Answer
The speed of body B relative to body A is $~~\sqrt{\frac{2Gm}{R_i}}$
Work Step by Step
We can use conservation of energy to find the kinetic energy of body B:
$K_f+U_f = K_i+U_i$
$K_f = 0+U_i - U_f$
$K_f = U_i - U_f$
$K_f = (-\frac{Gm^2}{R_i}) - (-\frac{Gm^2}{0.5~R_i})$
$K_f = \frac{Gm^2}{R_i}$
The kinetic energy of body B is $~~\frac{Gm^2}{R_i}$
We can find the speed of body B relative to body A:
$K = \frac{1}{2}mv^2 = \frac{Gm^2}{R_i}$
$v^2 = \frac{2Gm}{R_i}$
$v = \sqrt{\frac{2Gm}{R_i}}$
The speed of body B relative to body A is $~~\sqrt{\frac{2Gm}{R_i}}$
