"Published in Newark, California, USA"
The radius of the moon is roughly 1080 miles. The acceleration of gravity at the surface of the moon is about 0.165g, where g is the acceleration of gravity at the surface of the earth. Determine the velocity of escape for the moon. (Hint: g = 32.16 ft/sec2 and 1 mile = 5280 ft)
Solution:
Consider the problem of determining the velocity of a particle projected in a radial direction outward from the earth and acted upon by only one force, the gravitational attraction of the earth.
We shall assume an initial velocity in a radial direction so that the motion of the particle takes place entirely on a line through the center of the earth.
According to the Newtonian Law of Gravitation, the acceleration of the particle will be inversely proportional to the square of the distance from the particle to the center of the earth. Let r be that variable distance, and let R be the radius of the earth. If t represents time, v is the velocity of the particle, a its acceleration, and k is the constant of proportionality in the Newtonian Law, then
The acceleration is negative because the velocity is decreasing. Hence the constant k is negative. When r = R, then a = - g, the acceleration of gravity at the surface of the earth. Thus
from which
We wish to express the acceleration in terms of the velocity and the distance. We have a = dv/dt and v = dr/dt. Hence
so the differential equation for the velocity is now seen to be
The method of separation of variables applies to equation above and leads at once to the set of solutions
Suppose that the particle leaves the earth's surface with the velocity v0. Then v = v0 when r = R, from which the constant C is easily determined to be
Thus, a particle projected in a radial direction outward from the earth's surface with an initial velocity v0 will travel with a velocity v given by the equation
If v becomes zero, the particle would stop, the velocity would change from positive to negative, and the particle would return to the earth.
If r becomes very large, then the first term will be very small and even equal to zero.
To avoid that v becomes zero, then the next two terms in the equation above must be
If
then we can solve for the value of v0 as follows
The above equation is called the velocity of escape. This is applicable at any planets and moons in a solar system including earth.
Now, let's go back to the given problem that if the radius of a moon is 1080 miles and the acceleration of gravity is 0.165g, then the velocity of escape at the moon will be equal to
