Physics 201: Lecture 25, Pg 1 Lecture 25 Jupiter and 4 of its moons under the influence of gravity Goal: To use Newton’s theory of gravity to analyze the

Physics 201: Lecture 25, Pg 1

Lecture 25

Jupiter and 4 of its moons under the influence of gravity

Goal: To use Newton’s theory of gravity to analyze the motion of satellites and planets.


Exam results

Mean/Median 59/100 Nominal exam curve

A 81-100

AB 71-80

B 61-70

BC 51-60

C 31-50

D 25-30

F below 25


Newton proposed that every object in the universe attracts every other object.The Law: Any pair of objects in the Universe attract each other with a force that is proportional to the products of their masses and inversely proportional to the square of their distance

Newton’s Universal “Law” of Gravity


Newton’s Universal “Law” of Gravity

2by1on12221

1by 2on r̂ Fr

mmGF

“Big” G is the Universal Gravitational Constant

G = 6.673 x 10-11 Nm2 / kg2

Two 1 Kg particles, 1 meter away Fg= 6.67 x 10-11 N(About 39 orders of magnitude weaker than the electromagnetic force.)

The force points along the line connecting the two objects.


Suppose an object of mass m is on the surface of a planet of mass M and radius R. The local gravitational force may be written as

Little g and Big G

where we have used a local constant acceleration:

On Earth, near sea level, it can be shown that gsurface ≈ 9.8 m/s2.

surfacemgFG

2surface RGM

g

mMMm R

mGMF ,2by on r̂


g varies with altitude

With respect to the Earth’s surface

2)/()( hRGMhg EE


At what distance are the Earth’s and moon’s gravitational forces equal

Earth to moon distance = 3.8 x 108 m Earth’s mass = 6.0 x 1024 kg Moon’s mass = 7.4 x 1022 kg

22 // MMEE rGMrGM

EMEM MMrr // 22

11.0// EMEM MMrr

m108.3 8 EM rr

m104.3m/1.11 108.3 88 Er


The gravitational “field”

To quantify this invisible force (action at a distance) even when there is no second mass we introduce a construct called a gravitational “force” field.

The presence of a Gravitational Field is indicated by a field vector representation with both a magnitude and a direction.

r̂2r

GMg

M


Compare and contrast Old representation r̂gg

M

New representation

r̂2r

GMg

r̂)( 2hR

GMg

E

222 )1(

1)0(

)/1(

1)(

xg

RhR

GMhg

EE

...

23

21)0()(2

EE Rh

Rh

ghg


When two isolated masses m1 and m2 interact over large distances, they have a gravitational potential energy of

Gravitational Potential Energy

The “zero” of potential energy occurs at r = ∞, where the force goes to zero. Note that this equation gives the potential energy of masses m1

and m2 when their centers are separated by a distance r.

if

rrr G r

mGmrmGm

rdFW f

i

21211 )( 2

Recall for 1D: dxxdUxF /)()( f

i

xx dxxFW )(

rmGmrU /)( 21


A plot of the gravitational potential energy

Ug= 0 at r = ∞ We use infinity as reference

point for Ug(∞)= 0 This very different than U(r) = mg(r-r0) Referencing infinity has

some clear advantages All r allow for stable orbits

but if the total mechanical energy, K+ U > 0, then the object will “escape”.

rmGmrU /)( 21


It is the same physics

Shifting the reference point to RE. hRGmM

hRUE

EE

)(

E

E

E

EEE R

GmM

hR

GmMRUhRU

)()(

)(

)()(21 hRR

hRhRR

RmGm

EE

E

EE

E

hR

GmMhRR

hGmM

E

E

EE

E 22

mghhR

GMm

E

E 2


Gravitational Potential Energy with many masses

If there are multiple particles231312Total UUUU

||||||

)(32

32

31

31

21

21

rrmm

rrmm

rrmm

GrU

Neglects the potential energy of the many mass elements that make up m1, m2 & m3 (aka the “self energy”


Dynamics of satellites in circular orbits

Circular orbit of mass m and radius r Force: FG= G Mm/r2 =mv2/r Speed: v2 = G M/r (independent of m) Kinetic energy: K = ½ mv2 = ½ GMm/r Potential energy UG = - GMm/r

UG = -2 K

This is a general result

Total Mechanical Energy: E = K + UG = ½ GMm/r -GMm/r

E = - ½ GMm/r


Changing orbits

With man-made objects we need to change the orbit

Examples:

1. Low Earth orbit to geosynchronous orbit

2. Achieve escape velocity 1. We must increase the potential energy but one can

decrease the kinetic energy consistent with

rGMmrU )(G r

GMmrUrK G 2)()( 2

1

rGMm

rKrUrE2

)()()( G


Changing orbits Low Earth orbit to geosynchronous orbit

rGMm

rKrUrE2

)()()( G

A 470 kg satellite, initially at a low Earth orbit of hi = 280 km is to be boosted to geosynchronous orbit hf = 35800 km

What is the minimum energy required to do so?

(rE = 6400 km, M = 6 x 1024 kg

if

if rrGMmrErEE

2

1

2

1)()(

2/107.6

1102.41

470)106(1067.667

2411

E

J 102.1 10E


Escaping Earth orbit Exercise: suppose an object of mass m is

projected vertically upwards from the Earth’s surface at an initial speed v, how high can it go ? (Ignore Earth’s rotation)

221

G )( iEi mvRUE

0)(G hRUE Ef

0)(G hRUEE Eif

221/)/( iEE mvRGmMhRGmM )/(/1)/(1 2

21 GmMmvRhR iEE




)/(/1)/(1 221 GmMmvRhR iEE

)2/()( 22 GMvhRRh iEE

)2/())2/(1( 222 GMvRGMvRh iEiE

)2( 2

22

iE

iE

vRGMvR

h




)2( 2

22

iE

iE

vRGMvR

h

02 2 iEvRGM implies infinite height

km/s 2.112

Escape ER

GMv


Some interesting numbers

First Astronautical Speed Low Earth orbit: v = 7.9 km/s

Second Astronautical Speed Escaping the Earth: v = 11 km/s

Third Astronautical Speed Escaping Solar system: v= 42 km/s


Next time

Chapter 14 sections 1 to 3, Fluids

Documents

Physics 201: Lecture 25, Pg 1 Lecture 25 Jupiter and 4 of its moons under the influence of gravity Goal: To use Newton’s theory of gravity to analyze the