Solutions to problems in Leighton and Vogt's "Exercises in Introductory Physics" textbook

1

Solutions to problems in Leighton and Vogt’s textbook

"Exercises in Introductory Physics" (Addison-Wesley, 1969)

Pier F. Nali

(Revised April 10, 2016)

CHAPTER 1

Atoms in Motion

B-3

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics 2

a) 23

19 30

3 3

6.025 10 molecules / mole10 molecules/cm

22.41 10 cm / moleG

Nn

Molar_vol._at_STP

×

×∼ ∼ ∼ .

3

3 22 3

3

_ _ 1.0 g cm10 10 molecules/cm

_ 0.001 g cm

LL

G

n liquid air densityn

n air density

−

−⇒∼ ∼ ∼ ∼ .

b) 3

23

19 3

_ 0.001 g cm10 g/molecule

2.7 10 molecules / cmG

air densitym

n

−−

×∼ ∼ ∼ .

c) Said σ the (approx.) diameter of an air molecule the inverse of the mean free path is

21G

nσ ⋅∼�

, where 73

110 cm

Lnσ −∼ ∼ , assuming for a rough estimate that the molecules are condensed

in liquid air in such a way that each molecule occupies a cubic cell of side σ . Thus

( )2

7 19 3 5 1110 cm 10 molecules/cm 10 cm− −×∼ ∼

� or, calculating the reciprocal, the mean free path

510 cm−� ∼ .

d) From ( ) ( ) ( )

: : :STP STP STP

G GP P n n∼ ∼ � � ,

( )

( ) 5

1 atm,

10 cm,

1 m,

STP

STP

P

−

∼

� ∼

� ∼

one gets 710 atmP−

∼ .

B-4

Said P the pressure of argon (A) gas spread in a layer of volume V (at constant temperature), we have

( )

( ) 3 3

4

10 3

760 mm Hg 22.41 10 cm /mole

6.0 10 mm Hg

2.839 10 cm /mole

STP

STP

P Molar_vol._at_STP P Molar_vol._at_STPV

V PP−

× × ×= ⇒ = = =

×

= ×


and

( )23

3 13 30

10 3

6.025 10 atoms/mole_ A _ _ _ 2.123 10 atoms / cm

2.839 10 cm /mole

Nnumber density atoms per cm

V

×= = = ×

×.

Said A a constant of proportionality representing the effective target area per A atom and expressing (on

experimental basis) the percent intensity reduction of the atomic beam of K atoms as

_ 3.0 %A number density thickness× × = one gets

2

14 2

13 3 1

3.0 % 3.0 101.4 10 cm /atom

_ 2.123 10 atoms/cm 10 cmA

number density thickness

−−

−

×= = ≈ ×

× × ×.

B-5

The cubic lattice has N N N× × points and ( ) ( ) ( )1 1 1N N N− × − × − spacings between nearest

neighbours (Cl and Na ions in alternance). Then ( )3 3 3 3_ . 1lattice vol N s N s= − ⋅ ≈ (for large N ) and

3 33

33 3 3 24

1 1 1_ / A cm

22.4258 102.820 A

Nnumber density atoms

N s s

°−

−°

≈ = = =

×

.

Since the NaCl molecule is diatomic 1

2

0

_ _ _ _ NaCl_ _ NaCl

number density molecular weight ofdensity of

N

×= ,

Cl Na

Cl

Na

Cl….

Na….

2 x interatomic spacing s

Cl

Na


where: 3_ _ NaCl 2.165 g/cm ,

_ _ _ NaCl 58.454 g/mole,

density of

molecular weight of

=

=

and hence

1

232

0

_ _ _ _ NaCl6.02 10 molecules/mole

_ _ NaCl

number density molecular weight ofN

density of

×= = × .

B-6

The number of helium atoms expelled per second by one gram of radium in equilibrium with its

disintegration products is 10 1 1 14 1 113.6 10 atoms g s 117.504 10 atoms g day− − − −× = × . The corresponding

volume of helium produced per day by one gram of radium is 3 3 1

4 3 1 1

3

0.0824 10 cm day4.291 10 cm g day

192 10 g

− −− − −

−

×= ×

× and thus the number density of helium atoms at STP

is

a) ( )14 1 1

19 3

4 3 1 1

117.504 10 atoms g day2.7 10 atoms/cm

4.291 10 cm g day

STP

Hen

− −

− − −

×= ≈ ×

×

and the calculated Avogadro’s number

b) ( ) 19 3 3 3 23

0 2.7 10 atoms cm 22.41 10 cm /mole 6.1 10 atoms/moleSTP

HeN n Molar_vol._at_STP

−= × ≈ × × × ≈ × .


C-1

The molecular weight of H(CH2)18COOH is approximately 298 g/mole and the molar volume is 1

3

3

_ 298 g mole372.5 cm /mole

0.8 g cm

molecular weight

density

−

−≈ ≈ .

The volume of oil is deduced by dividing 0.81 mg by its density, i.e. 3

3 3

3

0.81 10 g1.0 10 cm

0.8 g cm

−−

−

×≈ × .

The surface area over which this volume of oil is spread is 2 2 3 214

84 cm 5.542 10 cmπ × ≈ × , so the

thickness of the oil film (calculated as if its density were the same as in non spread state) is 3 3 3 3

6

2 2 3 214

0.81 10 g 0.8 g cm 1.0 10 cm0.2 10 cm

84 cm 5.542 10 cml

π

− − −−× ×

≈ ≈ ≈ ×× ×

.

If the oil molecules were arranged in linear chains 20 atoms high ( thickness l≈ of the oil film) × 4

atoms wide ( 15

l≈ × ) so that a molecule occupies a volume roughly estimated (assuming in liquid state a

rotational degree of motion around the main axis of the chain) as

( )3

6322 31 1

5 5

0.2 10 cm3.2 10 cm

25 25

ll l l

−

−×

× × = ≈ ≈ × .

The reciprocal of the latter ( 22 30.3 10 molecules/cm× ) is the number density 0

_ .

N

Molar vol. So

22 3 22 3 3 24

0 0.3 10 molecules/cm _ . 0.3 10 mol./cm 372.5 cm /mole 10 molecules/mole.N Molar vol≈ × × ≈ × × ∼


C-2

.

Said m the mass of a gas molecule the gas density is ( )gas gm Nρ = ⋅ . If one assume for sake of

simplicity that in a solid the molecules are tightly packed in cells of side σ then ( ) 3solid mρ σ≈ (1).

Thus

( )

( )( )

( )( )

( )

23

3 233

gas gas gas1

solid solid solidg g g

g

N N NN

ρ ρ ρσ σ σ

ρ ρ ρ

≈ ⋅ ⇒ ≈ ⋅ ⇒ ≈ ⋅

.

