačica




2 24 4

2 2 2 2

2 2

M m r M T rm r

F FG M

G m G mg cpr T r mT


sdotsdot


sdotsdot rArr

2 3 2 3 2 34 4 42 2

2r r r r


π π ππ


sdot sdot sdot sdot

664 106 3

2 64 10 5085 102

11 24667 10 6 10

2

mm s

N mkg

kg

πsdot



35085 10 3600 141 h= sdot =

2inačica



2 2 2

4 4

2 2

r rG F m g m m

Tg mcp

T T m g


sdotsdot rArr

2 2 24 4 42 2

2r r r r

T T T Tg g g g

π π ππ


664 10 3 3

5082 55085 10 141

9

10 3600

812

ms h

m

s

πsdot





21 t)

Rezultat 141 h

14




Rješenje 111


G = 667 middot 10-11

(N middot m2) kg

2 ρ =




1 22

m mF G

r

sdot= sdot




m

Vρ =


3

4

3V r π= sdot sdot

RRRR

rrrr rrrr


2

1 2 2 2 22

1 2 2 22

m m m

m m R FF G F G mm m

GF G R R

R

R

G

= =sdot


sdot

2 22

2 R F R F F F

m m m R m rG G G G



6metoda

3supstitucij

2 6

234 3 4

34 3

3

e 4

m

V F F Fr r

F G G Gm r

G r rr

V r

r

ρ

ρ ρ ρπ ππ

π

=



=

sdot

sdot sdot

15

( )

3 43 3 10

233851 2 2 2 2 3

2 2 2 05 11667 10

2

F

F N kgG

Gr r N m mm

kg

ρ ρπ π π

sdot minus



Vježba 111







T

4 88 4


Rješenje 112

r1 = r T1 = T 1

2 8

T T= sdot r2 =

Keplerovi zakoni




Sunca 2 3

1 12 3

2

2

T r

T r

=

2 3 2 32 3 2 3 2 3 2 3 31 1 2 1

1 2 2 1 1 2 2 1 2

1

21

2 3 22 2 1


r TT T


21 3 1 12 3 3

183 3 3 3 364 642 2 2 22 2 2 6

2

4

T r T r r

r r r r r

T T

T

T



1 1 13 3 3 33 2 2 26 4 4

4 6


Odgovor je pod B

Vježba 112



T

3 99 81


Rezultat A

16




Rješenje 113


h =




1 22

m mF G

r

sdot= sdot




( )2

m M

F G

R h

sdot= sdot

+




Newtonovom poučku

G m g= sdot


na Zemlji jednaka



( ) ( )

( )2

2

2


R h R h

R h

m g


+ +

+sdot

sdot

( ) ( ) 2 2G M G M G M G M



211 24

667 10 6 102

664 10 98901511 989

7332

N mkg

kgh m m km

m

s



Vježba 113


6 middot 1021


-11 (N middot m

2) kg

2)

Rezultat 989 km





(N middot m2) kg

2)

Rješenje 114

v = 652 km s = 6520 m s M = 596 middot 1024

kg G = 667 middot 10-11

(N middot m2) kg

2

T =


17



1 22

m mF G

r

sdot= sdot


tijela


2v

F mcpr

= sdot





2

rv

T

πsdot sdot=




2 2

2 2

2

2 2


r

mr vvrr r

sdotsdot


2

G Mr

v

sdotrArr =


metod

2 222

2

2 2

a

zamj2

ene

G Mr rrv Tv

T vT vTG M

T

v

G M G M vr r rv v v

π πππ



sdot

[ ]

211 24

2 667 10 596 102

2901175 1502 min

3 3

6520

901175 60

N mkg

G M kgT s

v m

s

ππ



Vježba 114



Rezultat 12016 min





Rješenje 115


1 2



18



1 22

m mF G

r

sdot= sdot




( )

( )( ) ( )


težina tijela mase

2 2

1

1 2m na visini h

22

m M m MG G G G

R R

m M m MG G G G

R h R h


rArr rArrsdot sdot

= sdot sdot = sdot

+ +

( ) ( ) ( )

podijelimo

jednadžbe

1

2 2 22

1 1 1

2 2 22 2

m MG

G R R Rm

m MG

G

m MMG G

R h R h R

G

h

G

sdot sdotsdot sdot


sdot sdot sdot sdot

+ + +

( ) ( ) ( )( )

22 2 2

2 22 2 2 2

2

2 2

R h R h R hR h R

R R

R

R


( )2 2 2



Odgovor je pod C

Vježba 115




Rezultat A




N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rješenje 116


v =




1 22

m mF G

r

sdot= sdot



19


2v

F mcpr

= sdot




v 2 r

T

πsdot sdot=




2 22

2 2

v m M v m M M


r

m rr r



pomnožimo

jednadžbe

2

2 22 3

2

rv

r M MTv v G v G

M T r Tv

r

rG

r

π

π π

sdot sdot=


= sdot

3

2 2 23 3 3G M G M G Mv v v

T T T


211 24

2 667 10 596 1023

652284 652 9 000

N mkg

m kmkg

s s s

πsdotminus


= = asymp

Odgovor je pod B

Vježba 116



N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rezultat B





Rješenje 117

F1 = Z r1 = R 50 51

2 50 1 50 50 50

R R R R Rr R R

sdot += + = + = = sdot F2 =


20



1 22

m mF G

r

sdot= sdot




1

1

1 2 2

1

F Z

r R

m M m MF G Z G

r R


=

=


51

22 2 22 2 2601 250 512 250050


r R

r

R

Rsdot sdot sdot


= sdot

2500

2 22601

m MF G

R


sdot


2podijelimo

jednadžbe

2500 2500

222601 26012 2

25

2

002 22

2601

m M m MZ G G

F FRR

m

m MG

R

M m MZ Z m MGF G

RG

RR



sdot

2500 2500 25002 2 096 2 22601 2601 2 1

60

F FZ F Z F Z


Odgovor je pod D

Vježba 117




Rezultat C




Rješenje 118


6 m M = 6 middot 10

24 kg v =




1 22

m mF G

r

sdot= sdot


tijela

21

211

6 7 6 102

N mG

kg

sdotminus= sdot



centripetalna sila

2v

F mcpr

= sdot




( ) ( )

2

2

2 2


R h R

R h

mhR h R h


+ ++

+sdot

+

2 2G M G M G M

v v vR h R h R h


+ + +

211 24

667 10 6 102

77693 6 5

64 10 23 10

N mkg

mkg

sm m


= =sdot + sdot

Vježba 118



kg)

Rezultat 7000 m

sasymp





Rješenje 119

M = 6 middot 1024


6 m

g = 981 m s2 g1 =




1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot


22




Newtonovom poučku

G m g= sdot


na Zemlji jednaka

1inačica


( ) ( )1 2

1

2

mm M m M

G F m g G m g G

R h R h


+ +

( ) ( )

