Question

Consider two solid uniform spherical objects of the same density $\rho$. One has radius $R$ and the other has radius $2R$. They are in outer space where the gravitational fields from other objects are negligible. If they are arranged with their surfaces touching, what is the contact force between the objects due to their gravitational attraction?

• A. $G{\pi }^2{R}^4$
• B. $\frac{128}{81}G{\pi }^2{R}^4{\rho }^2$
• C. $\frac{128}{81}G{\pi }^2$
• D. $\frac{128}{81}{\pi }^2{R}^{4}G$

Answer 1

The correct option is B $\frac{128}{81}G{\pi }^2{R}^4{\rho }^2$

We know that, a Gravitational force is a non-contact, attractive force. For the situation given in the question to calculate force at a point outside, uniform spherical mass can be considered as a point mass placed at its center.


Free body diagram of $m_1$:


Since $m_1$ is in equlibrium, so the reaction on $m_1$, $$N=F_g$$
Mass of sphere, $m_1=\frac{4}{3}\pi R^3\rho$

Mass of sphere, $m_2=\frac{4}{3}\pi (2R{)}^3\rho$

Distance between the centres $=3R$

So, the force acting between the masses,$$F_g=\frac{G{m}_1{m}_2}{(3R{)}^2}$$
Substituting the values of $m_1$ and $m_2$,

$\Rightarrow F_g=\frac{G\times \frac{4}{3}\pi R^3\rho \times \frac{4}{3}\pi (2R{)}^3\rho }{(3R{)}^2}$

$\Rightarrow F_g=\frac{128}{81}G{\pi }^2{R}^4{\rho }^2$

Hence, option (b) is the correct answer.

Why this question? Note: For calculation of gravitational force at a point outside, a uniform spherical mass can be considered to be a point mass placed at its center.