(1)

For our purposes here we may ignore that in closed packaging configuration the proportion of space occupied by spherical

particles is ( )

( )

_ _ _ _ 74, 048%

3 2

particle

unitcell

nr of particles per cell V

V

π×∆ = = ≈ and therefore the number of particles per

unit cell is ( )

( )

3

32 particles/cell

3 2 6

unitcell

particle

V

V

π σπ σ

∆ ⋅ = ⋅ = rather than 1 particle/cell, ( )solidρ is 3

2 m σ⋅ rather than

3m σ and so on, so the subsequent results should contain

2

Ngin locus of

gN . This is the densest packing of equal spheres

in three dimensions, as proved by Gauss in 1831. See Gauss, C. F. "Besprechung des Buchs von L. A. Seeber:

Intersuchungen über die Eigenschaften der positiven ternären quadratischen Formen usw." Göttingsche Gelehrte Anzeigen

(1831, July 9) 2, 188-196, 1876 = Werke, II (1876), 188-196. For a proof of this result see, e.g., J. W. S. Cassels, —An

Introduction to the Geometry of Numbers, Springer-Verlag, 1971 [Chap. II, Th. III]


Comparing ( )21 2g

Nπ σ=� and ( )13 gas vη ρ= ⋅ ⋅� one also gets

( ) ( ) ( )2 1 2 gas 2 3g

N vσ π ρ π η= = ⋅ ⋅� .

Comparing again the right-hand sides of latter two expressions for 2

gN σ and cubing the second

members one gets

( )( )

( ) ( )2

3 3 3 3gas

gas 54 2solid

gN v

ρρ π η

ρ

⋅ ≈ ⋅ ⋅

and finally – assuming on experimental basis(2)

( ) ( )solid liquidρ ρ≈ -

( ) ( )2 3

19 3

3 3

gas liquid0.7 10 molecules/cm

54 2g

vN

ρ ρ

π η

⋅ ≈ ×⋅ ⋅

∼ .

(2)

This assumption is valid where we are dealing only whit orders of magnitude and is based on the observed condensation

behaviour of many substances. Unfortunately, at the times of Loschimdt’s work (1865) was not yet known how to liquefy

the air, so he could not directly observe condensation behaviour for air. However, the density of a substance in the

condensed state can be estimated by its chemical composition, as Loschmidt did. See Porterfield, William W., and Walter

Kruse. "Loschmidt and the Discovery of the Small.", J. Chem. Educ. , 1995, 72 (10), p 870

1



Pier F. Nali


CHAPTER 2

Conservation of Energy, Statics

1

According to the principle of virtual work (PVW) “the sum of the heights times the weights does not

change”, said 1h and 2h the heights of the weights 1W and 2W respectively, then

( )1 1 2 2 1 1 2 2 0W h W h W h W h∆ + = ∆ + ∆ = .

If we assume that the weight 1W goes down 1h∆ as the weight 2W rises 2h∆ , then

11 2 1 1 2 2

2

h h W W∆ = − ∆ ⇒ =�

� ��

.


2

The “sum of the heights times the weights” for an arbitrary number of weights becomes 0i i

i

W h∆ =∑

and considering the distances positive or negative depending on the side of the fulcrum, so that

ii j

j

h h∆ = ∆�

� (or i ih∆ ∝ � for any i, being

j

j

h∆

� independent of j), it becomes thus 0

i i

i

W =∑ � .

3

The PVW can be expressed in a more general form as 0N n

ij i

i j

∆ =

∑ ∑F ri , where the sum over j is

extended to all the n forces acting on the mass i and the sum over i is extended to all the N masses.

In the present case (a single body) 1N = , so the above formula reduces to 0n

i

i

∆ =

∑F ri .

a) For 1n = is 0∆ =F ri ; so, for the arbitrariness of the choice of the vector ∆r , 0=F .


b) For 2n = is ( )1 2 0+ ∆ =F F ri and, again for the arbitrariness of the choice of the vector ∆r , is

1 2 0+ =F F , whence 1 2= −F F ; so, the two forces are equal in magnitude, opposite in direction and

collinear.

c) For 3n = is, again, 1 2 3 0+ + =F F F . If we consider the plane on which lie the two vectors 1F and

2F (apart from the trivial case in which the two vectors are parallel, which degenerates to the case of

2n = ), 3F also lies in the same pane, being expressible as a linear combination of the other two

vectors. Let 0

X the vector joining the origin to the point of intersection of the lines of action of the

two vectors 1F and 2F and X the vector joining the origin to an arbitrary point of one or the other

line of action. Then the two lines of action are defined by the system of equations

( )( )

0 1

0 2

0

0

− ∧ =

− ∧ =

X X F

X X F.

By adding the latter two equation, immediately follows that 0

X also lies on the line of action of 3F .

0X defines a single point (through which the three lines pass) as each line of action cannot intersect

each other in more than one point (provided the lines are not parallel).

d) From r cosi i i

F∆ = ∆ ⋅ ∆F ri it follows that 1 1

cos 0n n

i i i

i i

F r= =

∆ = ∆ ⋅∆ =

∑ ∑F ri , and – again for the

arbitrariness of the choice of r∆ - 1

cos 0n

i i

i

F=

∆ =∑ .


A-1

Suppose the weight W goes down x as the centre of mass of the plank rises y. Therefore (refer to the

above figure) 2

sinLy θ= ∆ , ( )x c c′= − − and

( )2 2

sin ,

1 cos ,

.

a a L

b b L

c a b

θ

θ

′ = − ∆

′ = + − ∆

′ ′ ′= +

So – assuming θ∆ small - ( ) ( )2 2 2

2

21 2 sin 4 sin 1 sin

L L baL aL

c c cc c cθθ θ

+ ∆′ = − ∆ + ≅ − ∆ and sinaLc

x θ≅ − ∆ ,

where we have used the trigonometric double-angle formula 21 cos 2sin2

αα− = and the approx.

1 12

xx+ ≅ + for x small.

a

W

W

45°

a’

b

b’ c

c’

x

L

y

θ∆


Being also 2c a= ⋅ it follows 2

sin 2Lx yθ≅ − ∆ = − ⋅ . From PVW then we have

( ) ( )plankW x W y⋅ − = ⋅ , whence ( )

( ) 3kg-wts

2 2

plank

plank

WyW W

x= − = = .

(note: the length of the plank doesn't affect this derivation).

A-2

Let W the weight of the ball, P

F and W

F the forces exerted on the ball by the plane and by the wall

respectively and made use of PVW in the form 1

cos 0n

i i

i

F=

∆ =∑ .

Thus, with respect to an horizontal line joining the point in which the ball touches the wall with the

centre of mass of the ball, ( ) ( )2cos 0 cos 0w PF F π α+ + = , that is sin 0

w PF F α− = ; and, with respect to

a vertical line lying in the plane of the sheet and passing through the centre of mass of the ball,

2cos cos cos 0

W PF W Fπ π α+ + = , that is cos 0

PW F α− + = .

