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We saw earlier, in Sect. 5.5, that gravity is a conservative force, and, therefore,
has an associated potential energy. Let us obtain a general formula for this energy.
Consider a point object of mass , which is a radial distance from another point
object of mass . The gravitational force acting on the first mass is
of magnitude , and is directed towards the second mass. Imagine that the
first mass moves radially away from the second mass, until it reaches infinity. What is
the change in the potential energy of the first mass associated with this shift?
According to Eq. (155),

(550) 
There is a minus sign in front of because this force is oppositely directed to the motion.
The above expression can be integrated to give

(551) 
Here, we have adopted the convenient normalization that the potential energy at infinity is
zero. According to the above formula, the gravitational potential energy of a mass
located a distance from a mass is simply .
Consider an object of mass moving close to the Earth's surface. The potential
energy of such an object can be written

(552) 
where and are the mass and radius of the Earth, respectively, and
is the vertical height of the object above the Earth's surface. In the limit that
, the above expression can be expanded using the binomial theorem
to give

(553) 
Since potential energy is undetermined to an arbitrary additive constant, we could just as
well write

(554) 
where
is the acceleration due to gravity at
the Earth's surface [see Eq. (548)]. Of course, the above formula is equivalent
to the formula (125) derived earlier on in this course.
For an object of mass and speed , moving in the gravitational field of a fixed object
of mass , we expect the total energy,

(555) 
to be a constant of the motion. Here, the kinetic energy is written
, whereas
the potential energy takes the form
. Of course, is the distance between the
two objects. Suppose that the fixed object is a sphere of radius . Suppose,
further, that the second object is launched from the surface of this sphere
with some velocity which is such that it only just escapes the sphere's gravitational
influence. After the object has escaped, it is a long way away from the sphere, and hence
. Moreover, if the object only just escaped, then we also expect , since the
object will have expended all of its initial kinetic energy escaping from the sphere's
gravitational well. We conclude that our object possesses zero net energy:
i.e., . Since is a constant of the motion, it follows
that at the launch point

(556) 
This expression can be rearranged to give

(557) 
The quantity is known as the escape velocity. Objects launched from
the surface of the sphere with velocities exceeding this value will eventually
escape from the sphere's gravitational influence. Otherwise, the objects
will remain in orbit around the sphere, and may eventually strike its surface.
Note that the escape velocity is independent of the object's mass and launch
direction (assuming that it is not straight into the sphere).
The escape velocity for the Earth is

(558) 
Clearly, NASA must launch deep space probes from the surface of the Earth with velocities which exceed this value if they are
to have any hope of eventually reaching their targets.
Next: Satellite orbits
Up: Orbital motion
Previous: Gravity
Richard Fitzpatrick
20060202