Newtonian Gravitation

Newton realized that the force which holds the moon

Newton's Law of Gravitation

it its orbit is of the same nature with the force that makes

an apple drop near the surface of the earth. Newton concluded

that the earth attracts both apples as well as the moon but also

that every object in the universe attracts every other object.

The tendency of objects to move towards each other is known

as gravitation.

m1

m2

Newton formulated a force law known as

Every particle any other particle with a gravitational force

Newton's law of gravitat

attracts

ion.

Every particle any other particle with a gravitational force

that has the following characteristics:

attracts

1.

1 2

1 2

2

The force acts along the line that connects the two particles

2. Its magnitude is given by the equation:

Here and are the masses of the two particles, is their

separation and i

m m

F G

m m r

G

r=

11 2 2

s the gravitational constant. Its value is:

6.67 10 N.m /G kg−= ×

m1

m2

F12

F21

r

12 1 2

21

2 1

12 21

The gravitational force exerted on by

is equal in magnitude to the force exerted

on by but opposite in direction. The two forces

obey Newton's third law: 0

F m m

F

F

m m

F+ =� �

1 21 2

Newton proved that a uniform shell attarcts a particle

that is outside the shell as if the shell's mass were

concetrated at the shell center mm

F Gr

=

m1

m2r

F1

Note: If the particle is inside the shell, the

r

net force is zero

Consider the force F the earth (radius R, mass M) exerts

on an apple of mass m. The earth can be thought of as consisting

of concentric shells. Thus from the apple's point of view

the earth behaves li

2

ke a point mass at the earth center

The magnitude of the force is given by the equation:

mMF G

R=

m2

m1

The net gravitational force exerted by a group of

particles is equal to the vect

Gravitat

or sum o

ion and the Principle of Su

f the contribution

from each

perposition

particle.

1 1 2 3

12 13 1

1 12 13

2 3

1

For example the net force exerted on by and is equal to:

Here and are the forces exerted on by and , respectively.

In general the force exerted

on

FF m m m

F F m m m

F

m

F= +� �

� � ��

1 12 13 14 1 1

2

by n particles is given by the equation:

...n

n i

i

F F F F F F=

= + + + + =∑� � � � � �

m1

dm

r

dF�

The gravitation force exerted by a continuous extended

1object on a particle of mass can be calculated using the

principle of superposition. The object is divided into elements

of mass . the net f

m

dm 1

1 1

1

2

orce on is the vector sum of the forces

exerted by each element. The sum takes the form of an integral

Here is the force exerted on by

m

F dF dF m dm

Gm dmdF

r

=

=

∫� � �

( )( )( )

( )

( )

( )11 2 2 11

g 2 2

5.00 kg 10.0 kg6.673 10 N m /kg 0.100 kg 2.32 10

0.400 m 0.600 mxF

− −

= × ⋅ − + = − × Ν

with the minus sign indicating a net force to the left.

The force from the 5.00 kg sphere is greater than for the 10.0 kg sphere even

though its mass is less, because r is smaller for this mass.

(b) No, the force found in part (a) is the net force due to the other two spheres.

If we assume that the earth is a sphere of mass , the magnitude

of the force exerted by the earth on an object of mass

Gravitation near the e

placed at a distance

arth's Su

from t

rf

he

ace

center

M

F m

r2

of the earth is: GMm

Fr

=r

2

The gravitational force results in an acceleration known as

. Using New

gravitational

acceleration

free fall accele

ton's second law we have that

Up to this point we have assuned that the rati

g

g

a

GMa

r=

�

near the

surface of the earth is constant. However if we measure g at various points

on the surface of the earth we find that its value is not constant.

This is attributed to three reas

on

ons:

g

1. The earth's mass is not unirormly distributed

. Earth is approximately an ellipsoid flattened at

the poles and bulging at the equator. Its equatorial radius is larger than the polar

radius

2. The eart

by 21 km.

h is n

Thus t

ot a sphe

he va

re

lue of g at sea level increases as one goes from the

equator to the poles

The density of the earth varies radially as shown in the figure

The density of the outer section varies f

1. The earth's mass is not

rom region to region

over t

unirormly di

he earth's su

stribu

rfa

ted

ce. Thus g varies from point to point

2

Consider the crate shown in the figure.