So

,cos

tan ,

P

W

WF

F W

α

α

=

=

and the forces on the planes are

1 kg-wt,

cos

F tan kg-wt.

P

W

Fα

α

=

=


A-3

Again make use of PVW, this time in the form 1

0n

i i

i

W h=

=∑ .

- Without the loads: ( )C AA CD BB GHW W AP W PB′ ′+ ⋅ = ⋅ , that is 2C AA CD BB GH

W W W′ ′+ = .

- With the loads, assume the weight 1W goes down x as the weight 2W rises y, being 1

2x y= − .

Thus ( ) ( ) ( )1 2C AA CD BB GHW W W x W W y′ ′+ + ⋅ − = + ⋅ , that is ( )1 22C AA CD BB GHW W W W W′ ′+ + = + , and

subtracting to this equation the one without the loads above one gets 1 22W W= , so

12 12

0.25 kg-wtsW W= = .

A-4

l1 l2

L

xA

yA

Ay−∆

ϕ


The weight A can move both vertically and horizontally.

- Vertical displacement of A (assumed downwards): ( )1 kg1 kgA A B B

W y W y W y⋅ −∆ = ⋅∆ + ⋅∆ .

- Horizontal displacement of A (assuming B rises B

y∆ as A moves to the left):

( )1 kg1 kg B BW y W y⋅ −∆ = ⋅ ∆

Case 1 – vertical displacement

For the flattened triangle � (l1, A

y∆ , 1 1l l+ ∆ ), using Carnot’s theorem and assuming A

y∆ small,

( ) ( ) ( )2

21 1

2 2 2 2

1 1 1 1 1 1 122 cos 2 sin 1 2 sinA Ay y

A A A A l ll l l l y y l l y y lπ ϕ ϕ ϕ∆ ∆

+ ∆ = + ∆ + + ∆ = + ∆ − + ∆ = − + .

Neglecting the term of order 2

Ay∆ and using the approx. 1

21 1x x+ ≅ + , then

( ) ( )1 1

1 1 1 11 2 sin 1 sinA Ay y

l ll l l lϕ ϕ∆ ∆

+ ∆ ≅ − ≅ − and ( )11 1 11 sin sinAy

All l l yϕ ϕ∆

∆ ≅ − − ≅ −∆ , where

1 1 1

sin A Ay y

l l lϕ = ≅

+ ∆. So 1

1

AA

yl y

l∆ ≅ −∆ .

Similarly, for the triangle (not shown) � (l2, A

y∆ , 2 2l l+ ∆ ), 2

2

AA

yl y

l∆ ≅ −∆ . But

1 kg1l y∆ = ∆ and

2 Bl y∆ = ∆ , whence

( ) ( ) ( )1 kg 1 kg

1 2 1 2

1 1 1 11 kg 1 kg2 22 2

sin 30 sin 45 .

A A A AA A A B A A B

B B B

y y y yW y W y W y W W W

l l l l

W W W W W

⋅ −∆ ≅ ⋅ −∆ + ⋅ −∆ → ≅ ⋅ + ⋅ =

= ⋅ ° + ⋅ ° = + = +

Case 2 – horizontal displacement

In the same way, for the triangle (not even shown as well) � ( 1 1l l− ∆ , Ax∆ , 1l ) with Ax∆ small,

( )2

21 1

2 2

1 1 1 1 12 cos30 1 2 cos30A Ax x

A A l ll l l l x x l

∆ ∆− ∆ = + ∆ ° + ∆ = + ° + , and, with the above approximations,

( ) ( ) ( )1 1 1

1 1 1 1 1 1 11 2 cos30 1 cos30 1 cos30 cos30A A Ax x x

Al l ll l l l l l l x

∆ ∆ ∆− ∆ ≅ + ° ≅ + ° → ∆ ≅ − + ° ≅ −∆ ° .

Substituting 1

cos30 Ax

l° = in the above expressions one gets 1

1

AA

xl x

l∆ ≅ −∆ , and similarly, for triangle

� ( 2 2l l+ ∆ , Ax∆ , 2l ): 2

2

cos 45 AA A

L xl x x

l

−∆ ≅ ∆ ° = ∆ .

So from PVW


( ) ( )1 kg1 kg 1 kg 1 2 1 kg

1 2

3 21 kg 1 kg 1 kg 2 2

1 2

31 kg2

cos30 cos 45

.

A AB B B A B A

A AB B B

B

x L xW y W y W l W l W x W x

l l

x L xW W W W W W

l l

W W

−⋅ −∆ = ⋅ ∆ → ⋅ −∆ = ⋅ ∆ → ⋅ ∆ = ⋅ ∆

−→ ⋅ = ⋅ → ⋅ ° = ⋅ ° → ⋅ = ⋅

→ = ⋅

Whence finally 31 1 11 kg2 2 22

.A BW W W≅ + = + In summary:

( )312 2

32

kg-wts,

kg-wts.

A

B

W

W

= +

=

A-5

From the PVW (in the form1

cos 0n

i i

i

F=

∆ =∑ ):

- Case 1 – vertical fixed line:

( ) ( )

( ) ( ) ( )

2

1 2 1 2 1 2

2AB2 1

1 2 1 2AC 2

cos cos cos 0 cos 1 sin

1 .

T T W T T W T T W

T T W T T W

ϕ ϕ π ϕ ϕ+ + = → + = → + − = →

+ − = → + =

- Case 2 – horizontal fixed line:

( ) ( )1 2 1 22 2 2cos cos cos 0T T W T Tπ π πϕ ϕ− − + − + = → = .

Thus 1 2

50 lb-wts

2 2

WT T= = =

ϕ


A-7

Let V the reaction force of the edge in contact with the wheel and assume null the reaction force of the

wall at the ground soon as the force F is applied to the axle. Then from PVW 1

cos 0n

i i

i

F=

∆ =∑ :

- Case 1 – vertical fixed line: cos cos 0 cosR h

V W V W V WR

ϕ π ϕ−

+ = → = → = .

- Case 2 – horizontal fixed line: ( ) ( )2 2cos 0 cos cos 0 sin

xF V W F V V

R

π πϕ ϕ+ + + = → = = ,

where ( ) ( )22 2x R R h h R h= − − = − .

Thus ( )2 )h R hx

F W WR h R h

−= =

− −.

V

x

ϕ V

x

ϕ


A-9

Using 1

cos 0n

i i

i

F=

∆ =∑ we have:

- horizontal line AB :

( ) ( )2cos 45 cos cos 0 cos 0 cos 45 cos 0O P Q x xF F F F F F Fππ π ϕ ϕ+ ° + + ° + + = ⇒ − ° + − = ,

- vertical line OP :

( ) ( )2 2 2cos 45 cos0 cos cos 0 sin 45 sin 0O P Q x xF F F F F F Fπ π π ϕ ϕ+ ° + ° + + + = ⇒ − ° + − = ;

and confronting:

cos sinϕ ϕ= , whence 45ϕ = ° or 225ϕ = ° .

xF is positive when 45ϕ = ° therefore it is parallel to, and has the same direction of,

OF . Its magnitude

is ( )2 1 20.7 Nx

F F= − = , 45° down from AB .

