Chapter 13
Gravitation
Problems: 1, 4, 5, 7, 18, 19, 25, 29,
31, 33, 43
Every object in the universe attracts every
other object. This is called gravitation.
We’re use to dealing with falling bodies near
the Earth. For example a rock dropped off of a
cliff.
However the same rules apply for all objects,
no matter how large or how far apart.
(apples falling out of trees, planets, moons,
stars, galaxies…)
4 fundamental types of forces
1) Strong nuclear force
2) Weak nuclear force
3) Electric force
4) Gravitational force
Out of the 4 fundamental types of forces, the
gravitational force is the weakest of the 4.
The strong nuclear force is by far the strongest,
however it only acts over very short distances
(comparable to the size of a nucleus).
There is a gravitational attraction between every
object in the universe. However, this attraction is
only noticeable when either the objects are huge,
or when they are very close together.
Newton’s Law of Gravitation F
m1m2
G 2
r
r is the center to center distance between the
two objects.
G = 6.67x10-11 Nm2/kg2
Newton’s law of gravitation is a vector equation.
The direction of the force is in the direction of a line that
connects the two objects.
2
r̂
r

F
m1m2
G 2 rˆ
r
1
The force is attractive.
Object 1 is pulled towards Object 2
Because of Newton’s 3rd Law, object 2 is pulled towards
object 1.
G is the Newton’s universal gravitational constant.
It applies everywhere.
We have been using g = 9.8 m/s2
These two values are related but different.
g is not constant everywhere.
Find gravitational force between a person of mass
100 kg (standing at sea level) and the Earth.
F
F
m1m2
G 2
r
982 N
6.67 x10
11
Nm 2 (5.98 x10 24 kg)(100 kg)
kg 2
(6.37 x10 6 m) 2
F
m1m2
G 2
r
F
6.67 x10
F
N
9.8
kg
6.67 x10
11
11
Nm 2 (5.98x1024 kg)(100kg)
kg 2
(6.37 x106 m) 2
Nm 2 (5.98x1024 kg)
(100kg)
2
6
2
kg (6.37 x10 m)
m
9.8 2 100kg (100kg) g
s
at sea level (distance equal to radius of Earth)
Me
G 2
re
g
At different elevations, g has different values.
All semester we have been under the assumption
that g is a constant. This works as long as objects
stay near the surface of the Earth.
For noticeable changes in elevation, g varies.
Find the weight of a 100 kg person standing at:
Sea level.
r = re = 6.38x106m
On top of Mt. Everest
r = re + 10000m
In spacecraft 5x106m above Earth, r = 11.38x106m
Sea level: r = re = 6.38x106m
w = mpg = mp (GmE/r2) = 100 kg(9.8m/s2)
w = 980 N
(220.2 lbs)
Mt. Everest: r = re +105m = 6.39x106m
w = mpg = mp (GmE/r2) = 100 kg(9.77m/s2)
w = 977 N
(219.5 lbs)
Spacecraft: r = 11.38x106m
w = mpg = mp (GmE/r2) = 100 kg(3.08m/s2)
w = 308.0 N
(69.2 lbs)
On the Moon
To find g on the moon, we use Mmoon and rmoon
On the surface of the moon, g = gmoon =1.62m/s2
What would a 100 kg person weigh on the moon?
w = (100kg)(1.62m/s2) = 162 N
or 36.4 pounds
There are other reasons why g is not constant. They
have to do with the Earth itself.
1) The Earth’s mass is not uniformly distributed. The
thickness and makeup of the Earth’s crust varies.
2) The Earth is not a perfect sphere. The Earth is
flattened out a bit at the poles and bulges out at
the equator. The poles are closer to the center of
the Earth than the equator.
3) The Earth is rotating.
Take an object sitting on the equator. As the Earth rotates, the
object moves in a circle.
Applying circular motion:
FN points outward along the radial direction.
FN is the measured weight of the object (mg).
ag and v2/R point inward toward the center of the Earth.
FN
mg mag
v2/R
mag
mg
g
m
mag
ag
2
m
2
2
R
R
R
Using the values of and R at the equator, g varies from ag by
only about 0.034m/s2.
Gravitation inside the Earth
Newton’s Shell Theorem: A uniform shell of matter exerts no net
gravitational force on a particle inside it.
Break the Earth into uniform shells, only the shells that have a
radius smaller than the location of the object produce a net
gravitational force on the object.
When you take Physics 2 and see Gauss’ Law, think of this.
r
Gravitational Potential Energy
Earlier we introduced the potential energy of a particleEarth system and saw that the amount of potential energy
depended on the separation distance.
Gravitational potential energy is produced by the existence
of two or more particles separated by some distance(s).
If g was constant U = mgh where h is the separation.
We would pick as a manner of convenience the surface of
the Earth to be where U = 0.
As particles were raised above the surface, the gravitational
potential energy increased.
As particles were brought closer to the center of the Earth,
the gravitational potential energy decreased.
Since location of the zero of potential energy is arbitrarily picked,
only changes in gravitational potential energy were meaningful.
Let’s now allow the separation distance to be infinite.
Now let’s pick the separation that produces zero gravitational
potential energy to be infinity.
Now as particles get closer, they have a finite, but negative amount
of gravitational potential energy.
We will now see what happens when r varies noticeably and thus g
is not constant.
Use the work done by a variable force:
We are going from R away from the center of an object (for
example: Earth) to infinity away.


