Answer
$6.09\times 10^{24}Kg$
Work Step by Step
We know that
$K.E_a+U_a=K.E_p+U_p$
$\implies \frac{1}{2}mv_a^2-G\frac{mM}{r_a}=\frac{1}{2}mv_p^2-G\frac{mM}{r_p}$
$\implies v_a^2-2G\frac{M}{r_a}=v_p^2-2G\frac{M}{r_p}$
This simplifies to:
$M=\frac{v_a^2-v_p^2}{2G(\frac{1}{r_a}-\frac{1}{r_p})}$
We plug in the known values to obtain:
$M=\frac{(3990)^2-(4280)^2}{2(6.67\times 10^{-11})(\frac{1}{2.41\times 10^7}-\frac{1}{2.25\times 10^7})}$
$M=6.09\times 10^{24}Kg$