When the plate is in equilibrium the torque, with respect to the corner O, is

( )sin 0 0 0 sin 0

50 N 0,100 m0,34 m,

sin 45 20.7 N 0.7071

x Q P O x

x

F x OP F F F F x F OP

F OPx

F

π ϕ ϕ⋅ ⋅ − + ⋅ + ⋅ + ⋅ = ⇒ ⋅ ⋅ + ⋅ = ⇒

⋅ ×= − = − = −

° ×

i.e. .34 m left of O.

x φ

F

F

F

Fx


A-10

Assume 1 2W W> . Then as 1W goes down sinD θ− the weight 2W rises sinD θ+ . From the

conservation of energy ( )21 211 22

sin sin 0W W

v W D W Dg

θ θ +

⋅ + ⋅ − + ⋅ =

.

Thus 1 2

1 2

2 sinW W

v gDW W

θ−

=+

.

A-11

See the previous problem. In the present case,

( ) ( )212

2sin sin 0 sin sin

Wv W D W D v gD

gθ φ φ ϑ⋅ ⋅ + ⋅ + ⋅ − = ⇒ = − .


B-1

At 0t = ( ) ( ) ( ) ( )1 2 1 2

. . . . 0, . . . .2

HK E K E P E P E Mg= = = = , being 1 2M M M= =

At 0t > ( ) ( ) ( ) ( )2121 2 1 2

. . . . , . . , . .K E K E Mv P E Mgy P E Mgx= = = =

x y


Length of the cord: ( )1 22

H+

Length of the cord on the side of M2: H x−

Length of the cord on the side of M1: ( )2 12

Hx− +

Height of M2: x

Height of M1: 1

12 2 2

H xy

= − −

Then at 0t > ( )1

1. . 1

2 2 2

H xP E Mgy Mg

= = − −

and from the conservation of energy

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )1, 0 2, 0 1, 0 2, 0 1, 0 2, 0 1, 0 2, 0

2 2

2

. . . . . . . . . . . . . . . .

1 1 10 0 122 2 2 2 2 2

1 11 1 .2 22

t t t t t t t tK E K E P E P E K E K E P E P E

H H H xMg Mg Mv Mv Mg Mgx

HMgH Mv Mgx Mg

= = = = > > > >+ + + = + + + ⇒

+ + + = + + − − + ⇒

= + − + −

Taking the derivative with respect to t of each member of the above expression

10 2 1 02

Mva Mgv = + − +

, where we used dx

vdt

= and 2

2 2dv dv

v vadt dt

= = .

a) So the vertical acceleration of M2 at 0t > is 1 1

12 2

a g

= − −

.

b) Being 0a < the mass M2 will move downward. The eight of M2 is 2

Hx = and its velocity

( )0 0v = at 0t = whereas 21

2 2

Hx at= + at 0t > ; then it will strike the ground ( 0x = ) at

time 1

2

11

2

HHta

g

−= =

−

.

c) Being 1

12 2 2

H xy

= − −

the height of M1, then for 0x = is

11

2 2

Hy H

= − <

; so the

other mass (M1) will no strike the pulley.


B-2

When the boom rotates an angle θ∆ (assumed small, i.e. neglecting terms of the order if ( )2

θ∆ or

higher) around the pivot the length y of the horizontal cable will change of cosy x θ θ∆ ∆� . This can be

seen using Carnot’s theorem and the approx. 12

1 1x x+ ≅ + to find the length of the side opposite to

the angle θ :

( )

( )

2 2 2 2cos 2 cos cos sin

sin 1 2ctg 1 sin ctg cos .

y y y x x x x

x x x

θ θ θ θ θ

θ θ θ θ θ θ θ θ

′∆ = − = + − + ∆ − ≅

+ ∆ − ⋅ ∆ = ∆�

Or, in a more straightforward way,

( )sin sin cos cos sin sin cos

cos .

y x x x x x

y y y x

θ θ θ θ θ θ θ θ θ

θ

′ ≅ + ∆ = ∆ + ∆ ≅ + ∆ ⇒

′∆ = − ≅

The height of the weight W changes of ( )cos cos sinZ L L Lθ θ θ θ θ∆ = + ∆ − ≅ − ∆ and the weight w of

the boom (assumed applied to the centre of mass of the boom) moves of

( )2 2 2cos cos sinL L Lz θ θ θ θ θ∆ = + ∆ − ≅ − ∆ .

Then from PVW:

20 cos sin sin 0

tan .2

LT y W Z w z T x W L w

L wT W

x

θ θ θ θ θ θ

θ

∆ + ∆ + ∆ = ⇒ ⋅ ∆ − ⋅ ∆ − ⋅ ∆ ≅ ⇒

= +

y


B-3

Let L = 10 ft the length of the ladder, r = 8 ft the initial height of the rollers at the top end and suppose

that, slightly tilting the ladder leaning against the vertical wall, the rollers would go down r∆ .

Then (assuming r∆ small)

( ) ( ) ( ) ( ) ( )2 22 2 2 2 2 2 22 2 2 2

0.

L r r s s L r r r s s s r s r r s s L r r s s

r r s s

= + ∆ + + ∆ ⇒ ≅ + ∆ + + ∆ = + + ∆ + ∆ = + ∆ + ∆ ⇒

⇒ ∆ + ∆ =

Thus r

s rs

∆ = − ∆ .

Furthermore, said ,W wh the heights of the weights W and w, hangings from rungs at ,W wL feet from the

bottom end, then ,

,

W w

W w

Lh r

L= and

,

,

W w

W w

Lh r

L∆ = ∆ .

W is hung 2.5 feetl = from the top end and thereforeW

L L l= − whereas 2

w

LL = .

Let R, H and V, respectively, the force of the roller on the wall, the horizontal and vertical forces of the

ladder on the ground, then, from PVW,

( ) ( )

0 0

0 .2

W wW w

W wW w

L LrH s W h w h H r W r w r

s L L

L Lr s s LH W w H WL wL W L l w

s L L Lr Lr

∆ + ∆ + ∆ = ⇒ − ∆ + ∆ + ∆ = ⇒

− + + = ⇒ = + = − +

s

r L


Now, from PVW in the form 1

cos 0n

i i

i

F=

∆ =∑ one gets R H= and V W w= + , whence

a) 45 lb-wtsR =

b) 45 lb-wts, 90 lb-wtsH V= =

B-4

Let 3L R= the length of the plank. Then we have3

sin 602 2

L

Rα α= = ⇒ = ° .