211 24

667 10 6 102

731 1 2 2 2

6 664 10 1 10

N mkg

G M mkgg

sR h m m


sdotrArr = = =

+ sdot + sdot

2inačica


2 2

2


R

m

R R



( ) ( ) ( )

1 1 12 2

2


R h R h R h


+ + +


( )

( ) ( ) ( )

1

2 2 221 1 1

1

podijelimo

jednadž2

e2 2 2

b1

G MG Mg

g g gR h R h R hR

G M G Mg g g

G M

G Mg

R R RR h

sdot sdotsdot=


sdot sdot sdot=

+

( )

2 221 1 1

2

g g gR R R

g g R h g R hR

g

h


sdot++

22 664 10

981 734 1 2 6 6 2

64 10 1 10

R m m mg g

R h s m m s


+ sdot + sdot


Vježba 119




Rezultat 734 2

m

s

23





Rješenje 120





1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot





Newtonovom poučku

G m g= sdot


na Zemlji jednaka


2v

F mcpr

= sdot



m

m VV

ρ ρ= rArr = sdot


3

4

3V r π= sdot sdot



2 2

2

2

2


R R G mR R


sdotrArr

4 3

3

4 3

2 24 3

3

3

v R v RM R

G G

M V V R

M R

ρ π

ρ

ρ

π

π

= sdot = sdot sdot

= sdot

sdot sdot


sdot

sdot

2 24 33

23 4

3

34

v R vR

G GR R

ρ π ρππ

sdotsdot



24

( )

2

3 2920

5130 2 32

11 64 667 10 2439 10

2

m

kgs

N m mm

kgπ

sdot

= =sdotminus



2 2 21

v v v



2

2920

350 6 2

2439 10

m

ms

m s

= =sdot

Vježba 120




Rezultat 5130 350 3 2

kg m

m s

14




Rješenje 111


G = 667 middot 10-11

(N middot m2) kg

2 ρ =




1 22

m mF G

r

sdot= sdot




m

Vρ =


3

4

3V r π= sdot sdot

RRRR

rrrr rrrr


2

1 2 2 2 22

1 2 2 22

m m m

m m R FF G F G mm m

GF G R R

R

R

G

= =sdot


sdot

2 22

2 R F R F F F

m m m R m rG G G G



6metoda

3supstitucij

2 6

234 3 4

34 3

3

e 4

m

V F F Fr r

F G G Gm r

G r rr

V r

r

ρ

ρ ρ ρπ ππ

π

=



=

sdot

sdot sdot

15

( )

3 43 3 10

233851 2 2 2 2 3

2 2 2 05 11667 10

2

F

F N kgG

Gr r N m mm

kg

ρ ρπ π π

sdot minus



Vježba 111







T

4 88 4


Rješenje 112

r1 = r T1 = T 1

2 8

T T= sdot r2 =

Keplerovi zakoni




Sunca 2 3

1 12 3

2

2

T r

T r

=

2 3 2 32 3 2 3 2 3 2 3 31 1 2 1

1 2 2 1 1 2 2 1 2

1

21

2 3 22 2 1


r TT T


21 3 1 12 3 3

183 3 3 3 364 642 2 2 22 2 2 6

2

4

T r T r r

r r r r r

T T

T

T



1 1 13 3 3 33 2 2 26 4 4

4 6


Odgovor je pod B

Vježba 112



T

3 99 81


Rezultat A

16




Rješenje 113


h =




1 22

m mF G

r

sdot= sdot




( )2

m M

F G

R h

sdot= sdot

+




Newtonovom poučku

G m g= sdot


na Zemlji jednaka



( ) ( )

( )2

2

2


R h R h

R h

m g


+ +

+sdot

sdot

( ) ( ) 2 2G M G M G M G M



211 24

667 10 6 102

664 10 98901511 989

7332

N mkg

kgh m m km

m

s



Vježba 113


6 middot 1021


-11 (N middot m

2) kg

2)

Rezultat 989 km





(N middot m2) kg

2)

Rješenje 114

v = 652 km s = 6520 m s M = 596 middot 1024

kg G = 667 middot 10-11

(N middot m2) kg

2

T =


17



1 22

m mF G

r

sdot= sdot


tijela


2v

F mcpr

= sdot





2

rv

T

πsdot sdot=




2 2

2 2

2

2 2


r

mr vvrr r

sdotsdot


2

G Mr

v

sdotrArr =


metod

2 222

2

2 2

a

zamj2

ene

G Mr rrv Tv

T vT vTG M

T

v

G M G M vr r rv v v

π πππ



sdot

[ ]

211 24

2 667 10 596 102

2901175 1502 min

3 3

6520

901175 60

N mkg

G M kgT s

v m

s

ππ



Vježba 114



Rezultat 12016 min





Rješenje 115


1 2



18



1 22

m mF G

r

sdot= sdot




( )

( )( ) ( )


težina tijela mase

2 2

1

1 2m na visini h

22

m M m MG G G G

R R

m M m MG G G G

R h R h


rArr rArrsdot sdot

= sdot sdot = sdot

+ +

( ) ( ) ( )

podijelimo

jednadžbe

1

2 2 22

1 1 1

2 2 22 2

m MG

G R R Rm

m MG

G

m MMG G

R h R h R

G

h

G

sdot sdotsdot sdot


sdot sdot sdot sdot

+ + +

( ) ( ) ( )( )

22 2 2

2 22 2 2 2

2

2 2

R h R h R hR h R

R R

R

R


( )2 2 2



Odgovor je pod C

Vježba 115




Rezultat A




N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rješenje 116


v =




1 22

m mF G

r

sdot= sdot



19


2v

F mcpr

= sdot




v 2 r

T

πsdot sdot=




2 22

2 2

v m M v m M M


r

m rr r



pomnožimo

jednadžbe

2

2 22 3

2

rv

r M MTv v G v G

M T r Tv

r

rG

r

π

π π

sdot sdot=


= sdot

3

2 2 23 3 3G M G M G Mv v v

T T T


211 24

2 667 10 596 1023

652284 652 9 000

N mkg

m kmkg

s s s

πsdotminus


= = asymp

Odgovor je pod B

Vježba 116



N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rezultat B





Rješenje 117

F1 = Z r1 = R 50 51

2 50 1 50 50 50

R R R R Rr R R

sdot += + = + = = sdot F2 =


20



1 22

m mF G

r

sdot= sdot




1

1

1 2 2

1

F Z

r R

m M m MF G Z G

r R


=

=


51

22 2 22 2 2601 250 512 250050


r R

r

R

Rsdot sdot sdot


= sdot

2500

2 22601

m MF G

R


sdot


2podijelimo

jednadžbe

2500 2500

222601 26012 2

25

2

002 22

2601

m M m MZ G G

F FRR

m

m MG

R

M m MZ Z m MGF G

RG

RR



sdot

2500 2500 25002 2 096 2 22601 2601 2 1

60

F FZ F Z F Z


Odgovor je pod D

Vježba 117




Rezultat C




Rješenje 118


6 m M = 6 middot 10

24 kg v =




1 22

m mF G

r

sdot= sdot


tijela

21

211

6 7 6 102

N mG

kg

sdotminus= sdot



centripetalna sila

2v

F mcpr

= sdot




( ) ( )

