Gravitational Potential Energy
THEORY
Recall the change in potential energy of a system, V, is related to the work,
W, by the force one part of the system exerts on the other,
U  - W = Wext
where Wext is the work by an external force against the gravity force. We can
think of this external work as the energy needed to stretch the gravitational
bond between two masses.
Given two point masses, M and m, initially separated by a distance ri then M
attracts m with a force of magnitude,
F = GMm/r2 .
If an external force is applied with the above magnitude it can move m from
ri to a distance rf > ri . Then the external force does positive work on m given
by,
rf
Wext =
 GMm dr/r
2
(
= - GMm 1/rf - 1/ri
) > 0.
ri
Therefore, the change in gravitational potential energy of this two-particle
system equals Wext this result,
(
U = U(rr) – U(ri)  Wext = - GMm 1/rf - 1/ri
).
Setting rf → ∞ and choosing the point of zero potential energy at infinity,
U(rf → ∞)  0 gives us on replacing ri → r,
U(r) = -GMm/r.
Note: r is in the denominator, not r2 and the reference location for potential
energy is infinity, which then is the place of maximum potential energy.
If there is a collection of particles then there is an energy of this sort between
each distinct “bond” between pairs of particles,
U(r) = ½
i j
- G mi mj / rij ,
where in the second sum, j  i, and the ½ is to avoid double counting of
bonds.
APPLICATIONS OF THE ABOVE THEORY
 gravity inside earth r < R
To find the gravitational force on a point mass, m, at a distance r < R inside
earth use Newton’s shell theorem (i.e. treat that portion of the earth at r < R
as if it were concentrated at the center of the earth and ignore that portion at
r > R). Note: one will need mass density of earth (  M / V) and the volume
of a sphere (V  4R3/3).
Mass (between r = 0 and r) = earth density x volume of sphere of radius r.
= (M/4R3/3) x 4r3/3 = M (r/R)3
 F(r<R) = force on m by earth = - G M(r/R)3 m/r2
F(r<R) = -GMm r/R3
Conclusion, mass m moves like a particle attached to a spring, F  - r.
 escape velocity
Assume a particle starts with an initial radial speed v from the surface of a
sphere of mass M and radius R. The total energy of the system is
Ei = ½ mv2 - GMm/R
If v = vescape then the particle will “just make it to infinity” where it will be at
rest and so,
Ef = 0.
Conservation of energy (i.e. set Ei = Ef ) gives after a little algebra,
Ei = ½ mvescape2 - GMm/R = Ef = 0

vescape =
 2GM/R
Note: we have neglected air friction, so m is not in this formula.
 satellites in circular orbit
Using the universal law of gravitation with Newton’s second law and the
formula for centripetal acceleration, we find for a particle of mass m in a
circular orbit of radius r around a sphere of mass M an energy, kinetic
energy, and gravitational potential energy given by,
F=GMm/r2 =mv2/r 
K = ½ mv2 = ½ (GMm/r)
U = -GMm/r
E=K+U
Therefore,
E = -K = V/2 = - ½ GMm/r
Note: this equation only applies to circular orbits.
 Kepler’s three laws of planetary motion
The laws are results found by direct observation of the Solar system, they
are experimental facts.
Law I. The planets orbit the sun in elliptical orbits with the Sun at a focal
point.
Law II. The radius from the Sun to the Planet sweeps out equal areas in
equal times. [Note: this law is a direct consequence of the
conservation of angular momentum]
Law III. The square of the period, T, of a planet’s motion is proportional to
the cube of the semi-major axis, a, of the orbit, i.e.
T2 = 42 a3 / GM.
EXAMPLES[in class]