Question

As shown in the figure, two spherical cavities are made in a uniform solid sphere of radius \(R\). The boundaries of the cavities touch at the centre of the sphere. The centers of the cavities and the sphere lie on the \(x\)-axis. The mass of the solid sphere before the cavities were created was \(M\). The gravitational force on a point mass \(m\) at a distance \(d\) away from the center of the solid sphere is

• A. \(\frac{GMm}{d^2}\left[1-\frac{1}{8}\frac{1}{\left(1+\frac{R}{2d}\right)^2}-\frac{1}{8}\frac{1}{\left(1-\frac{R}{2d}\right)^2}\right]\)
• B. \(\frac{GMm}{d^2}\left[1-\frac{1}{8}\frac{1}{\left(1+\frac{R}{d}\right)^2}-\frac{1}{8}\frac{1}{\left(1-\frac{R}{d}\right)^2}\right]\)
• C. \(\frac{GMm}{d^2}\left[1-\frac{1}{8}\frac{1}{\left(1+\frac{d}{R}\right)^2}-\frac{1}{8}\frac{1}{\left(1-\frac{d}{R}\right)^2}\right]\)
• D. \(\frac{GMm}{d^2}\left[1-\frac{1}{8}\frac{1}{\left(1+\frac{d}{R}\right)^2}+\frac{1}{8}\frac{1}{\left(1-\frac{d}{R}\right)^2}\right]\)

Hint:

For cavities inside a uniform sphere, use superposition: \[ \text{Field of remaining body} = \text{Field of complete sphere} - \text{Field of removed parts}. \] Also, if the cavity radius is \(\frac{R}{2}\), then its mass is \[ \frac{M}{8} \] because mass is proportional to volume.

Answer 1

The Correct Option is A

Step 1: Use superposition — treat cavities as negative mass spheres.
Each cavity has radius \( R/2 \), so volume \( = \frac{1}{8}\) of sphere. Mass of each cavity \( = M/8 \). Centers are at \( x = \pm R/2 \).

Step 2: Net force = force from full sphere minus two cavity contributions.
\( F = \frac{GMm}{d^2} - \frac{G(M/8)m}{(d-R/2)^2} - \frac{G(M/8)m}{(d+R/2)^2} \)
\( = \frac{GMm}{d^2}\left[1 - \frac{1/8}{(1-R/2d)^2} - \frac{1/8}{(1+R/2d)^2}\right] \)

\[ \boxed{F = \frac{GMm}{d^2}\left[1 - \frac{1/8}{(1-\frac{R}{2d})^2} - \frac{1/8}{(1+\frac{R}{2d})^2}\right]} \]