Answer
$m=1.91\times 10^{30}Kg$
Work Step by Step
In the given scenario
$F_{gravity}=F_{centripetal}$
$\implies \frac{Gm^2}{d^2}=m\frac{v^2}{r}$ (as given that $m_A=m_B$)
This simplifies to:
$m=\frac{d^2v^2}{rG}$
substituting $v=\frac{\pi d}{T}$, we obtain:
$m=\frac{2d}{G}(\frac{\pi d}{T})^2$
$m=\frac{2\pi^2d^3}{GT^2}$
We plug in the known values to obtain:
$m=\frac{2\pi^2(3.45\times 10^{12})^3}{6.67\times 10^{-11}(2.52\times 10^9)^2}$
$m=1.91\times 10^{30}Kg$
