Answer
$7.39\times 10^{-5}m/s$
Work Step by Step
We can find the required speed as follows:
$U_i=3(-\frac{Gm^2}{r})$
$\implies U_i=3(-6.67\times 10^{-11}(\frac{5.95}{10.0}))=-7.08\times 10^{-10}J$
and $U_f=3(-6.67\times 10^{-11})(\frac{(5.95)^2}{0.143})=-4.95\times 10^{-8}J$
Now $K.E_i+U_i=K.E_f+U_f$
$\implies K.E_f=U_i-U_f+0$
$K.E_f=(-7.08\times 10^{-10})-(-4.95\times 10^{-8})$
$K.E_f=4.88\times 10^{-8}J$
We know that
$v^2=\frac{2K.E_f}{3m}$
This simplifies to:
$v=\sqrt{\frac{2K.E_f}{3m}}$
We plug in the known values to obtain:
$v=\sqrt{\frac{2(4.88\times 10^{-8})}{3(5.95)}}$
$v=7.39\times 10^{-5}m/s$