For the heights of the weights W and w one gets that ( )2

2 cos2W

Lh R R θ= − − ,

( )coswh R R α θ= − − and (for a small tilt θ∆ )

( ) ( )2 2

2 sin 1 sin cos sin sin2 2 2

W

RL Lh R R RR

θ θ θ θ α θ θ θ θ∆ ≅ − ∆ = − ∆ = ∆ = ∆ ,

being 12

cosα = , ( )sinwh R α θ θ∆ ≅ − ∆ .

From PVW, remembering that Wh∆ and wh∆ lie in opposite directions, we have W wW h w h∆ = ∆ , whence

( ) ( ) ( )sin sin sin sin sin 30 .2 2 2

R WW wR R

αθ θ α θ θ α θ θ θ α θ θ α θ θ∆ = − ∆ = − ∆ ⇒ = − ⇒ = − ⇒ = = °

α α θ

α θ−


B-5

From PVW we have sin 30 sin 60 0W x W y∆ ° + ∆ ° = .

From the rectangular triangle shown (see problem B-3) 0 tanx

x x y y y x xy

α∆ + ∆ = ⇒ ∆ = − ∆ = − ∆ .

Then

sin30cos30

sin 30 sin 60 0 sin 30 sin 60 0 sin 30 tan sin 60 0

sin 30 tan sin 60 0 sin 30 tan cos30 0 tan tan 30 ,

W x W y x y x xα

α α α °°

∆ ° + ∆ ° = ⇒ ∆ ° + ∆ ° = ⇒ ∆ ° − ∆ ° = ⇒

° − ° = ⇒ ° − ° = ⇒ = = °

and therefore 30α = ° .

B-6

Being L the length of the string and r the radius of the sphere, then, from Carnot’s theorem,

( ) ( ) ( ) ( )22 2 cosL r L R R r R R r α α+ + ∆ = + − − − + ∆ .

x

y

R R

α

L

r

h


If α∆ is small ( ) ( ) ( ) ( )22 2 cos 2 sinL r L R R r R R r R R rα α α+ + ∆ ≅ + − − − + − ∆ .

But ( ) ( ) ( )2 22 2 cosR R r R R r L rα+ − − − = + , thus

( ) ( ) ( ) ( )( )

( )

( )( )

( )( )

( )

2

2

22 sin 1 sin

sin sin .

R R rL r L L r R R r L r

L r

R R r R R rL r L

L r L r

α α α α

α α α α

−+ + ∆ ≅ + + − ∆ ≅ + + ∆

+

− −+ + ∆ ⇒ ∆ ∆

+ +

�

�

Furthermore, being h the height of the centre of the sphere,

( ) ( ) ( ) ( )sin sin cosh R r R r R rα α α α α∆ = − + ∆ − − ≅ − ∆ .

Let F the strength of the string, from PVW: coth L r

F L W h F W WL R

α∆ +

∆ = ∆ ⇒ = ≅∆

. One also gets

49 4.5 cm=44.5cm

40 4.5 cm=44.5cm

R r

L r

− = −

+ = +

So the triangle is isosceles so that 1cot2 2

L rF W

hα

+= ⇒ ≅ . Let h the height of the isosceles, then

( ) ( )22 2 244.5 24.5 1380

2Rh L r= + − = − = and

44.5 22.250.6

2 2 1380 1380

L rF W W W W

h

+≅ = =

×� .

B-7


The centres of the spheres lie on the vertices of a regular tetrahedron of edge a and height 6

3

a , then

cos6

3θ = .

Let aw the component of the weight w along each edge a. From symmetry considerations

3 cosaw wθ× = , whence 2 ton-wts3cos 6

a

w ww

θ= = = .

Along the horizontal surface of the tetrahedron aw has a component 3

sin3

a aw wθ = ⋅ and the two

welds at the contact points of each sphere with the two other exert along the same direction a force

2 cos30 3T T° = , being T the tension of each weld.

Then from PVW (in the form 1

cos 0n

i i

i

F=

∆ =∑ ) and fixing the horizontal direction, one gets

33

3 3

aa

wT w T= ⋅ ⇒ = , and thus, for a factor of safety of 3, ( )3

3 2 ton-wtsaT T w= × = = .

B-8

θ

θ

30°


Let l the length of the cord, 1 2,h h the heights of the two masses when they lie at distances 1 2,c c from

the top in downwards direction along the two legs of the right triangle.

Then from PVW 1 1 2 2T l m h m h∆ = ∆ + ∆ where: 1

1 1 12

32 2 22

sin 30

sin 60

h c c

h c c

∆ = ∆ ° = ∆

∆ = ∆ ° = ∆

Being

1

2

cos

sin

c l

c l

α

α

=

=

one gets 11 1 2 2

sin cos sin coslc l l h lα α α α α α∆ = − ∆ + ∆ ⇒ ∆ = − ∆ + ∆ ,

3 32 2 2 2

cos sin cos sin lc l l h lα α α α α α∆ = ∆ + ∆ ⇒ ∆ = ∆ + ∆

Thus

( ) ( )3 31 11 1 2 2 1 2 1 22 2 2 2

cos sin sin cosT l m h m h m m l m m lα α α α α∆ = ∆ + ∆ = + ∆ − − ∆ .

,l α∆ ∆ being independent of one another, tacking 0l∆ = one finds that the system is in static

equilibrium when 1 2sin 3 cos 0 tan 3 3m mα α α− = ⇒ = .

Reading the tables one gets 79 .1α = ° and, taking 0α∆ = , 31

1 22 2cos79 .1 sin 79 .1 265 g-wtsT m m= ° + ° = .


B-9

Being 0L∆ = we have 0 cot 30x

x x y y y x xy

∆ + ∆ = ⇒ ∆ = − ∆ = − °∆ .

But ( )cos30 cos30 cos30x r s r s r r′ ′∆ = ° + ∆ − ° = ∆ − − ° .

Assuming s∆ small we have cos30r r s′− ≅ ∆ ° .

Thus ( )2 21 cos 30 sin 30x s s∆ ≅ ∆ − ° = ∆ ° .

a) from PVW

2cot 30 cot 30 sin 300

3cos30 sin 30 .

4

y x sT s W y T W W W

s s s

T W W

∆ − °∆ °⋅ °∆ ∆ + ∆ = ⇒ = − = − = ⇒

∆ ∆ ∆

= ° ° =

s∆

x∆

y∆

L L

r

r’


b)

Using the method of the force components (FC) on cart we have:

3cos180 cos300 0

2T N T N° + ° = ⇒ = (horizontal).

Using FC on W: 1

cos0 cos 240 02

N W N W° + ° = ⇒ = .

So 3

4T W= .

B-10

Assume the centre of the bobbin goes down H∆ and m rises h∆ as the bobbin rolls downwards. Then

from PVW: M H m h∆ = ∆ .

horizontal

N

N

T

normal


For a rolling of the bobbin of L R α∆ = ∆ then M goes down sinH L θ∆ = ∆ , whereas m rises (assuming

α∆ small) r

l r LR

α∆ ≅ ∆ = ∆ and goes down H∆ .