2

2

2 2


R h R

R h

mhR h R h


+ ++

+sdot

+

2 2G M G M G M

v v vR h R h R h


+ + +

211 24

667 10 6 102

77693 6 5

64 10 23 10

N mkg

mkg

sm m


= =sdot + sdot

Vježba 118



kg)

Rezultat 7000 m

sasymp





Rješenje 119

M = 6 middot 1024


6 m

g = 981 m s2 g1 =




1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot


22




Newtonovom poučku

G m g= sdot


na Zemlji jednaka

1inačica


( ) ( )1 2

1

2

mm M m M

G F m g G m g G

R h R h


+ +

( ) ( )

211 24

667 10 6 102

731 1 2 2 2

6 664 10 1 10

N mkg

G M mkgg

sR h m m


sdotrArr = = =

+ sdot + sdot

2inačica


2 2

2


R

m

R R



( ) ( ) ( )

1 1 12 2

2


R h R h R h


+ + +


( )

( ) ( ) ( )

1

2 2 221 1 1

1

podijelimo

jednadž2

e2 2 2

b1

G MG Mg

g g gR h R h R hR

G M G Mg g g

G M

G Mg

R R RR h

sdot sdotsdot=


sdot sdot sdot=

+

( )

2 221 1 1

2

g g gR R R

g g R h g R hR

g

h


sdot++

22 664 10

981 734 1 2 6 6 2

64 10 1 10

R m m mg g

R h s m m s


+ sdot + sdot


Vježba 119




Rezultat 734 2

m

s

23





Rješenje 120





1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot





Newtonovom poučku

G m g= sdot


na Zemlji jednaka


2v

F mcpr

= sdot



m

m VV

ρ ρ= rArr = sdot


3

4

3V r π= sdot sdot



2 2

2

2

2


R R G mR R


sdotrArr

4 3

3

4 3

2 24 3

3

3

v R v RM R

G G

M V V R

M R

ρ π

ρ

ρ

π

π

= sdot = sdot sdot

= sdot

sdot sdot


sdot

sdot

2 24 33

23 4

3

34

v R vR

G GR R

ρ π ρππ

sdotsdot



24

( )

2

3 2920

5130 2 32

11 64 667 10 2439 10

2

m

kgs

N m mm

kgπ

sdot

= =sdotminus



2 2 21

v v v



2

2920

350 6 2

2439 10

m

ms

m s

= =sdot

Vježba 120




Rezultat 5130 350 3 2

kg m

m s

15

( )

3 43 3 10

233851 2 2 2 2 3

2 2 2 05 11667 10

2

F

F N kgG

Gr r N m mm

kg

ρ ρπ π π

sdot minus



Vježba 111







T

4 88 4


Rješenje 112

r1 = r T1 = T 1

2 8

T T= sdot r2 =

Keplerovi zakoni




Sunca 2 3

1 12 3

2

2

T r

T r

=

2 3 2 32 3 2 3 2 3 2 3 31 1 2 1

1 2 2 1 1 2 2 1 2

1

21

2 3 22 2 1


r TT T


21 3 1 12 3 3

183 3 3 3 364 642 2 2 22 2 2 6

2

4

T r T r r

r r r r r

T T

T

T



1 1 13 3 3 33 2 2 26 4 4

4 6


Odgovor je pod B

Vježba 112



T

3 99 81


Rezultat A

16




Rješenje 113


h =




1 22

m mF G

r

sdot= sdot




( )2

m M

F G

R h

sdot= sdot

+




Newtonovom poučku

G m g= sdot


na Zemlji jednaka



( ) ( )

( )2

2

2


R h R h

R h

m g


+ +

+sdot

sdot

( ) ( ) 2 2G M G M G M G M



211 24

667 10 6 102

664 10 98901511 989

7332

N mkg

kgh m m km

m

s



Vježba 113


6 middot 1021


-11 (N middot m

2) kg

2)

Rezultat 989 km





(N middot m2) kg

2)

Rješenje 114

v = 652 km s = 6520 m s M = 596 middot 1024

kg G = 667 middot 10-11

(N middot m2) kg

2

T =


17



1 22

m mF G

r

sdot= sdot


tijela


2v

F mcpr

= sdot





2

rv

T

πsdot sdot=




2 2

2 2

2

2 2


r

mr vvrr r

sdotsdot


2

G Mr

v

sdotrArr =


metod

2 222

2

2 2

a

zamj2

ene

G Mr rrv Tv

T vT vTG M

T

v

G M G M vr r rv v v

π πππ



sdot

[ ]

211 24

2 667 10 596 102

2901175 1502 min

3 3

6520

901175 60

N mkg

G M kgT s

v m

s

ππ



Vježba 114



Rezultat 12016 min





Rješenje 115


1 2



18



1 22

m mF G

r

sdot= sdot




( )

( )( ) ( )


težina tijela mase

2 2

1

1 2m na visini h

22

m M m MG G G G

R R

m M m MG G G G

R h R h


rArr rArrsdot sdot

= sdot sdot = sdot

+ +

( ) ( ) ( )

podijelimo

jednadžbe

1

2 2 22

1 1 1

2 2 22 2

m MG

G R R Rm

m MG

G

m MMG G

R h R h R

G

h

G

sdot sdotsdot sdot


sdot sdot sdot sdot

+ + +

( ) ( ) ( )( )

