Question 6.10: Escape from the Moon A body is projected from the surface of...
Escape from the Moon
A body is projected from the surface of the Moon with speed u in any direction. Show that the body cannot escape from the Moon if u² < 2MG/R, where M and R are the mass and radius of the Moon. [Assume that the Moon is spherically symmetric.]
If the Moon is spherically symmetric, then the force F that it exerts on the body is given by F =-\left(m M G / r^{2}\right) \widehat{ r }, where m is the mass of the body, and r is the position vector of the body relative to an origin at the centre of the Moon. This force is a conservative field with potential energy V = −mMG/r. Hence energy conservation applies in the form
\frac{1}{2} m|v|^{2}-\frac{m M G}{r}=E,
and, from the initial conditions, E=\frac{1}{2} m u^{2}-(m M G) / R . Thus the energy conservation equation is
| v |^{2}=u^{2}+2 M G\left(\frac{1}{r}-\frac{1}{R}\right).
Since the left side of the above equation is positive, the values of r that occur in the motion must satisfy the inequality
u^{2}+2 M G\left(\frac{1}{r}-\frac{1}{R}\right) \geq 0.
If the body is to escape, this inequality must hold for arbitrarily large r. This means that the condition
u^{2}-\frac{2 M G}{R} \geq 0
is necessary for escape. Hence if u² < 2MG/R, the body cannot escape. The interesting feature here is that the ‘escape speed’ is the same for all directions of projection from the surface of the Moon. (The special case in which the body is projected vertically upwards was solved in Chapter 4.)