The crate is resting on a scale at a point on the equator

The net force along the y-a

3. The earth

xis

Here and is t

is rotating.

ynet g N g N

g N

F F F ma F

GMa F mg

R

= − = −

= =

2

he normal force exerted

on the crate by the scale. The crate has an acceleration

due to the rotation of the earth about its acis

every 24 hours. If we apply Newton's second law we get:

a Rω=

y-axis every 24 hours. If we apply Newton's second law we get:

gma mg m− = 22 2

ggR mg ma gm a RRω ω ω=− → −→ =

y-axis

2 2 2

Free fall acceleration = gravitational acceleration - centripetal acce

The term 0.034 m/s which is much smaller than

le

9.

rat

8

ion

m/sRω =

m1

m2r

F1

m2

m1

Newton proved that the net gravitational force on a particle by a shell depends

on the postion of the particle with respect to the shell

If the pa

Gravitation

rticle is i

insid

nside

e the ear

the s

th

hell, the net force is zero

1 21 2

If the particle is outside the shell the force is given by:

Consider a mass inside the earth at a distance from the center of the earth

If we divide the earth in a

m r

m mF G

r=

2

series of concentric shells, only the shells with

radius less than exert a force on . The net force on is:

Here is the mass of the part of the earth inside a sphere os radius

ins

ins

i

GmMr m m F

r

M r

M

=

34 4 is linear with r

3 3ns ins

r GmV F r F

π π ρρ ρ= = → =

In chapter 8 we derived the potential energy of a mass

near the surface of the earth. W e will remove this restriction

and assume that t

Gravitational P

he mass can

otential E

move away

ne

fr

rgy

om th

U m

m e surface

of the earth, at a distance from the center of the earth as shown

in the figure. In this case the gravitational potential energy is:

The negative sign of expresses the fact thGmM

Ur

r

U= − at

the corresponding gravitational force is attractive

r

m

The gravitational potential energy is not only associated

with the mass but with as well i.e. with

N

ot

both o

e

b

:

s

jectm M

GmM

Ur

= −

1 2 3

1 3 2 31 2

12 13 23

If we have three masses , , and positioned as shown in the figure

the potential energy due to the gravitational forces among the objects is:

We take into

m m m

U

Gm m Gm mGm mU

r r r

= − + +

account each pair once

Gravitational potential energy of the earth and a mass

placed at a distance from the center of the earth. We

assume that we move the mass upwards so that it reaches

a great distance (practica

U m

R

m

2

lly infinite) from the earth.

The potential energy is equal to the work

that the gravitational force does on .

1( )

We chose an integration path that po

RR R

U W

m

GMm GMmU W F r dr dr GMm

r r R

∞∞ ∞

= = ⋅ = − = = − ∫ ∫�

�

ints radially outwardsWe chose an integration path that points radially outwards

we are allowed to do this because the gravitational force is

conservative. This can also be seen in the figure to the left.

In this case we follow a different path of arbotrary shape.

We then break the path into radial sections (AB, CD, and EF)

and tangential sections (BC, DE, and FG). The work in the

latter is equal to zero because the force is at right angles to

the displa2

cement. Thus we are left with the sum R

GMmdr

r

∞

−∫

m

r = ∞

B

v = 0

If a projectile of mass is fired upward

at point A as shown in the figure, the projectile

will stop momentarily and return t

Escape

o the e

Spe

.

ed

arth

m

There is however a minimum initial speed for which the projectile

will escape from the gravitational pull of the earth and will stop

at infinity (point B in the figure). This minimum speed is known as

m

v

A

at infinity (point B in the figure). This minimum speed is known as

es

2

2

We can determine the escape velocity using

energy conservation between point A and point B.

cape velo

02

02

The escape speed form the earth

2

city.

is

A B

A B

mv GMmE K U E K U

R

mv GMmE E

R

GMv

R

= + = − = + =

= → − = → =

11.2 km/s

The escape speed does depeNote: nd not on m

.

The orbits are described by two

pa

All planets move o

rameters: The sem

n elliptical orbits

with the sun at o

imajor axis and the eccentricity

Kepler's F

The orb

ne focus

it in th

irst

e fig

Law

.

ure h

a e

as = 0.74. The actual eccentricity

of the earth's orbit is only 0.0167

e

T he line that connects a p lanet

to the sun sw eeps ou t equal areas

in the p lane of the orb it in equal

K epler's S econd Law

A∆

dA

in the p lane of the orb it in equal

tim e in tervals . cons at n t

A

dAt

d t

∆

∆ =

2

2 2

2

1The area swept out by the planet (fig.a ) is given by the equation:

2

1 1 where is the planet's angualar speed. The planet's angular

2 2

momentum

A A r

dA dr r

dt dt

dAL rp rmv rm r mr

dt

θ

θω ω

ω ω⊥ ⊥

∆ ∆ ≈ ∆

= =

= = = = → = Kepler's second law2

is equivalent to the law of conservation of angular momentum

L

m

2. An imaginary line drawn from each planet to the Sun

sweeps out equal areas in equal times.

Kepler’s second law. The two shaded regions have equal areas. Theplanet moves from point 1 to point 2 in the same time as it takes to movefrom point 3 to point 4. Planets move fastest in that part of their orbitwhere they are closest to the Sun.

The square of the period of any planet is proportional to

the cube of the semimajor For the sake

of simplicity we will consider

Keple

the c

r's Third

irculr orb

axis

it sh

la

ow

of

n i

its

n

t

or

he

w

bit.

figure.

( )( )2 2 2

A planet of mass moves on a circular orbit of radius around a star of mass .