22 2 2

2 22 2 2 2

2

2 2

R h R h R hR h R

R R

R

R


( )2 2 2



Odgovor je pod C

Vježba 115




Rezultat A




N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rješenje 116


v =




1 22

m mF G

r

sdot= sdot



19


2v

F mcpr

= sdot




v 2 r

T

πsdot sdot=




2 22

2 2

v m M v m M M


r

m rr r



pomnožimo

jednadžbe

2

2 22 3

2

rv

r M MTv v G v G

M T r Tv

r

rG

r

π

π π

sdot sdot=


= sdot

3

2 2 23 3 3G M G M G Mv v v

T T T


211 24

2 667 10 596 1023

652284 652 9 000

N mkg

m kmkg

s s s

πsdotminus


= = asymp

Odgovor je pod B

Vježba 116



N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rezultat B





Rješenje 117

F1 = Z r1 = R 50 51

2 50 1 50 50 50

R R R R Rr R R

sdot += + = + = = sdot F2 =


20



1 22

m mF G

r

sdot= sdot




1

1

1 2 2

1

F Z

r R

m M m MF G Z G

r R


=

=


51

22 2 22 2 2601 250 512 250050


r R

r

R

Rsdot sdot sdot


= sdot

2500

2 22601

m MF G

R


sdot


2podijelimo

jednadžbe

2500 2500

222601 26012 2

25

2

002 22

2601

m M m MZ G G

F FRR

m

m MG

R

M m MZ Z m MGF G

RG

RR



sdot

2500 2500 25002 2 096 2 22601 2601 2 1

60

F FZ F Z F Z


Odgovor je pod D

Vježba 117




Rezultat C




Rješenje 118


6 m M = 6 middot 10

24 kg v =




1 22

m mF G

r

sdot= sdot


tijela

21

211

6 7 6 102

N mG

kg

sdotminus= sdot



centripetalna sila

2v

F mcpr

= sdot




( ) ( )

2

2

2 2


R h R

R h

mhR h R h


+ ++

+sdot

+

2 2G M G M G M

v v vR h R h R h


+ + +

211 24

667 10 6 102

77693 6 5

64 10 23 10

N mkg

mkg

sm m


= =sdot + sdot

Vježba 118



kg)

Rezultat 7000 m

sasymp





Rješenje 119

M = 6 middot 1024


6 m

g = 981 m s2 g1 =




1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot


22




Newtonovom poučku

G m g= sdot


na Zemlji jednaka

1inačica


( ) ( )1 2

1

2

mm M m M

G F m g G m g G

R h R h


+ +

( ) ( )

211 24

667 10 6 102

731 1 2 2 2

6 664 10 1 10

N mkg

G M mkgg

sR h m m


sdotrArr = = =

+ sdot + sdot

2inačica


2 2

2


R

m

R R



( ) ( ) ( )

1 1 12 2

2


R h R h R h


+ + +


( )

( ) ( ) ( )

1

2 2 221 1 1

1

podijelimo

jednadž2

e2 2 2

b1

G MG Mg

g g gR h R h R hR

G M G Mg g g

G M

G Mg

R R RR h

sdot sdotsdot=


sdot sdot sdot=

+

( )

2 221 1 1

2

g g gR R R

g g R h g R hR

g

h


sdot++

22 664 10

981 734 1 2 6 6 2

64 10 1 10

R m m mg g

R h s m m s


+ sdot + sdot


Vježba 119




Rezultat 734 2

m

s

23





Rješenje 120





1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot





Newtonovom poučku

G m g= sdot


na Zemlji jednaka


2v

F mcpr

= sdot



m

m VV

ρ ρ= rArr = sdot


3

4

3V r π= sdot sdot



2 2

2

2

2


R R G mR R


sdotrArr

4 3

3

4 3

2 24 3

3

3

v R v RM R

G G

M V V R

M R

ρ π

ρ

ρ

π

π

= sdot = sdot sdot

= sdot

sdot sdot


sdot

sdot

2 24 33

23 4

3

34

v R vR

G GR R

ρ π ρππ

sdotsdot



24

( )

2

3 2920

5130 2 32

11 64 667 10 2439 10

2

m

kgs

N m mm

kgπ

sdot

= =sdotminus



2 2 21

v v v



2

2920

350 6 2

2439 10

m

ms

m s

= =sdot

Vježba 120




Rezultat 5130 350 3 2

kg m

m s

16




Rješenje 113


h =




1 22

m mF G

r

sdot= sdot




( )2

m M

F G

R h

sdot= sdot

+




Newtonovom poučku

G m g= sdot


na Zemlji jednaka



( ) ( )

( )2

2

2


R h R h

R h

m g


+ +

+sdot

sdot

( ) ( ) 2 2G M G M G M G M



211 24

667 10 6 102

664 10 98901511 989

7332

N mkg

kgh m m km

m

s



Vježba 113


6 middot 1021


-11 (N middot m

2) kg

2)

Rezultat 989 km





(N middot m2) kg

2)

Rješenje 114

v = 652 km s = 6520 m s M = 596 middot 1024

kg G = 667 middot 10-11

(N middot m2) kg

2

T =


17



1 22

m mF G

r

sdot= sdot


tijela


2v

F mcpr

= sdot





2

rv

T

πsdot sdot=




2 2

2 2

2

2 2


r

mr vvrr r

sdotsdot


2

G Mr

v

sdotrArr =


metod

2 222

2

2 2

a

zamj2

ene

G Mr rrv Tv

T vT vTG M

T

v

G M G M vr r rv v v

π πππ



sdot

[ ]

211 24

2 667 10 596 102

2901175 1502 min

3 3

6520

901175 60

N mkg

G M kgT s

v m

s

ππ



Vježba 114



Rezultat 12016 min





Rješenje 115


1 2



18



1 22

m mF G

r

sdot= sdot




( )

( )( ) ( )


težina tijela mase

2 2

1

1 2m na visini h

22

m M m MG G G G

R R

m M m MG G G G

R h R h


rArr rArrsdot sdot

= sdot sdot = sdot

+ +

( ) ( ) ( )

podijelimo

jednadžbe

1

2 2 22

1 1 1

2 2 22 2

m MG

G R R Rm

m MG

G

m MMG G

R h R h R

G

h

G

sdot sdotsdot sdot


sdot sdot sdot sdot

+ + +

( ) ( ) ( )( )

22 2 2

2 22 2 2 2

2

2 2

R h R h R hR h R

R R

R

R


( )2 2 2



Odgovor je pod C

Vježba 115




Rezultat A




N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rješenje 116


v =




1 22

m mF G

r

sdot= sdot



19


2v

F mcpr

= sdot




v 2 r

T

πsdot sdot=




2 22

2 2

v m M v m M M


r

m rr r



pomnožimo

jednadžbe

2

2 22 3

2

rv

r M MTv v G v G

M T r Tv

r

rG

r

π

π π

sdot sdot=


= sdot

3

2 2 23 3 3G M G M G Mv v v

T T T


211 24

2 667 10 596 1023

652284 652 9 000

N mkg

m kmkg

s s s

πsdotminus


= = asymp

Odgovor je pod B

Vježba 116



N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rezultat B





Rješenje 117

F1 = Z r1 = R 50 51

2 50 1 50 50 50

R R R R Rr R R

sdot += + = + = = sdot F2 =


20



1 22

m mF G

r

sdot= sdot




1

1

1 2 2

1

F Z

r R

m M m MF G Z G

r R


=

=


51

22 2 22 2 2601 250 512 250050


r R

r

R

Rsdot sdot sdot


= sdot

2500

2 22601

m MF G

R


sdot


2podijelimo

jednadžbe

2500 2500

222601 26012 2

25

2

002 22

2601

m M m MZ G G

F FRR

m

m MG

R

M m MZ Z m MGF G

RG

RR



sdot

2500 2500 25002 2 096 2 22601 2601 2 1

60

F FZ F Z F Z


Odgovor je pod D

Vježba 117




Rezultat C




Rješenje 118


6 m M = 6 middot 10

24 kg v =




1 22

m mF G

r

sdot= sdot


tijela

21

211

6 7 6 102

N mG

kg

sdotminus= sdot



centripetalna sila

2v

F mcpr

= sdot




( ) ( )

