MATH 164, Spring 2001 Due Date: Name(s):
Honors Project 9: Escape Velocity
If a ball is thrown straight up into the air, it will travel upward, stop, and then fall back to earth. If a
bullet is shot straight up into the air, it will travel upward, stop (at an altitude higher than the ball, unless
the person who threw the ball was a real zealot), and then fall back to earth. In this project we discuss the
problem of determining the velocity with which an object needs to be propelled straight up into the air so
that the object goes into orbit; that is, so that it never falls back to earth. This is the escape velocity for
the earth. (We add the phrase "for the earth" since the escape velocity, as we will see, depends on the mass
of the body -- whether it be a planet, or moon, or ... -- from which the object is propelled.)
We make two simplifying assumptions in our derivation: First, that there is no air resistance. Second,
that the object has constant mass throughout its flight. This second assumption is significant since it means
our analysis does not apply to a rocket, for example, since the mass of the rocket changes as the rocket burns
fuel.
Narrative
Newton's Second Law of Motion states that if an object with mass m travels along a straight-line path
parametrized by x = x(t) then the force which it will exert on a particle at a point along its path is given
by F = ma where a = d2
x/dt2
. Since a may also be written
a =
dv
dt
=
dv
dx
dx
dt
= v
dv
dx
,
we may rewrite Newton's Second Law
F = mv
dv
dx
. (1)
Now the Universal Law of Gravitation states that the direction of the force of gravity is along a line joining
the centers of mass of the earth and the object, and that the magnitude of this force is
-
GMEm
(RE + x)2
where G is the universal gravitational constant, ME is the mass of the earth, RE is the radius of the earth,
and x is the distance of the object above the surface of the earth. The fact that the weight w (in pounds)
of an object near the surface of the earth is given by w = mg where m is the mass (in slugs) of the object
and g = 32 ft/sec2
, is based on an approximation. For an object of mass m a distance x above the surface
of the earth,
w =
GMEm
(RE + x)2

GMEm
R2
E
=
GME
R2
E
m = mg
where g = GME/R2
E and RE is the radius of the earth. This approximation is justified by the fact that --
in certain applications -- x is generally much smaller than RE  3963 mi, so RE + h  RE. In any case,
g =
GME
R2
E
implies that gR2
E = GME, so we can write the magnitude of gravitational force
-
GMEm
(RE + x)2
= -
gR2
Em
(RE + x)2
. (2)
Since the force given by (1) must be balanced by the gravitational force given by (2) for an object being
propelled vertically into the air, it must follow that
-
gR2
Em
(RE + x)2
= mv
dv
dx
or -
gR2
Em
(RE + x)2
dx = mv dv.
If we integrate this equation from ground level (at which x = 0 and v0 is initial velocity) to the point at
which the object is no longer acted upon by the gravitational force of the earth (at which x =  and v = 0)

0
-
gR2
Em
x2
dx =
0
v0
mv dv, (3)
we obtain an equation which we can solve for the escape velocity v0: Indeed, upon integrating, we find that
-gRE = -
1
2
v2
0, (4)
so
v0 = 2gRE.
Tasks
1. Verify equation (4).
2. Assuming that g = 9.8 m/sec2
, RE  6.38 × 106
meters, and G = 6.67 × 10-11
N  m2
/kg2
:
a) What is the mass ME of the earth in kg? In tons?
b) What is the escape velocity v0 from the earth in meters per second? In miles per hour?
3. Repeat the above computations to find the escape velocity for the moon. (You will have to use references
to find quantities such as the radius of the moon.)