3Forces
IA Dynamics and Relativity
3.4 Gravity
We’ll now study an important central force — gravity. This law was discovered
by Newton and was able to explain the orbits of various planets. However, we
will only study the force and potential aspects of it, and postpone the study of
orbits for a later time.
Law
(Newton’s law of gravitation)
.
If a particle of mass
M
is fixed at a origin,
then a second particle of mass m experiences a potential energy
V (r) = −
GMm
r
,
where G ≈ 6.67 × 10
−11
m
3
kg
−1
s
−2
is the gravitational constant.
The gravitational force experienced is then
F = −∇V = −
GMm
r
2
ˆ
r.
Since the force is negative, particles are attracted to the origin.
The potential energy is a function of the masses of both the fixed mass
M
and the second particle
m
. However, it is useful what the fixed mass
M
does
with reference to the second particle.
Definition
(Gravitaional potential and field)
.
The gravitational potential is the
gravitational potential energy per unit mass. It is
Φ
g
(r) = −
GM
r
.
Note that potential is confusingly different from potential energy.
If we have a second particle, the potential energy is given by V = mΦ
g
.
The gravitational field is the force per unit mass,
g = −∇Φ
g
= −
GM
r
2
ˆ
r.
If we have many fixed masses
M
i
at points
r
i
, we can add up their gravitational
potential directly. Then the total gravitational potential is given by
Φ
g
(r) = −
X
i
GM
i
r − r
i

.
Again, V = mΦ
g
for a particle of mass m.
An important (mathematical) result about gravitational fields is that we can
treat spherical objects as point particles. In particular,
Proposition.
The external gravitational potential of a spherically symmetric
object of mass
M
is the same as that of a point particle with the same mass at
the center of the object, i.e.
Φ
g
(r) = −
GM
r
.
The proof can be found in the Vector Calculus course.
Example.
If you live on a spherical planet of mass
M
and radius
R
, and can
move only a small distance z R above the surface, then
V (r) = V (R + z)
= −
GMm
R + z
= −
GMm
R
1 −
z
R
+ ···
≈ const. +
GMm
R
2
z
= const. + mgz,
where
g
=
GM/R
2
≈ 9.8 m s
−2
for Earth. Usually we are lazy and just say that
the potential is mgz.
Example.
How fast do we need to jump to escape the gravitational pull of the
Earth? If we jump upwards with speed v from the surface, then
E = T + V =
1
2
mv
2
−
GMm
R
.
After escape, we must have
T ≥
0 and
V
= 0. Since energy is conserved, we
must have E ≥ 0 from the very beginning. i.e.
v > v
esc
=
r
2GM
R
.
Inertial and gravitational mass
A careful reader would observe that “mass” appears in two unrelated equations:
F = m
i
¨
r
and
F = −
GM
g
m
g
r
2
ˆ
r,
and they play totally different roles. The first is the inertial mass, which
measures the resistance to motion, while the second is the gravitational mass,
which measures its response to gravitational forces.
Conceptually, these are quite different. There is no a priori reason why these
two should be equal. However, experiment shows that they are indeed equivalent
to each other, i.e. m
i
= m
g
, with an accuracy of 10
−12
or better.
This (philosophical) problem was only resolved when Einstein introduced his
general theory of relativity, which says that gravity is actually a fictitious force,
which means that the acceleration of the particle is independent of its mass.
We can further distinct the gravitational mass by “passive” and “active”, i.e.
the amount of gravitational field generated by a particle (
M
), and the amount
of gravitational force received by a particle (
m
), but they are still equal, and we
end up calling all of them “mass”.