#### Answer

$v_B=34.8Mm/h$

#### Work Step by Step

We can determine the required speed of the satellite as follows:
We know that
$v=\frac{-GM_em}{r}$
We plug in the known values to obtain:
$v=\frac{-66.73\times 10^{-12}\times 5.976(10^{24})\times 60}{20\times 10^6}$
$\implies v=-1.196\times 10^9J$
Similarly $v_2=\frac{-66.73\times 10^{-12}\times 5.97\times 10^{24}\times 60}{80\times 10^6}=-0.299\times 10^9J$N
We know that
$T_1=\frac{1}{2}mv_A^2$
$\implies T_1=\frac{1}{2}\times 60\times (\frac{40\times 10^6}{3600})^2=3.7\times 10^9J$
According to the conservation of energy equation
$T_1+v_1=T_2+v_2$
We plug in the known values to obtain:
$3.7(10)^9-1.196(10)^9=30v_B^2-0.299(10)^9$
This simplifies to:
$v_B=9666.1m/s=\frac{9666.1\times 3600}{10^6}=34.8Mm/h$