Question

Two identical solid copper spheres of radius $R$ are placed in contact with each other. The gravitational attraction between them is proportional to

• A. $R^2$
• B. $R^{-2}$
• C. $R^{-4}$
• D. $R^4$

Answer 1

The correct option is D $R^4$

Mass of each sphere $M_1=M_2=\rho \left(\frac{4}{3}\pi R^3\right)$
Gravitational force between them is given by $F=\frac{G{M}_1{M}_2}{(2R{)}^2}$
$F=\frac{G\times \frac{4}{3}\pi R^3\rho \times \frac{4}{3}\pi R^3\rho }{4R^2}=\frac{4}{9}{\pi }^{2}G{\rho }^2{R}^4$
So $F\propto R^4$.