Answer
a) The proof is below.
b) The proof is below.
c) $ T = 2\pi \sqrt{\frac{r_0^3}{GM}}$
d) 16 days
Work Step by Step
a) At equilibrium, the value of $\frac{dr^2}{dt^2}$ is 0, so we find:
$0=4\pi r_0^2p_0-\frac{GMm}{r_0^2}$
$p_0= \frac{GMm}{4\pi r_0^4} $
b) When x is small compared to r, we see that $r_0\approx r$.
We first find a value for $p$:
$p =\frac{p_0r_0^5}{r^5}=\frac{GMm}{4\pi r_0^4}\frac{r_0^5}{r^5}=\frac{GMmr_0}{4\pi r^5}$
We plug this into the original equation to find:
$=4\pi r^2 (\frac{GMmr_0}{4\pi r^5})-\frac{GMm}{r^2}$
$= (\frac{GMmr_0}{ r^3})-\frac{GMm}{r^2}$
We now have just r in the denominators. Using the binomial approximation, this becomes:
$=-\frac{GMmx}{r_0^3}$
c) Following the steps that the book outlines, we find:
$m\frac{d^2x}{dt^2}=-\frac{GMmx}{r_0^3}$
$\frac{d^2x}{dt^2}=-\frac{GMx}{r_0^3}$
Comparing this to the equations in the book that the problem asks us to look at, we find:
$ T = 2\pi \sqrt{\frac{r_0^3}{GM}}$
d) Plugging in the given values, we find:
$ T = 2\pi \sqrt{\frac{r_0^3}{GM}}$
$T=1,399,680 \ s \approx \fbox{16 days}$