2

2

2 2


R h R

R h

mhR h R h


+ ++

+sdot

+

2 2G M G M G M

v v vR h R h R h


+ + +

211 24

667 10 6 102

77693 6 5

64 10 23 10

N mkg

mkg

sm m


= =sdot + sdot

Vježba 118



kg)

Rezultat 7000 m

sasymp





Rješenje 119

M = 6 middot 1024


6 m

g = 981 m s2 g1 =




1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot


22




Newtonovom poučku

G m g= sdot


na Zemlji jednaka

1inačica


( ) ( )1 2

1

2

mm M m M

G F m g G m g G

R h R h


+ +

( ) ( )

211 24

667 10 6 102

731 1 2 2 2

6 664 10 1 10

N mkg

G M mkgg

sR h m m


sdotrArr = = =

+ sdot + sdot

2inačica


2 2

2


R

m

R R



( ) ( ) ( )

1 1 12 2

2


R h R h R h


+ + +


( )

( ) ( ) ( )

1

2 2 221 1 1

1

podijelimo

jednadž2

e2 2 2

b1

G MG Mg

g g gR h R h R hR

G M G Mg g g

G M

G Mg

R R RR h

sdot sdotsdot=


sdot sdot sdot=

+

( )

2 221 1 1

2

g g gR R R

g g R h g R hR

g

h


sdot++

22 664 10

981 734 1 2 6 6 2

64 10 1 10

R m m mg g

R h s m m s


+ sdot + sdot


Vježba 119




Rezultat 734 2

m

s

23





Rješenje 120





1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot





Newtonovom poučku

G m g= sdot


na Zemlji jednaka


2v

F mcpr

= sdot



m

m VV

ρ ρ= rArr = sdot


3

4

3V r π= sdot sdot



2 2

2

2

2


R R G mR R


sdotrArr

4 3

3

4 3

2 24 3

3

3

v R v RM R

G G

M V V R

M R

ρ π

ρ

ρ

π

π

= sdot = sdot sdot

= sdot

sdot sdot


sdot

sdot

2 24 33

23 4

3

34

v R vR

G GR R

ρ π ρππ

sdotsdot



24

( )

2

3 2920

5130 2 32

11 64 667 10 2439 10

2

m

kgs

N m mm

kgπ

sdot

= =sdotminus



2 2 21

v v v



2

2920

350 6 2

2439 10

m

ms

m s

= =sdot

Vježba 120




Rezultat 5130 350 3 2

kg m

m s

17



1 22

m mF G

r

sdot= sdot


tijela


2v

F mcpr

= sdot





2

rv

T

πsdot sdot=




2 2

2 2

2

2 2


r

mr vvrr r

sdotsdot


2

G Mr

v

sdotrArr =


metod

2 222

2

2 2

a

zamj2

ene

G Mr rrv Tv

T vT vTG M

T

v

G M G M vr r rv v v

π πππ



sdot

[ ]

211 24

2 667 10 596 102

2901175 1502 min

3 3

6520

901175 60

N mkg

G M kgT s

v m

s

ππ



Vježba 114



Rezultat 12016 min





Rješenje 115


1 2



18



1 22

m mF G

r

sdot= sdot




( )

( )( ) ( )


težina tijela mase

2 2

1

1 2m na visini h

22

m M m MG G G G

R R

m M m MG G G G

R h R h


rArr rArrsdot sdot

= sdot sdot = sdot

+ +

( ) ( ) ( )

podijelimo

jednadžbe

1

2 2 22

1 1 1

2 2 22 2

m MG

G R R Rm

m MG

G

m MMG G

R h R h R

G

h

G

sdot sdotsdot sdot


sdot sdot sdot sdot

+ + +

( ) ( ) ( )( )

22 2 2

2 22 2 2 2

2

2 2

R h R h R hR h R

R R

R

R


( )2 2 2



Odgovor je pod C

Vježba 115




Rezultat A




N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rješenje 116


v =




1 22

m mF G

r

sdot= sdot



19


2v

F mcpr

= sdot




v 2 r

T

πsdot sdot=




2 22

2 2

v m M v m M M


r

m rr r



pomnožimo

jednadžbe

2

2 22 3

2

rv

r M MTv v G v G

M T r Tv

r

rG

r

π

π π

sdot sdot=


= sdot

3

2 2 23 3 3G M G M G Mv v v

T T T


211 24

2 667 10 596 1023

652284 652 9 000

N mkg

m kmkg

s s s

πsdotminus


= = asymp

Odgovor je pod B

Vježba 116



N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rezultat B





Rješenje 117

F1 = Z r1 = R 50 51

2 50 1 50 50 50

R R R R Rr R R

sdot += + = + = = sdot F2 =


20



1 22

m mF G

r

sdot= sdot




1

1

1 2 2

1

F Z

r R

m M m MF G Z G

r R


=

=


51

22 2 22 2 2601 250 512 250050


r R

r

R

Rsdot sdot sdot


= sdot

2500

2 22601

m MF G

R


sdot


2podijelimo

jednadžbe

2500 2500

222601 26012 2

25

2

002 22

2601

m M m MZ G G

F FRR

m

m MG

R

M m MZ Z m MGF G

RG

RR



sdot

2500 2500 25002 2 096 2 22601 2601 2 1

60

F FZ F Z F Z


Odgovor je pod D

Vježba 117




Rezultat C




Rješenje 118


6 m M = 6 middot 10

24 kg v =




1 22

m mF G

r

sdot= sdot


tijela

21

211

6 7 6 102

N mG

kg

sdotminus= sdot



centripetalna sila

2v

F mcpr

= sdot




( ) ( )

2

2

2 2


R h R

R h

mhR h R h


+ ++

+sdot

+

2 2G M G M G M

v v vR h R h R h


+ + +

211 24

667 10 6 102

77693 6 5

64 10 23 10

N mkg

mkg

sm m


= =sdot + sdot

Vježba 118



kg)

Rezultat 7000 m

sasymp





Rješenje 119

M = 6 middot 1024


6 m

g = 981 m s2 g1 =




1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot


22




Newtonovom poučku

G m g= sdot


na Zemlji jednaka

1inačica


( ) ( )1 2

1

2

mm M m M

G F m g G m g G

R h R h


+ +

( ) ( )

