Question

As shown in the figure, two spherical cavities are made in the 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:

Use the principle of superposition: force due to body with cavity = force due to complete body - force due to mass in cavity.
Gravitational force by a sphere of mass $M_s$ on a point mass $m$ outside it at distance $r$ from its center is $G M_s m / r^2$.
Calculate the mass of the material that would fill the cavity based on the density of the original sphere.
Be careful with distances from the centers of the cavities to the point mass.

Answer 1

The Correct Option is A

Problem Summary: A solid sphere of mass \( M \) and radius \( R \) has two spherical cavities (each of radius \( R/2 \)) carved such that they touch at the center. A point mass \( m \) lies on the axis of symmetry at a distance \( d \) from the center of the sphere (\( d > R \)). Find the net gravitational force on \( m \).

Step 1: Use Superposition Principle

The net gravitational force is:

• The force due to the original full solid sphere
• Minus the force due to the two missing cavity masses (as if they were present and exerting gravitational pull)

Step 2: Compute Density and Cavity Mass

Density of original sphere:

\[ \rho = \frac{M}{\frac{4}{3}\pi R^3} \]

Volume of each cavity \( V_{\text{cav}} = \frac{4}{3}\pi \left(\frac{R}{2}\right)^3 = \frac{1}{8} \cdot \frac{4}{3}\pi R^3 \)

Mass of each cavity if it were filled:

\[ M_{\text{cav}} = \rho \cdot V_{\text{cav}} = \frac{M}{8} \]

Step 3: Gravitational Forces

1. Force from original sphere: \[ F_{\text{orig}} = \frac{GMm}{d^2} \]
2. Force from first cavity at \( x_1 = -\frac{R}{2} \): \[ d_1 = d + \frac{R}{2}, \quad F_{\text{fill1}} = \frac{G(M/8)m}{(d + R/2)^2} \]
3. Force from second cavity at \( x_2 = +\frac{R}{2} \): \[ d_2 = d - \frac{R}{2}, \quad F_{\text{fill2}} = \frac{G(M/8)m}{(d - R/2)^2} \]

Step 4: Net Force

By superposition:

\[ F_{\text{net}} = \frac{GMm}{d^2} - \frac{GMm}{8(d + R/2)^2} - \frac{GMm}{8(d - R/2)^2} \]

Factor \( \frac{GMm}{d^2} \):

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

Convert into normalized form:

\[ F_{\text{net}} = \frac{GMm}{d^2} \left[ 1 - \frac{1}{8\left(1 + \frac{R}{2d}\right)^2} - \frac{1}{8\left(1 - \frac{R}{2d}\right)^2} \right] \]

✅ Final Answer:

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

This matches option (a).