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Chapter 12 Universal Gravitation

- The earth exerts a gravitational force mg on a

mass m. - By the action-reaction law, the mass m exerts a

force mg on the earth. - By symmetry, since the force mg is proportional

to the mass m, the value of g must also be

proportional to the mass M of the earth. - Isaac Newton realized that the motion of

projectiles near the earth, the moon around the

earth, the planets around the sun, could be

described by a universal law of gravitation

Gravity

- If two particles of mass m1 and m2 are separated

by a distance r, then the magnitude of the

gravitational force is - G is a constant 6.67 ? 10-11 Nm2/kg2

The force is attractive The direction of the

force on one mass is toward the other mass. Why

1/r2? It works!

The gravitational force varies like 1/r2. It

decreases rapidly as r increases, but it never

goes to zero.

Example The gravitational force between two

masses is 10-10 N when they are separated by 6 m.

If the distance between the two masses is

decreased to 3 m, what is the gravitational force

between them? What is the force of gravity at

the surface of the earth?

Gravitational Attraction of Spherical Bodies

If you have an extended object, it behaves as if

all of its mass is at the center of mass.

Therefore, to calculate the gravitational force

between two objects, use the distance between

their centers of mass.

Gravitational force between the Earth and the

moon.

Gravitation of finite objects

- I. Newton invented Differential Calculus to

interpret his theory F ma - I. Newton invented Integral Calculus to prove

that the gravitational force of the earth and

motion of the moon is the same as if the earth

and moon were each concentrated in a single point.

Example

Calculate the gravitational force between a 70 kg

man and the Earth.

F m g (70 kg) (9.8 m/s2) 686 N, but

Variation of g with height

The gravitational force between the Earth and the

space shuttle in orbit is almost the same as when

the shuttle is on the ground.

Geo-synchronous orbit

- Find the radius of an orbit above the equator

such that the satellite completes one orbit in

one day.

Keplers Laws of Orbital Motion

1. Objects follow elliptical orbits, with the

mass being orbited at one focus of the ellipse.

A circle is just a special case of an ellipse.

Keplers Laws (cont.)

2. As an object moves in its orbit, it sweeps

out an equal amount of area in an equal amount of

time.

This law is just conservation of angular

momentum. Gravity does not exert a torque on the

planet because the force is directed toward the

axis of rotation.

perigee

apogee

Keplers Laws (cont.)

3. The period of an objects orbit, T, is

proportional to the 3/2 power of its average

distance from the thing it is orbiting, r

Note M is the mass that is being orbited. The

period does not depend on the mass of the

orbiting object. Thus the space station and the

astronauts in it go around the sun at the same

rate as the earth (even as they go around the

earth)!

Example

1. The space shuttle orbits the Earth with a

period of about 90 min. Find the average

distance of the shuttle above the Earths surface.

2. Rank the moon, the space shuttle and a

geosynchronous satellite in order of (a) smallest

period to largest period and (b) smallest

orbital velocity to largest orbital velocity.

Gravitational Potential Energy

The gravitational potential energy of a pair of

objects is

When we deal with astronomical objects, we

usually choose U 0 when two objects are

infinitely far away from each other. In this

case, gravitational potential energy is negative.

The formula we have used in the past, U mgy, is

valid only near the surface of the Earth (and has

a different location for U0).

Escape Speed

We can use conservation of energy to calculate

the speed with which an object must be launched

from Earth in order to entirely escape the

Earths gravitational field.

Initially, the object has kinetic (velocity v)

and potential energy. In order to escape, the

object must have just enough energy to reach

infinity with no speed left. In this case, M

mass of Earth and R radius of Earth.

Example

- A satellite is orbiting the Earth as shown below.

At what part of the orbit, if any, are the

following quantities largest? - Kinetic energy (b) Potential energy (c) Total

energy (d) Orbital velocity (e) Gravitational

force (f) Centripetal acceleration (g) Angular

momentum

C

A

B

Elliptic Orbit

An asteroid of mass m is orbiting the Sun (mass

M) as shown. E KU constant KAUA

KBUBKCUC KA (1/2) mvA2 KB (1/2) mvB2 UA

- GmM/rA UB - GmM/rB rA lt rC lt rB At what

part of the orbit, is the mechanical energy E

largest? A), B), C), or D) the energy E is the

same everywhere in the orbit.

Elliptic Orbit

An asteroid of mass m is orbiting the Sun (mass

M) as shown. E KU constant KAUA

KBUBKCUC KA (1/2) mvA2 KB (1/2) mvB2 UA

- GmM/rA UB - GmM/rB rA lt rC lt rB At what

part of the orbit, is the potential energy U

largest (remember 5 lt -3) ?

Elliptic Orbit

An asteroid of mass m is orbiting the Sun (mass

M) as shown. E KU constant KAUA

KBUBKCUC KA (1/2) mvA2 KB (1/2) mvB2 UA

- GmM/rA UB - GmM/rB rA lt rC lt rB At what

part of the orbit, is the potential energy K

largest?

Elliptic Orbit

An asteroid of mass m is orbiting the Sun (mass

M) as shown. Kinetic Energy is largest at A KA

(1/2) mvA2 KB (1/2) mvB2 UA - GmM/rA UB -

GmM/rB rA lt rC lt rB At what part of the

orbit, is the velocity v largest?

Elliptic Orbit

An asteroid of mass m is orbiting the Sun (mass

M) as shown. FA GmM/rA2 FB GmM/rB2 At

what point on the orbit is the gravitational

force on m greatest? Enter D) if the force is

equal everywhere

Circular Orbit

Virial Theorem U -2K All orbits (elliptical,

parabolic, hyperbolic) ltUgt - 2 ltKgt

Conic Section

e eccentricity e 0 Circular 0 lt e lt 1

Elliptical e 1 Parabolic e gt 1 Hyperbolic

r

q

Transfer Orbits

Gravitational LensingEmc2mE/c2

GPS