Chapter 12 - Gravity - Problems and Conceptual Exercises - Page 411: 50

$v_B=0.185m/s$

We can find the required speed as follows: $r_A=\sqrt{(3000m)^2+(1500)^2}=3350m$ We know that $v_B^2=4GM(\frac{1}{r_B}-\frac{1}{r_A})$ $\implies v_B=\sqrt{4GM(\frac{1}{r_B}-\frac{1}{r_A})}$ We plug in the known values to obtain: $v_B=\sqrt{4(6.67\times 10^{-11})(3.50\times 10^{11})(\frac{1}{1500}-\frac{1}{3350})}$ $v_B=0.185m/s$