Answer
$v_A=4.89\times 10^3m/s$, $v_B=3.26\times 10^3m/s$
Work Step by Step
The required speed can be determined as follows:
$v_p=\sqrt{\frac{2GM_e r_a}{r_p(r_p+r_A)}}$
We plug in the known values to obtain:
$v_p=\sqrt{\frac{2\times 66.73\times 10^{-12}\times 5.9676\times 10^{24}\times 30\times 10^6 }{20\times 10^6(20\times 10^6+30\times 10^6)}}$
This simplifies to:
$v_p=4891.49m/s=4.89\times 10^3m/s$
But $v_p=v_A$
$\implies v_A=4.89\times 10^3m/s$
Now $v_B=\frac{r_pv_p}{r_a}$
We plug in the known values to obtain:
$v_B=\frac{(20\times 10^6)(4891.49)}{30\times 10^6}=3261m/s=3.26\times 10^3m/s$
