Answer
$v_A=0.885m/s$
Work Step by Step
We know that
$r_A=\sqrt{r_B^2+r^2}$
$\implies r_A=\sqrt{(1.50\times 10^3m)^2+(3.00\times 10^3m)^2}$
$\implies r_A=3.35\times 10^3m$
Now $v_A=\sqrt{v_B^2+4GM(\frac{1}{r_A}-\frac{1}{r_B})}$
We plug in the known values to obtain:
$v_A=\sqrt{(0.905m/s)^2+4(6.67\times 10^{-11}Nm^2/Kg^2)(3.5\times 10^{11}Kg[\frac{1}{3.35\times 10^3m}-\frac{1}{1.5\times 10^3m}]}$
This simplifies to:
$v_A=0.885m/s$
