Kevin S. Huang

2008: Problem 25

Topic: Gravity
Concepts: Circular orbits, Elliptical orbits

Solution:

Recall for the circular orbit, we have

v_0=\sqrt{\frac{GM}{R}}

For the elliptical orbit, the starting point is the apogee so

R=a+c

\frac{1}{2}v_0=v_{\text{apogee}}=\frac{a-c}{b}\sqrt{\frac{GM}{a}}=\sqrt{\frac{a-c}{a+c}}\sqrt{\frac{GM}{a}}

using our results for elliptical orbits (note $b^2=a^2-c^2$ ). Substituting these into the first equation,

2\sqrt{\frac{a-c}{a+c}}\sqrt{\frac{GM}{a}}=\sqrt{\frac{GM}{a+c}}

2\sqrt{1-\frac{c}{a}}=1

\frac{c}{a}=\frac{3}{4}

We have

\frac{r}{R}=\frac{a-c}{a+c}=\frac{1-(c/a)}{1+(c/a)}=\frac{1/4}{7/4}=\frac{1}{7}

so the answer is E.