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Gravitation Notes
Kepler’s Laws
1. Planets move in elliptical orbits with the sun at one focus.
2. As a planet moves in its orbit, a line drawn from the sun to the planet
sweeps out equal areas in equal time intervals.
3. If T is the period and a is the length of a the semi-major axis, then
T2/a3 is a constant
Newton’s Universal Law of Gravitation- any massive objects in the
universe exert an attractive force on each other called the gravitational
force.
Fg= 
Gm1m 2
r2
G = 6.67 x 10-11 Nm2/kg2
Where m1 is the mass of the first object, m2 is the mass of the second
object and r is the distance between their centers of masses. And Fg is the
gravitational force between them. G is the universal gravitation constant.
The force is negative because it is an attractive force.
Using this equation, you can determine the acceleration of gravity on the
surface of a planet by setting m2g=
Gm1m 2
and solving for g.
r2
m2 will cancel out, m1 is the mass of the planet and r is the distance from the
center of mass of the planet.
If there are more than 2 objects, you can find the net force on one object
by finding its gravitational attraction to each of the masses surrounding it
and then performing a vector summation of those forces. Here it could be
helpful to write the forces as unit vectors ( i and j).
In the equation, the masses are directly proportional to the force. This
means that if one mass is doubled, the force between the two objects is
doubled and if both masses double, the force between the two objects is
____________.
An important part of this equation comes from the fact that r2 appears in
the denominator. This is known as an inversed-square law. From this you can
see that as the distance between the two objects is doubled, the force
between them is actually ¼ of what it was before. Likewise, if the distance
is tripled, the force becomes ______ of what it was before.
This also means that as you go up in altitude, the acceleration due to gravity
lessens, which also lessens your weight.
The gravitational acceleration of a planet follows the same rules apply.
Planet X has a mass M, radius R and acceleration due to gravity g. Planet Y
has a mass of 2M and radius 3R. What is the acceleration due to gravity of
this planet? ____g
Circular Orbits
If we can assume a circular orbit for a satellite in orbit around a planet (or a
planet around a star) we know that the gravitational force between them is
causing the circular motion. In other words, centripetal force = gravitational
force.
Fcentripetal = Fgravitation
m1v2/r=
Gm1m 2
r2
where m1 is the mass of the satellite in orbit and m2 is the mass of the
planet, and r is the distance between the objects’ centers.
Using these equations, you can obviously solve for any of the variables
included, but you can also solve for the satellite’s period. Period, T, is the
time it takes for the satellite to complete one full revolution.
T=
2r
v
Gravitational Fields
Spherical Shell – When you are outside of the shell, it can be treated as if
all the mass is at the center, a distance r away.
When you are inside a spherical shell, you experience no net gravitational
force.
g(r) = 0
Fg = 0
for r< R
g(r) =
Gm
r2
Fg =
Gm1m 2
r2
for r  R
Gravitational Force vs. Distance for A Spherical
Shell
Fgravitation
R
r
Solid Sphere – When you are outside of the sphere, it can be treated as if
all the mass is at the center, a distance r away.
When you are inside a solid sphere, the mass in the shell above you exerts no
net gravitational force (like above). However, the mass in the sphere
“below” you does exert a force. So when you are a distance r away from the
center, you experience a gravitational force due to the mass and radius of
the mass “beneath” you acting as its own planet.
g(r) = Gm r
Fg =
R3
g(r) =
Gm
r2
Fg =
Gm1m 2
r
R3
Gm1m 2
r2
for r< R
for r  R
The equations for r< R come from a more general equation
Fg=
Gm1 min side
r2
In this space, solve for
Fg =
Gm1m 2
r
R3
using Newton’s Universal Law of gravitation. Hint: use density to put minside
in terms of known variables such as the mass of the planet and the radius of
the planet.
Gravitational Force vs. Distance for A Solid
Sphere
Fgravitation
R
r
Gravitational Potential Energy
U =-

-
U=-
Gm1m 2
dr
r2
Gm1m 2
r
This means that the gravitational potential energy is negative at the surface
of the planet and becomes closer and closer to zero until you are an infinite
distance away, in which case the gravitational potential energy is zero.
You will want to use this equation instead of U= mgh when you are dealing
with a problem where the projectile is far from the surface of the Earth
(when g is no longer 9.8 m/s2).
Gravitational potential energy is also helpful when solving problems involving
the conservation of mechanical energy, such as when a projectile is launched
away from the surface of the Earth or when a projectile in space comes
towards the Earth.
Total mechanical energy
E= Uo + Ko
Uo + Ko = Uf + Kf
Instead of using mgh here, make sure you use the new form.
-
Gm1m 2
Gm1m2
+ ½ m vo2 = + ½ m vf2
ro
rf
(make sure that you use the distance from their center of masses for r).
Escape Speed- the minimum speed required to escape from the Earth’s
gravity.
If Ko > Uo then it will escape (unbounded)
If Ko < Uo then it won’t escape (bounded)
So if E = Ko + Uo  0, than the objects can escape!
-
Gm1m 2
+ ½ m vo2  0
ro
then
Gm1m 2
= ½ m vo2
ro
so vescape =
2Gmearth
R