F (r ) dr
W
R
F (r )dr cos
R
F is attractive so F will be in the opposite direction of dr.
= 1800


F (r ) dr
GMm
dr
2
r
putting this in the integral we get: W
R
W
GMm
dr
2
r
GMm
0
R
GMm
r
GMm
R
R
W
R
W
GMm
dr
2
r
GMm
r
GMm
0
R
R
GMm
R
Using: W = - U
U
U
setting U
0
U
W
W
GMm
r
Relation between force and potential energy
GMm
U (r )
r
GMm
Fg
The minus sign here tells us the force is attractive.
2
r
From chapter 8 we found: F(x) = -dU(x)/dx
dU
dr
dU
dr
d
GMm
dr
r
GMm
F (r )
2
r
Graphs of gravitational force and gravitational potential energy
F(N)
U(J)
r(m)
1
2
r
r(m)
1
r
For objects that are infinitely far away the gravitational
potential energy is zero.
Escape Speed
Throw a projectile upwards. it will slow down because of the
gravitational pull of the Earth until it comes to a stop, then will fall
back to the ground.
The escape speed is the speed necessary for an object to have at
the surface of a planet (or other object) so that is has enough
kinetic energy to escape the gravitational pull of the planet.
Since the gravitational field extends to infinity, the launched
particle will theoretically come to rest after being infinitely far
from the planet.
On the surface of the planet, the projectile-planet system has
gravitational potential energy:
U
GMm
r
The particle is launched so it has some velocity and some kinetic
energy.
At the surface of the planet the total energy is:
1 2
mv
2
K U
GMm
R
When the particle is infinitely far away, it comes to a rest, so the total
energy will be zero.
1 2
mv
2
GMm
R
0
The necessary velocity at the launch to produce this is:
v
2GM
R
Escape velocity depends on the mass and radius of
the planet you are trying to escape from.
See table 13-2
Kepler’s Laws
People use to think that the Earth was the center of the solar
system. One problem with this is the retrograde motion of Mars.
Kepler used data (observations of stars and planets) to come up
with 3 laws of planetary motion. We will apply the laws to
planets orbiting the Sun but they work for any satellites, natural
or artificial that orbit a massive central body.
The central body will be the sun(M) and the planet will be mass
(m).
1) All planets move in elliptical orbits, with the Sun at one
focus.
Ellipses are ‘flattened’ circles. The eccentricity of an ellipse
tells how ‘flattened’ the ellipse is.
A circle would have an eccentricity of zero.
The two foci would be located at the same point.
The eccentricity of the Earth’s orbit is only 0.0167.
nothing at second focus
The eccentricity of this picture is greatly exaggerated.
2) The Law of Areas. A line that connects the planet to the Sun
sweeps out equal areas in the plane of the orbit during equal
time intervals.
4
1
3
2
If the time interval to go from points 1 to 2 is the same as to go
from point 3 to 4, the areas of the two wedges are equal.
As a result, the planet moves faster when it is closer to the Sun
and moves slower when it is further away.
Planet is fastest at perihelion (nearest point to the Sun)
Planet is slowest at aphelion (farthest point to the Sun)
3) Law of Periods: The square of the period of any planetary
orbit proportional to the cube of the semimajor axis of its orbit.
Since the planets have orbits of low eccentricity, we can replace
the semimajor axis with the average distance from the Sun.
Use Newtons’s 2nd law: F = mac
GMm mv
r2
r
GM
v2
r
2
Use: v
2 r
T
GM
r
T2
2 r
T
2
4 2 3
r
GM
Problems: 8, 9, 16, 24, 52