211 24

667 10 6 102

731 1 2 2 2

6 664 10 1 10

N mkg

G M mkgg

sR h m m


sdotrArr = = =

+ sdot + sdot

2inačica


2 2

2


R

m

R R



( ) ( ) ( )

1 1 12 2

2


R h R h R h


+ + +


( )

( ) ( ) ( )

1

2 2 221 1 1

1

podijelimo

jednadž2

e2 2 2

b1

G MG Mg

g g gR h R h R hR

G M G Mg g g

G M

G Mg

R R RR h

sdot sdotsdot=


sdot sdot sdot=

+

( )

2 221 1 1

2

g g gR R R

g g R h g R hR

g

h


sdot++

22 664 10

981 734 1 2 6 6 2

64 10 1 10

R m m mg g

R h s m m s


+ sdot + sdot


Vježba 119




Rezultat 734 2

m

s

23





Rješenje 120





1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot





Newtonovom poučku

G m g= sdot


na Zemlji jednaka


2v

F mcpr

= sdot



m

m VV

ρ ρ= rArr = sdot


3

4

3V r π= sdot sdot



2 2

2

2

2


R R G mR R


sdotrArr

4 3

3

4 3

2 24 3

3

3

v R v RM R

G G

M V V R

M R

ρ π

ρ

ρ

π

π

= sdot = sdot sdot

= sdot

sdot sdot


sdot

sdot

2 24 33

23 4

3

34

v R vR

G GR R

ρ π ρππ

sdotsdot



24

( )

2

3 2920

5130 2 32

11 64 667 10 2439 10

2

m

kgs

N m mm

kgπ

sdot

= =sdotminus



2 2 21

v v v



2

2920

350 6 2

2439 10

m

ms

m s

= =sdot

Vježba 120




Rezultat 5130 350 3 2

kg m

m s

18



1 22

m mF G

r

sdot= sdot




( )

( )( ) ( )


težina tijela mase

2 2

1

1 2m na visini h

22

m M m MG G G G

R R

m M m MG G G G

R h R h


rArr rArrsdot sdot

= sdot sdot = sdot

+ +

( ) ( ) ( )

podijelimo

jednadžbe

1

2 2 22

1 1 1

2 2 22 2

m MG

G R R Rm

m MG

G

m MMG G

R h R h R

G

h

G

sdot sdotsdot sdot


sdot sdot sdot sdot

+ + +

( ) ( ) ( )( )

22 2 2

2 22 2 2 2

2

2 2

R h R h R hR h R

R R

R

R


( )2 2 2



Odgovor je pod C

Vježba 115




Rezultat A




N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rješenje 116


v =




1 22

m mF G

r

sdot= sdot



19


2v

F mcpr

= sdot




v 2 r

T

πsdot sdot=




2 22

2 2

v m M v m M M


r

m rr r



pomnožimo

jednadžbe

2

2 22 3

2

rv

r M MTv v G v G

M T r Tv

r

rG

r

π

π π

sdot sdot=


= sdot

3

2 2 23 3 3G M G M G Mv v v

T T T


211 24

2 667 10 596 1023

652284 652 9 000

N mkg

m kmkg

s s s

πsdotminus


= = asymp

Odgovor je pod B

Vježba 116



N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rezultat B





Rješenje 117

F1 = Z r1 = R 50 51

2 50 1 50 50 50

R R R R Rr R R

sdot += + = + = = sdot F2 =


20



1 22

m mF G

r

sdot= sdot




1

1

1 2 2

1

F Z

r R

m M m MF G Z G

r R


=

=


51

22 2 22 2 2601 250 512 250050


r R

r

R

Rsdot sdot sdot


= sdot

2500

2 22601

m MF G

R


sdot


2podijelimo

jednadžbe

2500 2500

222601 26012 2

25

2

002 22

2601

m M m MZ G G

F FRR

m

m MG

R

M m MZ Z m MGF G

RG

RR



sdot

2500 2500 25002 2 096 2 22601 2601 2 1

60

F FZ F Z F Z


Odgovor je pod D

Vježba 117




Rezultat C




Rješenje 118


6 m M = 6 middot 10

24 kg v =




1 22

m mF G

r

sdot= sdot


tijela

21

211

6 7 6 102

N mG

kg

sdotminus= sdot



centripetalna sila

2v

F mcpr

= sdot




( ) ( )

2

2

2 2


R h R

R h

mhR h R h


+ ++

+sdot

+

2 2G M G M G M

v v vR h R h R h


+ + +

211 24

667 10 6 102

77693 6 5

64 10 23 10

N mkg

mkg

sm m


= =sdot + sdot

Vježba 118



kg)

Rezultat 7000 m

sasymp





Rješenje 119

M = 6 middot 1024


6 m

g = 981 m s2 g1 =




1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot


22




Newtonovom poučku

G m g= sdot


na Zemlji jednaka

1inačica


( ) ( )1 2

1

2

mm M m M

G F m g G m g G

R h R h


+ +

( ) ( )

211 24

667 10 6 102

731 1 2 2 2

6 664 10 1 10

N mkg

G M mkgg

sR h m m


sdotrArr = = =

+ sdot + sdot

2inačica


2 2

2


R

m

R R



( ) ( ) ( )

1 1 12 2

2


R h R h R h


+ + +


( )

( ) ( ) ( )

1

2 2 221 1 1

1

podijelimo

jednadž2

e2 2 2

b1

G MG Mg

g g gR h R h R hR

G M G Mg g g

G M

G Mg

R R RR h

sdot sdotsdot=


sdot sdot sdot=

+

( )

2 221 1 1

2

g g gR R R

g g R h g R hR

g

h


sdot++

22 664 10

981 734 1 2 6 6 2

64 10 1 10

R m m mg g

R h s m m s


+ sdot + sdot


Vježba 119




Rezultat 734 2

m

s

23





Rješenje 120





1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot





Newtonovom poučku

G m g= sdot


na Zemlji jednaka


2v

F mcpr

= sdot



m

m VV

ρ ρ= rArr = sdot


3

4

3V r π= sdot sdot



2 2

2

2

2


R R G mR R


sdotrArr

4 3

3

4 3

2 24 3

3

3

v R v RM R

G G

M V V R

M R

ρ π

ρ

ρ

π

π

= sdot = sdot sdot

= sdot

sdot sdot


sdot

sdot

2 24 33

23 4

3

34

v R vR

G GR R

ρ π ρππ

sdotsdot



24

( )

