Answer
a) $F=\frac{4}{9}G\pi ^2a^4\rho^2$
b) $r=(\frac{12M_E}{\pi \rho})^{\frac{1}{3}}$
c) $r=8.67\times 10^7m$
Work Step by Step
(a) We can write the expression for the gravitational force between the masses 'm' as
$F=G\frac{m^2}{(2a)^2}$
Given that $m=(\frac{4}{3}\pi a^3\rho)^2$
$\implies F=G\frac{(\frac{4}{3}\pi a^3\rho)^2}{4a^2}$
$\implies F=\frac{4}{9}G\pi ^2a^4\rho^2$
(b) The required distance can be determined as
$G\frac{m^2}{4a^2}=\frac{4GmM_Ea}{r^3}$
This simplifies to:
$r=(\frac{16a^3M_E}{m})^{\frac{1}{3}}$
$r=(\frac{12M_E}{\pi \rho})^{\frac{1}{3}}$
(c) We know that
$r_s=\frac{12M_s}{\pi \rho}$
We plug in the known values to obtain:
$r_s=\frac{12(95.1\times 5.97\times 10^{24}Kg)}{\pi(3330Kg/m^3)}$
$r=8.67\times 10^7m$
