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 surface touching, what is the contact force between the objects due to their traditional 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.
  $\frac{128}{87}{\pi }^2{R}^{2}G$

Answer 1

The correct option is B

$\frac{128}{81}G{\pi }^2{R}^4{\rho }^2$
The arrangement of the given spherical objects is shown below.


Let $m_1$ and $m_2$ be the masses of the smaller and larger spheres, respectively.

Using mass- density relation,

$m_1=\frac{4}{3}\pi R^3\rho$, and $m_2=\frac{4}{3}\pi (2R{)}^3\rho$

Free body diagram of mass $m_1$ is given below.


In the figure, $F_g$ is gravitational force of attraction on $m_1$ due to $m_2$ and $N$ is normal reaction on $m_1$.

Since, $m_1$ is in equilibrium, so from FBD, we get

$N=F_g=\frac{G{m}_1{m}_2}{r^2}$

Substituting the values, from figure

$\Rightarrow F_g=\frac{G\times (\frac{4}{3}\pi R^3\rho )(\frac{4}{3}\times 8\pi R^3\rho )}{(3R{)}^2}$

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

Hence, option $\left(b\right)$ is the correct choice.