We apply Newton's second law to the motion:

(eqs.1) The period can be expressed

m r M

GMm GMF ma m r m r Tω ω ω= = = = → =( )( )2 2 2

2 3 (eqs.1) The period can be expressed

in

gF ma m r m r Tr r

ω ω ω= = = = → =

22

2

2

2

3

2 2

3

2 4terms of the angular speed . = (eqs.2)

If we substitute form eqs.1 into eqs.2 we get:

: The ratio does

4

not deNote pend on the mass of the

1

p

T

r M

T

m

G

T

T

r

π πω

ω

π

ω

ω

= →

=

2

3Note 2:

lanet but only on the mass of the central star.

For elliptical orbits the ratio remains cons a t nt

M

T

a

Kepler’s third law:

2 2

3

4

T

r MG

π=

3. The square of a planet’s orbital period is proportional to the

cube of its mean distance from the Sun.

Kepler’s third law:

The ratio of the squares of the periods of any two planets revolvingabout the Sun is equal to the ratio of the cubes of their semimajoraxes.

3 / 2 3 / 2

22 1

48,000 km(6.39 days) 24.5 days

rT T

= = = 2 1

1

(6.39 days) 24.5 days19,600 km

T Tr

= = =

3 / 2

2

64,000 km(6.39 days) 37.7 days

19,600 kmT

= =

For the other satellite

T increases when r increases.

2

g S H/ ,F Gm m r= 2

rad /a v r= 2

g H / ,F F m v r= =∑2 2

S H H/ /Gm m r m v r=

(a)

S H H/ /Gm m r m v r=

11 2 2 30 9

S/ (6.673 10 N m /kg )(1.99 10 kg)/43 10 mv Gm r−= = × ⋅ × ×

45.6 10 m/s 56 km/s= × =

The actual speed is 71 km/s, so the orbit cannot be

circular.

(b) The orbit is a circle or an ellipse if it is closed, a parabola orhyperbola if open. The orbit is closed if the total energy is negative, sothat the object cannot reach .r → ∞

2 3 2 9 2 21 1H H H2 2

(71 10 m/s) (2.52 10 m /s )K m v m m= = × = ×

11 2 2 30

S H H/ ( (6.673 10 N m /kg )(1.99 10 kg)/(43U Gm m r m −= − = − × ⋅ × ×

9 9 2 2

H10 m)) (3.09 10 m /s )m= − ×

9 2 2 9 2 2

H H(2.52 10 m /s ) (3.09 10 m /s )E K U m m= + = × − ×8 2 2

H(5.7 10 m /s )m= − ×

The total energy E is negative, so the orbit is closed. We know frompart (a) that it is not circular, so it must be elliptical.

22

2

Consider a satellite that follows a circular orbit of radius

around a planet of mass . We apply Newton's second

law and hav

Sate

e:

llites: Orbits and Ener

The kinetic ene

gy

rgy

r

M

GMm v GMma m v

r r r= = → =

2

(eqs.1)2 2

The potential energy (eqs.2)

mv GMmK

r

GMmU

r

= =

= −

If we compare eqs.1 with eqs.2 we have: (eqs.3)2

The total energy 2 2

The energies ,

r

UK

GMm GMm GMmE K U K

r r r

E K

= −

= + = − = − = −

, and are plotted as function of r

in the figure to the left.

For elliptical orbits 2

Note:

U

GMmE

a= −

2

GMmE

r= −

The kinetic energy is and the potential energy is 21

2K mv= GMm

Ur

= −

The mass of the earth is .

(a)

(b) .

24

E 5.97 10 kgM = ×

3 2 91

2(629 kg)(3.33 10 m/s) 3.49 10 JK = × = ×

11 2 2 247E

9

(6.673 10 N m /kg )(5.97 10 kg)(629 kg)8.73 10 J

2.87 10 m

GM mU

r

−× ⋅ ×= − = − = − ×

×

Satellites and “Weightlessness”

Satellites are routinely put into orbit around the Earth. The

tangential speed must be high enough so that the satellite

does not return to Earth, but not so high that it escapes

Earth’s gravity altogether.

A satellite, the International Space Station, circling the Earth.

Artificial satellites launched at different speeds.

The satellite is kept in orbit by its speed—it is continually

falling, but the Earth curves from underneath it.

Objects in orbit are said to experience weightlessness. They

do have a gravitational force acting on them, though!

The satellite and all its contents are in free fall, so there is no

normal force. This is what leads to the experience of

weightlessness.

(a) An object in an elevator at rest exerts a force on a spring scale equal to its

weight. (b) In an elevator accelerating upward at ½ g, the object’s apparent

weight is 1 ½ times larger than its true weight. (c) In a freely falling elevator, the

object experiences “weightlessness”: the scale reads zero.

More properly, this effect is called apparent weightlessness,

because the gravitational force still exists. It can be

experienced on Earth as well, but only briefly:

Experiencing “weightlessness” on Earth