Problem 3:

One of the Apollo 14 astronauts hit a golf ball while on the moon. It was said to have gone for ``miles and miles''.
a) What would the initial speed have to have been to allow the ball to escape completely from the moon?
b) Assume the golf ball started off with less than this initial speed, but instead reached a maximum height of 200 km above the moon's surface. What is the acceleration due to the moon's gravity at that height?
Possibly useful data: G = 6.67 x 10-11 N m2/kg2, the moon's radius is 1.74 x 103 km, its mass is 7.35 x 1022 kg, the radius of the moon's orbit about the Earth is 3.84 x 105 km, the mass of a typical golf ball is 75 gm.
Solution:

a) We are looking for the escape speed ve (same thing as in homework problem 12-19). If we choose the gravitational potential energy to be zero at infinity, we have
Etotal = K + U = ½mve2 - GMmm/Rm = 0
where the total energy is zero, since this means the object can just reach an infinite distance (as the kinetic energy goes to zero). Here Mm and Rm are the mass and radius of the moon respectively. Solving for ve gives
ve = [2GMm/Rm]1/2
plugging in gives
ve = [ 2(6.67 x 10-11 N m2/kg2)(7.35 x 1022 kg)/(1.74 x 106 m)]1/2
= 2374 m/s
It is not very likely that the golf ball was hit with anything near the escape speed! Some of the data supplied were not needed (such as the mass of the golf ball...)

b) Apply Newton's law of gravitation:
F = GMmm/(Rm+h)2
where h is height above the Moon's surface. Since F = ma, the acceleration due to gravity is just
a = F/m = GMm/(Rm+h)2
= (6.67 x 10-11 N m2/kg2)(7.35 x 1022 kg)/(1.74 x 106 m + 200 x 103 m)2
= 1.30 m/s2
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last updated: Nov. 23 1998