Radius of moon is $$\frac{1}{4}$$ times that of earth and mass is $$\frac{1}{81}$$ times that of earth. The point at which gravitational field due to earth becomes equal and opposite to that of moon, is (Distance between centres of earth and moon is $$60R,$$  where $$R$$ is radius of earth)

Solution :

$$\eqalign{ & {E_{{\text{earth}}}} = {E_{{\text{moon}}}} \cr & \Rightarrow \frac{{GM}}{{{x^2}}} = \frac{{\frac{{GM}}{{81}}}}{{{{\left( {60R - x} \right)}^2}}} \cr & \Rightarrow \frac{1}{x} = \frac{1}{{9\left( {60R - x} \right)}} \cr & \Rightarrow x = 54\,R\,{\text{from centre of earth}}{\text{.}} \cr} $$