Thus

( )

sin sin sin

sin sin sin sin 0,5

30 .

r rh l H L M L m L

R R

r r m rM m m M m m

R R M m R

θ θ θ

θ θ θ θ

θ

∆ = ∆ − ∆ = ∆ − ⇒ ∆ = ∆ − ⇒

= − ⇒ + = ⇒ = = ⇒+

= °

B-11

If the chain rises h∆ resting in a horizontal circle, then its radius varies ofr

x hh

∆ = ∆ and its length

of 2l xπ∆ = ∆ .

From PVW: 2 22

r WhW h T l W h T x T h T

h rπ π

π∆ = ∆ ⇒ ∆ = ⋅ ∆ = ⋅ ∆ ⇒ = .

x∆

h∆


B-12

Assuming w rises wh∆ whereas W goes down Wh∆ , being:

( )1 2 3

0W W Wl l l∆ + + = ,1 2 1 3

12W W W Wl l l l∆ = ∆ ⇒ ∆ = − ∆ , ( )

1 20w wl l∆ + = ,

3 1 2W w wl l l∆ = −∆ = ∆ ,

then 2 3 2

2w w W wh l l l∆ = ∆ + ∆ = ∆ , 1 3 2

1 1 12 2 4

sin sin sin sinW W W w wh l l l hθ θ θ θ∆ = ∆ = − ∆ = − ∆ = − ∆ .

From PVW: 0w Ww h W h∆ + ∆ = . So 14

4sin 0

sinw w

ww h W h Wθ

θ∆ − ⋅ ∆ = ⇒ = .

B-13

1Wl

2Wl

1wl 2wl

3Wl

y x

EF

FG

EG

FG

DF


From PVW: 1 2F x F y= , but 2 1: 2 :1 2x y F F= ⇒ = .

Using cos 0i i

i

F ∆ =∑ then, with respect to the vertical direction, one gets

1 2 1 2 1 10 33

WF W F F F W F W F− + = ⇒ + = ⇒ = ⇒ = , 2

2

3

WF = .

Using FC at the joint G:

. vertical components: 22 2

23cos 0 cos150 0 0

2 3FG FG FG

FF T F T T° + ° = ⇒ − = ⇒ = .

Using FC at the joint F:

- horizontal components: ( )12

cos120 cos 240 cos0 0FG EF FG EFT T F F T T° + ° + ° = ⇒ = +

- vertical components: ( )3

cos30 cos150 0 002

FG EF FG EF EF FGT T T T T T° + ° = ⇒ − ⇒ =

whence 22 4

3 3 3FG

F WF T= = = .


B-14

a) members AB, BD, DF, FG should be rigid in order that A remains in fixed position; CD also

rigid to fix C and D. The elements that could be flexible are therefore AC, CE, EG, BC, EF, ED.

b) meanwhile note that 35

ˆcosA = , 45

ˆsin A= . From the result of the preceding problem is then

13

WF = , 2

2

3

WF = .

Using FC at the joint A:

- horizontal: ÂBcos A=AC

- vertical: 1

5ÂBsin A= AB3 12

W WF = ⇒ =

Using FC at the joint B:

- horizontal: ( ) ÂB+BC cos A=BD


- vertical: ÂB=BC BD=2ABcosA=2

W⇒

Using FC at the joint C:

- vertical: BC CD=

Using FC at the joint D:

- vertical: 5

DE CD DE=CD=BC=AB=12

W= ⇒

So BD=2

W ,

5DE=

12

W .

B-15

At its maximum swing the pendulum has zero velocity. Then from the conservation of energy

0 wW w

W

hW h w h W w

h

∆∆ + ∆ = ⇒ = −

∆. At the maximum swing (pendulum at the ceiling) 3wh∆ = ,

4Wh∆ = − . Thus 34

W w= .


B-16

As the third mass reaches the table top, the two equal masses rise ( )50 2 1 cm− ; when the third mass

has descended a distance h∆ the two equal masses have risen by

2

50 1 150

h ∆ + −

.

Then, said V and v the final velocities (in module) of the third mass and of the two equal masses

respectively, 2

Vv = .

The kinetic energy varies of ( ) ( )2 2 2 2 231 12 2 2

2 2m V mv m V v mV+ ⋅ = + = and the potential energy of

( ) ( ) ( ) ( )2 50 2 1 2 50 2 50 2 2mg mg mg⋅ − − ⋅ = − ⋅ − .

From the conservation of energy: ( )( )

232

200 2 22 50 2 2

3

gmV mg V

−= ⋅ − ⇒ = .

With 2981 cm sg −= thus one obtains 1196 cm sV−= .


B-17

Let’s consider a small layer of liquid of thickness z∆ a distance h up from the hole.

Its potential energy of position is Ag z hρ ∆ ⋅ .

The internal tension force due to the weight of the overlying liquid, ( )Ag H hρ − , corresponds to a

pressure ( )p g H hρ= − and an internal energy ( )A Ap z g H h zρ∆ = − ∆ .

The total potential energy is the sum of the position energy and of the internal energy:

( )A A Ag z h g H h z g H zρ ρ ρ∆ ⋅ + − ∆ = ∆ (independent of h, and hence also valid for 0h =

corresponding to the hole position).

Soon as the hole is open the above energy is converted to kinetic energy, so

( ) 212

A A z 2g H z v v gHρ ρ∆ = ∆ ⇒ = .


C-1

60 90 30

60 30

β θ β θ

α β θ

+ ° + = °⇒ = ° −

= ° − = ° +

From FC:

- vert. cos cosT X Wα β+ =

- horiz. sin sinT Xα β=

- obliq. sin cos 602

TW Tθ = ° =

(Note that in the shown arrangement ( )limit2 sinT W θ= ).

In terms of cot β :

cos sin cot sin cot cos sin cotX X T T T Wβ β β α β α α β= = ⇒ + = .

W

W

W

T

T

X

α

β

60° θ


But

( )

( )

( ) ( ) ( ) ( )

3 12 2

312 2

cos cos 30 cos30 cos sin 30 sin cos sin

sin sin 30 sin 30 cos cos30 sin cos sin

cot cot 30 cot 1 cot cot 30 3 cot 1 cot 3

α θ θ θ θ θ

α θ θ θ θ θ

β θ θ θ θ

= ° + = ° − ° = −

= ° + = ° + ° = +

= ° + − ° = + −

Thus we have

( ) ( )3 31 12 2 2 2

3 cot 1cos sin cot cos sin cos sin

2sincot 3

TT W T W

θα α β θ θ θ θ

θθ

++ = ⇒ − + + ⋅ = =

−

( )2 2 3 cot 13 cos sin sin cos sin 3 sin 1

cot 3

θθ θ θ θ θ θ

θ

+⇒ − + + ⋅ =

−.

