Answer
(a) $3.32\times 10^8m$
(b) The answer to part(a) is independent of the mass of the spaceship.
Work Step by Step
(a) We can find the required distance as follows:
$F_E=2F_M$
$\implies \frac{GM_Em}{x^2}=\frac{2GM_Mm}{(r-x)^2}$
$\implies \frac{(r-x)^2}{x^2}=\frac{2M_M}{M_E}$
This simplifies to:
$x=\frac{r}{1+\sqrt{\frac{2M_M}{M_E}}}$
We plug in the known values to obtain:
$x=\frac{3.84\times 10^8}{1+\sqrt{\frac{2\times 7.35\times 10^{22}}{2.57\times 10^{24}}}}$
$x=3.32\times 10^8m$
(b) From part(a), we can see that $x=\frac{r}{1+\sqrt{\frac{2M_M}{M_E}}}$. This equation shows that the answer to part(a) is independent of the mass of the spaceship.