2

3 2920

5130 2 32

11 64 667 10 2439 10

2

m

kgs

N m mm

kgπ

sdot

= =sdotminus



2 2 21

v v v



2

2920

350 6 2

2439 10

m

ms

m s

= =sdot

Vježba 120




Rezultat 5130 350 3 2

kg m

m s

19


2v

F mcpr

= sdot




v 2 r

T

πsdot sdot=




2 22

2 2

v m M v m M M


r

m rr r



pomnožimo

jednadžbe

2

2 22 3

2

rv

r M MTv v G v G

M T r Tv

r

rG

r

π

π π

sdot sdot=


= sdot

3

2 2 23 3 3G M G M G Mv v v

T T T


211 24

2 667 10 596 1023

652284 652 9 000

N mkg

m kmkg

s s s

πsdotminus


= = asymp

Odgovor je pod B

Vježba 116



N middot m2 kg


24 kg)

1580 652 815 1120km km km km

A B C Ds s s s

Rezultat B





Rješenje 117

F1 = Z r1 = R 50 51

2 50 1 50 50 50

R R R R Rr R R

sdot += + = + = = sdot F2 =


20



1 22

m mF G

r

sdot= sdot




1

1

1 2 2

1

F Z

r R

m M m MF G Z G

r R


=

=


51

22 2 22 2 2601 250 512 250050


r R

r

R

Rsdot sdot sdot


= sdot

2500

2 22601

m MF G

R


sdot


2podijelimo

jednadžbe

2500 2500

222601 26012 2

25

2

002 22

2601

m M m MZ G G

F FRR

m

m MG

R

M m MZ Z m MGF G

RG

RR



sdot

2500 2500 25002 2 096 2 22601 2601 2 1

60

F FZ F Z F Z


Odgovor je pod D

Vježba 117




Rezultat C




Rješenje 118


6 m M = 6 middot 10

24 kg v =




1 22

m mF G

r

sdot= sdot


tijela

21

211

6 7 6 102

N mG

kg

sdotminus= sdot



centripetalna sila

2v

F mcpr

= sdot




( ) ( )

2

2

2 2


R h R

R h

mhR h R h


+ ++

+sdot

+

2 2G M G M G M

v v vR h R h R h


+ + +

211 24

667 10 6 102

77693 6 5

64 10 23 10

N mkg

mkg

sm m


= =sdot + sdot

Vježba 118



kg)

Rezultat 7000 m

sasymp





Rješenje 119

M = 6 middot 1024


6 m

g = 981 m s2 g1 =




1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot


22




Newtonovom poučku

G m g= sdot


na Zemlji jednaka

1inačica


( ) ( )1 2

1

2

mm M m M

G F m g G m g G

R h R h


+ +

( ) ( )

211 24

667 10 6 102

731 1 2 2 2

6 664 10 1 10

N mkg

G M mkgg

sR h m m


sdotrArr = = =

+ sdot + sdot

2inačica


2 2

2


R

m

R R



( ) ( ) ( )

1 1 12 2

2


R h R h R h


+ + +


( )

( ) ( ) ( )

1

2 2 221 1 1

1

podijelimo

jednadž2

e2 2 2

b1

G MG Mg

g g gR h R h R hR

G M G Mg g g

G M

G Mg

R R RR h

sdot sdotsdot=


sdot sdot sdot=

+

( )

2 221 1 1

2

g g gR R R

g g R h g R hR

g

h


sdot++

22 664 10

981 734 1 2 6 6 2

64 10 1 10

R m m mg g

R h s m m s


+ sdot + sdot


Vježba 119




Rezultat 734 2

m

s

23





Rješenje 120





1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot





Newtonovom poučku

G m g= sdot


na Zemlji jednaka


2v

F mcpr

= sdot



m

m VV

ρ ρ= rArr = sdot


3

4

3V r π= sdot sdot



2 2

2

2

2


R R G mR R


sdotrArr

4 3

3

4 3

2 24 3

3

3

v R v RM R

G G

M V V R

M R

ρ π

ρ

ρ

π

π

= sdot = sdot sdot

= sdot

sdot sdot


sdot

sdot

2 24 33

23 4

3

34

v R vR

G GR R

ρ π ρππ

sdotsdot



24

( )

2

3 2920

5130 2 32

11 64 667 10 2439 10

2

m

kgs

N m mm

kgπ

sdot

= =sdotminus



2 2 21

v v v



2

2920

350 6 2

2439 10

m

ms

m s

= =sdot

Vježba 120




Rezultat 5130 350 3 2

kg m

m s

20



1 22

m mF G

r

sdot= sdot




1

1

1 2 2

1

F Z

r R

m M m MF G Z G

r R


=

=


51

22 2 22 2 2601 250 512 250050


r R

r

R

Rsdot sdot sdot


= sdot

2500

2 22601

m MF G

R


sdot


2podijelimo

jednadžbe

2500 2500

222601 26012 2

25

2

002 22

2601

m M m MZ G G

F FRR

m

m MG

R

M m MZ Z m MGF G

RG

RR



sdot

2500 2500 25002 2 096 2 22601 2601 2 1

60

F FZ F Z F Z


Odgovor je pod D

Vježba 117




Rezultat C




Rješenje 118


6 m M = 6 middot 10

24 kg v =




1 22

m mF G

r

sdot= sdot


tijela

21

211

6 7 6 102

N mG

kg

sdotminus= sdot



centripetalna sila

2v

F mcpr

= sdot




( ) ( )

2

2

2 2


R h R

R h

mhR h R h


+ ++

+sdot

+

2 2G M G M G M

v v vR h R h R h


+ + +

211 24

667 10 6 102

77693 6 5

64 10 23 10

N mkg

mkg

sm m


= =sdot + sdot

Vježba 118



kg)

Rezultat 7000 m

sasymp





Rješenje 119

M = 6 middot 1024


6 m

g = 981 m s2 g1 =




1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot


22




Newtonovom poučku

G m g= sdot


na Zemlji jednaka

1inačica


( ) ( )1 2

1

2

mm M m M

G F m g G m g G

R h R h


+ +

( ) ( )

211 24

667 10 6 102

731 1 2 2 2

6 664 10 1 10

N mkg

G M mkgg

sR h m m


sdotrArr = = =

+ sdot + sdot

2inačica


2 2

2


R

m

R R



( ) ( ) ( )

1 1 12 2

2


R h R h R h


+ + +


( )

( ) ( ) ( )

1

2 2 221 1 1

1

podijelimo

jednadž2

e2 2 2

b1

G MG Mg

g g gR h R h R hR

G M G Mg g g

G M

G Mg

R R RR h

sdot sdotsdot=


sdot sdot sdot=

+

( )

