Chapter 12: Universal Gravitation

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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

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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!
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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?
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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.
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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.

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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
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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.
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Geo-synchronous orbit

• Find the radius of an orbit above the equator
  such that the satellite completes one orbit in
  one day.

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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.
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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
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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)!
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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.
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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).
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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.
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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
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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.
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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) ?
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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?
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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?
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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
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Circular Orbit
Virial Theorem U -2K All orbits (elliptical,
parabolic, hyperbolic) ltUgt - 2 ltKgt
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Conic Section
e eccentricity e 0 Circular 0 lt e lt 1
Elliptical e 1 Parabolic e gt 1 Hyperbolic
r
q
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Transfer Orbits
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Gravitational LensingEmc2mE/c2
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GPS