In terms of cotθ :

2

2

2

cotcos sin

1 cot

1sin

1 cot

θθ θ

θ

θθ

=+

=+

Thus 2 2

3 cot 1 cot 3 3 cot 1 2 3 11 1 cot 3 3 tan

1 cot 1 cot cot 3 cot 3 3 3

θ θ θθ θ

θ θ θ θ

− + ++ ⋅ = ⇒ = ⇒ = ⇒ =

+ + − − and

finally 1 1tan 10 ,9

3 3θ −= = ° .


C-2

As w moves of a (small) distance wh∆ the spool rotates an angle wh

rα

∆∆ ≅ and W moves in the same

direction by the same distance wh∆ due to the vertical displacement of the spool, and in the opposite

direction of a distance Wh R α∆ ≅ ∆ due to the rotation of the spool.

Then from PVW:

( ) ( ) ( ) ( )w W ww h W h h W R r W R r wr W R rα α α α α∆ = ∆ − ∆ = ∆ − ∆ = − ∆ ⇒ ∆ = − ∆ , whence wr

WR r

=−

.


C-3

Define B-A

tan9 m

t θ′ ′= = , A-C

tan9 m

t θ′′ ′′= = and 34.8 10 kg-wts

6w

×= .

From FC at the ends of B, A, C:

B)

Horiz. 1. 2

1

2 1

TT

t′=

′+

Vert. 2. 22 1

T tT w

t

′′= +

′+

A)

θ ′

θ ′′

T

T ′ T ′′ T ′′′

C


Horiz. 3. 2 2

1 1

1 1T T

t t′ ′′=

′ ′′+ +

Vert. 4. 2 21 1

t tT T w

t t

′ ′′′ ′′= +

′ ′′+ +

C)

Horiz. 5. 2

1

1T T

t′′ ′′′=

′′+

Vert. 6. 21

tT w

t

′′′′ =

′′+

From 3. and 4.: ( )

2 2 21 1 1

t tt tT T w T w

t t t

′ ′′−′ ′′′′ ′′ ′′= + ⇒ =

′′ ′′ ′′+ + +.

From 6. and the latter: ( )

2 21 1

t ttT T

t t

′ ′′−′′′′ ′′=

′′ ′′+ +.

Thus 2

tt t t t

′′′ ′ ′′ ′′= − ⇒ = .

From 6. and 4.: 2

21

tT w

t

′′ =

′+, whence from 2. 3

2

Tw= .

Finally, from 1., 2 13 3

3 2wt w t t′ ′ ′′= ⇒ = ⇒ = . Thus B-A 2 A-C 1

, .9 m 3 9 m 3

= =

But C = 2.00 m, so A=5 m, B=11 m.

To determine Tmax let’s return back to 1. – 6.:

1’. 3

2 13

T T ′= , whence T T ′>

2’. 2

2 13

T Tw

′= +

3’. 3 3

13 10

T T′ ′′= , whence T T′ ′′>


4’. 2

213 10

T Tw w

′ ′′= + = using 6’. below, from which 13T w′ =

5’. 3

10

TT

′′′′′= , whence T T′′ ′′′>

6’. 10

Tw

′′=

Finally, therefore, 3

3

max

3 2 4.8 10 kg-wts3 2 34 10 kg-wts

13 2T T T w

×′= = = = ≈ × .

C-4

FA


a) Given the condition 0BF =��

, from PVW (in the form 0i i

i

W =∑ � ), with respect to A, one gets

LFh WL F W

h= ⇒ = .

b) From FC:

h. cos

v. sin

A

A

F F

W F

θ

θ

=

=

��

��

whence 1tan tanW h h

F L Lθ θ −

= = ⇒ =

and

2 2

211 tan 1 1

cosA

L L L h LF W W W W

h h h L hθ

θ

= ⋅ = + = + = +

��.

1



Pier F. Nali


CHAPTER 3

Kepler’s Laws and Gravitation


x=0 x=a a

b

In general, for a curve having parametric equations ( ) ( ),x t y tϕ ψ= = , the curvilinear trapezoid area

in the figure below

where ( ) ( )1 2,a t b tϕ ϕ= = and ( ) 0tψ ≥ in the interval ( )1 2,t t , is ( ) ( )2

1

A

t

t

t t dtψ ϕ′= ⋅∫ .

We can use this formula to calculate the area of an ellipse having parametric equations

cos , sinx a y bθ θ= = .

From the symmetry of the ellipse we can calculate the area of a quarter ellipse between the limits

( )20 ,x πϑ= = ( )0x a ϑ= = .

Hence ( )2

2

0

2

4

0

Asin sin sin A= ab

4b a d ab d ab

π

π

πθ θ θ θ θ π= ⋅ − = = ⇒∫ ∫ .

A-1

b x

a


For the moon ( )12

405,500 km + 363,300 km 384,500 km

2a pa r r= + = =

For the Earth sat.

( ) ( ) ( )1 1 12 2 2

12,756 km + 935 km710 km + 225 km 935 km 6,845.5 km

2a pa r r r r d⊕ ⊕ ⊕= + = + + = + = =

Thus

3 32 26,845.5 km

27 .322 0 .065 1.6 hr384,500 km

d dat t

a

= = × ≈ ≈

A-2

From the Kepler’s second law 1 1 0.0167 1.0167

1.0331 1 0.0167 0.9833

p aa a p p

a p

v r ea cr v r v

v r a c e

⊕

⊕

++ += ⇒ = = = = = =

− − −.

A-3


2 2 2

2

2 2

2

1.000 6,378 km, 9.80 m s

81.3 1,738 km

1.6 m s6

M R RM g M Mg G g G g g

R R g M R M R

g

−⊕ ⊕ ⊕⊕ ⊕

⊕ ⊕ ⊕ ⊕

− ⊕

= = ⇒ = ⇒ = = × × =

�

A-4

a) From the Kepler’s third law

23

2 3

comet comet cometcomet

T a Ta a

T a T⊕

⊕ ⊕ ⊕

= ⇒ = ⋅

. Thus

( )( )

23

23

comet

1986 1456 / 71 A.U. 75.7 A.U. 17.9 A.U.

1 yra

−= × ≈ ≈

and

comet cometcomet2 2 17.9 A.U. - 0.60 A.U. 35.8 A.U. - 0.60 A.U. 35.2 A.U.a pr a r= − ≈ × ≈ ∼

b) From the Kepler’s second law comet

comet comet

comet

maxmax min

min

35.2 A.U.59

0.60 A.U.

a

p a

p

rvv r v r

v r= ⇒ = = ∼ .


A-5

From the Kepler’s third law

( )

2 23 3

23

2 3

24 h 24 h 24 h24 h

100 min 100 min

24 60 min14.4 5.9

100 minE E E E

E

T R TR R R R R

T R T

× = ⇒ = ⋅ = ⋅ = ⋅

∼ .

