Answer
$6.5\times 10^{23}\ kg$
Work Step by Step
Given that:
The radius of circular path $R=9.4\times 10^{6}\ m$
Time period $T =7h\ 39min = (7\times 3600\ s)+(39\times 60\ s) =27540\ s$
Let mass of mars be M.
Then, from Keplers third law,
$T^2 =\frac{4\pi^2}{GM}R^3$
$\frac{T^2}{R^3}=\frac{4\pi^2}{GM}$
$M=\frac{4\pi^2}{G}(\frac{R^3}{T^2})$
$M =\frac{4\pi^2(9.4\times 10^{6})^3}{6.67\times 10^{-11}\times 27540^2}$
$M=6.5\times 10^{23}\ kg$
