In the figure masses $400 kg$ and $100kg$ are fixed. What is the minimum kinetic energy with which the $1kg$ mass must be projected from $A$to the right to reach the point $B$ ?

First let us find the point (say) $C$ between $A$ and $B$ where field strength due to $400kg$ and $100kg$ masses is zero. Let its distance from $400kg$ is $r$.

Then, $\frac{G\left(400\right)}{r^2}=\frac{G\left(100\right)}{(10-r{)}^2}$

                         $r=\frac{20}{3}m$

and       $(10-r)=\frac{10}{3}m$

So, we have to move the body only from $A$ to $C$. After that $100kg$ mass will pull $1kg$ mass by its own.

$\therefore$ Minimum kinetic energy $=\Delta U=U_C-U_A$

                                         $=m\left(V_c-V_A\right)$

                                         $=\left(1\right)\left[-\frac{400}{20/3}-\frac{100}{10/3}+\frac{100}{8}+\frac{400}{2}\right]\left(6.67\times {10}^{-11}\right)$  

                                                  $=8.17\times {10}^{-4}J$