B-1


a) The geographic latitude of P is restricted to 0λ = (equator). Although a synchronous orbit

doesn’t need to be equatorial, a body in a synchronous non equatorial orbit could not remain in a

fixed position with respect to P: it would appear to oscillate between the north and south of the

Earth's equator. Moreover, a body in a synchronous non circular (elliptical) orbit would appear

to oscillate East-and-West. The combination of the two latter (synchronous non-circular non-

equatorial) would appear as a path in the shape of eight.

b) From the Kepler’s third law

3 22 31 1

( 1 day)= ( 27 days)27 9

ss s

rt T r r r

r⊕ ⊕

⊕

= = ⇒ = =

.

B-2

a) From the Kepler’s third law

222 2 2

3 3

3

4 4,

a MTT T

a GM a GM T a M

π π ⊕⊕

⊕ ⊕ ⊕ ⊕ ⊕ ⊕

= = ⇒ ⋅ =

� �

� �

.

From Earth and Moon data:

11 11

8 11

1.47 10 m, 1.52 10 m, 365 .242

3.63 10 m, 4.06 10 m, 27 .322

d

P A

d

P A

r r T

r r T

⊕ ⊕ ⊕= × = × =

= × = × =


( )

( )

1112

812

1.50 10 m

3.84 10 m

P A

P A

a r r

a r r

⊕ ⊕⊕ = + = ×

= + = ×

Thus

2 311

5

8

27 .322 1.50 10 m3.33 10

365 .242 3.84 10 m

d

d

M

M⊕

×= ⋅ = ×

×

�

b)

22 22 2

3 3

3

4 4,

T T T a M

a GM a GM T a M

π π

⊕ ⊕ ⊕ ⊕

= = ⇒ ⋅ =

421,800 km, 384,000km, 1 .769da a T⊕= = =

Thus

2 3

27 .322 421,800 km318

1 .769 384,000 km

d

d

M

M⊕

= ⋅ =

.

B-3

For elliptic orbits 2 3

2

4a b

Rm m

GT

π+ = .

For the Earth-Sun system ( )( )( )

324 1 A.U.

1 yrm m m

G

π⊕+ ≅ =

� �.


Thus 3

2a b

Rm m m

T+ =

� with R expressed in A.U. and T in years.

B-4

II: unchanged.

The Kepler’s second low says that the areal velocity is constant, i.e., said d� the arc length of the

trajectory of m and R the distance from M :

( )A 1 1 1

R R R const.2 2 2 2

d d Lv mv

dt dt m m= = = = =

�, being L the orbital angular momentum of m .

But L remains constant for central forces, so the areal velocity also doesn’t change. Indeed,

( ) ( ) ( )3 30

a a

d L GMm GMmR F R R R R

dt R R+ +

= × = × − = − × =

��

, being 0R R× ≡��

.

Thus also L cost. and Ad

dtso well.

III: For a circular orbit 2F m Rω= −��

, where 2

T

πω = . Hence

( ) ( ) ( )( )

2 232 2 2

23 3 3

4 4 a

a a a

GMm GM GMR m R T R

T GMR R R

π πω ω +

+ + +− = − ⇒ = ⇒ = ⇒ =

�� .

d� R


C-1

Ignoring the moon’s gravitation one can assume2

GMg

R

⊕

⊕

= .

Moon at zenith: ( )

2z

GMg g

R R⊕ ⊕

= −−

, at nadir:( )

2n

GMg g

R R⊕ ⊕

= ++

.

Thus 2 2

1 1

1 1

n zg gg M

g g M R R

R R

⊕ ⊕ ⊕

⊕ ⊕

−∆ = = +

+ −

.

But 8 8 8

6 6 6

3.84 10 m 3.90 10 m 3.78 10 m, 1 , 1 ,

6.38 10 m 6.38 10 m 6.38 10 m

R R R

R R R

⊕ ⊕ ⊕

⊕ ⊕ ⊕

× × ×= + = − =

× × ×

22

24

7.34 10 kg

5.98 10 kg

M

M⊕

×=

×, so

( )2 2

6 6 6

6

7.34 6.38 6.3810 1.23 2.68 2.85 10 1.23 5.53 10

5.98 3.90 3.78

(approx.) 7 10 .

g

g

− − −

−

∆ = × + × = × + × = × × =

= ×


C-2

32 3 22 3

2 2

44,

a Ta aM M M

GT GT a T

ππ⊕

⊕

⊕

⊕

= = ⇒ = ⋅

�

�

�

.

Being T in days and V in km s-1

,

[ ]

[ ]

[ ][ ]

2 km 2 km 86, 400

86, 400 s 2secondss days

day

a aV a TV

TT T

π π

π= = ⇒ =

× = ×

.

Thus ( ) ( )3

2 3 7 38

86, 400 km 365 days 1.02 102 1.50 10 km

M TV M TV Mπ

−⊕= × × ⋅ = ×

× × �.


C-3

For the ellipse the radius of curvature is ( )

32

p a

c

R RR

ab= ,

where ( ) ( ) 21 > 1 , 1 .a pR a e R a e b a e= + = − = −

Thus ( ) ( )1 1c p aR R e R e= + = − and ( )1 1

1 cosc

er R

ϕ= − .


a) From the conservation of energy: 2 21 1

2 2a

p a

GM GMv v

R R− = −� � (simplifying the comet mass),

where v is the velocity at perihelion, and from the Kepler’s second law p a avR v R= .

Thus ( )2 21 1 12

2a

p a c

GMv v GM e

R R R

− = − =

�

�,

1 12 , 2

1 1 1

p

a a a

a

R e ev v v v v v v v v

R e e e

−= = ⇒ − = + =

+ + + ,

so ( )( )

2 2 2

2

12 2

2 1a

c

GMev v v e

Re− = =

+

� , whence

( ) ( )222 1 6

6

2 11 3 2

500.0 km s 1.00 10 km1.88 10 km

1.33 10 km s

pcc c

p

vRGM RR R

v R GM

−

−

× × = ⋅ ⇒ = = = × ×

�

�

.

b) 6 6

61.00 10 km 10 km1 1, 8.33 10 km

1.881 0.1222

1.00

p pc c

cp p

p

R RR Re e a

RR R e

R

×+ = ⇒ = − = = = = = ×

− −−

.

c) From the Kepler’s third law 2 3

2 4 2c c

a a aT T

GM GM

π π= ⇒ =

� �

.

From a)

6 6 6

1 6

2 2 8.33 10 km 8.33 10 km 1.88 10 km414,000 s

500.0 Km s 1.00 10 km

p c

c

pc

vR a aRGM T

vRR

π π−

× × × × ×= ⇒ = =

× ×�

�

[ ] [ ] [ ]( )

1

1

414,000 sdays secondsdays s s 4.8 days

s day 86, 400 s dayc c c

T T T

−

−

= × = × = ≈

�.

Documents

Solutions to problems in Leighton and Vogt's "Exercises in Introductory Physics" textbook