Question

Two balls, each of radius $R$, with equal mass and density are placed in contact, then the force of gravitation between them is proportional to

• A. $\begin{array}{l}F\propto \frac{1}{R^2}\end{array}$
• B. $\begin{array}{l}F\propto R\end{array}$
• C. $\begin{array}{l}F\propto R^4\end{array}$
• D. $\begin{array}{l}F\propto \frac{1}{R}\end{array}$

Answer 1

The correct option is C

$\begin{array}{l}F\propto R^4\end{array}$

Step 1: Given data

The radius of each ball $=R$

Their equal masses $=m$

Their equal density $=\rho$

Therefore their equal volumes $=V$

Universal gravitational constant $=G$

The distance between the centers of the two balls is $=R+R=2R$

Step 2: Formula used

Density $\rho =\frac{m}{V}$

Gravitational force $F=G\frac{m_1{m}_2}{R^2}$ where $m=m_1=m_2$

Step 3: Calculate the mass

We know that the density is given as,

$$\begin{gathered} \rho =\frac{m}{V} \\ =\frac{m}{{\displaystyle \frac{4}{3}}{\mathrm{\pi R}}^3} \\ \therefore m=\rho \times \frac{4}{3}{\mathrm{\pi R}}^3 \end{gathered}$$

Step 4: Calculate the Gravitational force

Therefore, the gravitational force between the two balls can be given as,

$$\begin{gathered} F=G\frac{{\left({\displaystyle \frac{4}{3}}{\mathrm{\pi R}}^3\mathrm{\rho }\right)}^2}{{\left(2R\right)}^2} \\ =G\frac{{\left(\frac{4}{3}\pi \mathrm{\rho }\right)}^2{R}^6}{4R^2} \\ =G\frac{{\left(\frac{4}{3}\pi \mathrm{\rho }\right)}^2{R}^4}{4} \\ \therefore F\propto R^4 \end{gathered}$$

Hence, the correct answer is an option (C).