Chapter 8 Problem 52

Problem

(a) Show that the total mechanical energy of a satellite (mass m) orbiting at a distance r from the center of the Earth (mass M ) is

$E=-{\frac {1}{2}}{\frac {GmM_{\text{E}}}{r}}$

if $U=0$ at $r=\infty$.

(b) Show that although friction causes the value of $E$ to decrease slowly, kinetic energy must actually increase if the orbit remains a circle.

Solution

(a)

With the condition$U=0$ at $r=\infty$ the potential energy is given by

$U=-{\frac {GmM_{\text{E}}}{r}}$

the kinetic energy is found from the fact that for a circular orbit, the gravitational force is a centripetal force.

${\frac {GmM_{\text{E}}}{r^{2}}}={\frac {mv_{\text{orbit}}^{2}}{r}}$ $\rightarrow mv_{\text{orbit}}^{2}={\frac {GmM_{\text{E}}}{r}}$ $\rightarrow K={\frac {1}{2}}mv_{\text{orbit}}^{2}={\frac {1}{2}}{\frac {GmM_{\text{E}}}{r}}$

$E=K+U={\frac {1}{2}}{\frac {GmM_{\text{E}}}{r}}-{\frac {GmM_{\text{E}}}{r}}=-{\frac {1}{2}}{\frac {GmM_{\text{E}}}{r}}$

(b)

As the value of $E$ decreases, since $E$ is negative, the radius $r$ must get smaller. But as the radius gets smaller, the kinetic energy increases, since $K\propto {\frac {1}{r}}$. If the total energy decreases by 1 Joule, the potential energy decreases by 2 Joules and the kinetic energy increases by 1 Joule.