2 221 1 1

2

g g gR R R

g g R h g R hR

g

h


sdot++

22 664 10

981 734 1 2 6 6 2

64 10 1 10

R m m mg g

R h s m m s


+ sdot + sdot


Vježba 119




Rezultat 734 2

m

s

23





Rješenje 120





1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot





Newtonovom poučku

G m g= sdot


na Zemlji jednaka


2v

F mcpr

= sdot



m

m VV

ρ ρ= rArr = sdot


3

4

3V r π= sdot sdot



2 2

2

2

2


R R G mR R


sdotrArr

4 3

3

4 3

2 24 3

3

3

v R v RM R

G G

M V V R

M R

ρ π

ρ

ρ

π

π

= sdot = sdot sdot

= sdot

sdot sdot


sdot

sdot

2 24 33

23 4

3

34

v R vR

G GR R

ρ π ρππ

sdotsdot



24

( )

2

3 2920

5130 2 32

11 64 667 10 2439 10

2

m

kgs

N m mm

kgπ

sdot

= =sdotminus



2 2 21

v v v



2

2920

350 6 2

2439 10

m

ms

m s

= =sdot

Vježba 120




Rezultat 5130 350 3 2

kg m

m s

21

211

6 7 6 102

N mG

kg

sdotminus= sdot



centripetalna sila

2v

F mcpr

= sdot




( ) ( )

2

2

2 2


R h R

R h

mhR h R h


+ ++

+sdot

+

2 2G M G M G M

v v vR h R h R h


+ + +

211 24

667 10 6 102

77693 6 5

64 10 23 10

N mkg

mkg

sm m


= =sdot + sdot

Vježba 118



kg)

Rezultat 7000 m

sasymp





Rješenje 119

M = 6 middot 1024


6 m

g = 981 m s2 g1 =




1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot


22




Newtonovom poučku

G m g= sdot


na Zemlji jednaka

1inačica


( ) ( )1 2

1

2

mm M m M

G F m g G m g G

R h R h


+ +

( ) ( )

211 24

667 10 6 102

731 1 2 2 2

6 664 10 1 10

N mkg

G M mkgg

sR h m m


sdotrArr = = =

+ sdot + sdot

2inačica


2 2

2


R

m

R R



( ) ( ) ( )

1 1 12 2

2


R h R h R h


+ + +


( )

( ) ( ) ( )

1

2 2 221 1 1

1

podijelimo

jednadž2

e2 2 2

b1

G MG Mg

g g gR h R h R hR

G M G Mg g g

G M

G Mg

R R RR h

sdot sdotsdot=


sdot sdot sdot=

+

( )

2 221 1 1

2

g g gR R R

g g R h g R hR

g

h


sdot++

22 664 10

981 734 1 2 6 6 2

64 10 1 10

R m m mg g

R h s m m s


+ sdot + sdot


Vježba 119




Rezultat 734 2

m

s

23





Rješenje 120





1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot





Newtonovom poučku

G m g= sdot


na Zemlji jednaka


2v

F mcpr

= sdot



m

m VV

ρ ρ= rArr = sdot


3

4

3V r π= sdot sdot



2 2

2

2

2


R R G mR R


sdotrArr

4 3

3

4 3

2 24 3

3

3

v R v RM R

G G

M V V R

M R

ρ π

ρ

ρ

π

π

= sdot = sdot sdot

= sdot

sdot sdot


sdot

sdot

2 24 33

23 4

3

34

v R vR

G GR R

ρ π ρππ

sdotsdot



24

( )

2

3 2920

5130 2 32

11 64 667 10 2439 10

2

m

kgs

N m mm

kgπ

sdot

= =sdotminus



2 2 21

v v v



2

2920

350 6 2

2439 10

m

ms

m s

= =sdot

Vježba 120




Rezultat 5130 350 3 2

kg m

m s

22




Newtonovom poučku

G m g= sdot


na Zemlji jednaka

1inačica


( ) ( )1 2

1

2

mm M m M

G F m g G m g G

R h R h


+ +

( ) ( )

211 24

667 10 6 102

731 1 2 2 2

6 664 10 1 10

N mkg

G M mkgg

sR h m m


sdotrArr = = =

+ sdot + sdot

2inačica


2 2

2


R

m

R R



( ) ( ) ( )

1 1 12 2

2


R h R h R h


+ + +


( )

( ) ( ) ( )

1

2 2 221 1 1

1

podijelimo

jednadž2

e2 2 2

b1

G MG Mg

g g gR h R h R hR

G M G Mg g g

G M

G Mg

R R RR h

sdot sdotsdot=


sdot sdot sdot=

+

( )

2 221 1 1

2

g g gR R R

g g R h g R hR

g

h


sdot++

22 664 10

981 734 1 2 6 6 2

64 10 1 10

R m m mg g

R h s m m s


+ sdot + sdot


Vježba 119




Rezultat 734 2

m

s

23





Rješenje 120





1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot





Newtonovom poučku

G m g= sdot


na Zemlji jednaka


2v

F mcpr

= sdot



m

m VV

ρ ρ= rArr = sdot


3

4

3V r π= sdot sdot



2 2

2

2

2


R R G mR R


sdotrArr

4 3

3

4 3

2 24 3

3

3

v R v RM R

G G

M V V R

M R

ρ π

ρ

ρ

π

π

= sdot = sdot sdot

= sdot

sdot sdot


sdot

sdot

2 24 33

23 4

3

34

v R vR

G GR R

ρ π ρππ

sdotsdot



24

( )

2

3 2920

5130 2 32

11 64 667 10 2439 10

2

m

kgs

N m mm

kgπ

sdot

= =sdotminus



2 2 21

v v v



2

2920

350 6 2

2439 10

m

ms

m s

= =sdot

Vježba 120




Rezultat 5130 350 3 2

kg m

m s

23





Rješenje 120





1 22

m mF G

r

sdot= sdot


tijela

211

6 7 6 102

N mG

kg

sdotminus= sdot





Newtonovom poučku

G m g= sdot


na Zemlji jednaka


2v

F mcpr

= sdot



m

m VV

ρ ρ= rArr = sdot


3

4

3V r π= sdot sdot



2 2

2

2

2


R R G mR R


sdotrArr

4 3

3

4 3

2 24 3

3

3

v R v RM R

G G

M V V R

M R

ρ π

ρ

ρ

π

π

= sdot = sdot sdot

= sdot

sdot sdot


sdot

sdot

2 24 33

23 4

3

34

v R vR

G GR R

ρ π ρππ

sdotsdot



24

( )

2

3 2920

5130 2 32

11 64 667 10 2439 10

2

m

kgs

N m mm

kgπ

sdot

= =sdotminus



2 2 21

v v v



2

2920

350 6 2

2439 10

m

ms

m s

= =sdot

Vježba 120




Rezultat 5130 350 3 2

kg m

m s

24

( )

2

3 2920

5130 2 32

11 64 667 10 2439 10

2

m

kgs

N m mm

kgπ

sdot

= =sdotminus



2 2 21

v v v



2

2920

350 6 2

2439 10

m

ms

m s

= =sdot

Vježba 120




Rezultat 5130 350 3 2

kg m

m s

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