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 the principle of superposition.
The given body can be treated as a complete solid sphere of mass \(M\), from which two smaller spherical masses are removed.
So, the gravitational force due to the remaining body is \[ F=F_{\text{complete sphere}}-F_{\text{left cavity}}-F_{\text{right cavity}}. \]

Step 2: Find the radius and mass of each cavity.
Since the two spherical cavities touch at the centre of the original sphere and also fit symmetrically inside the sphere, each cavity has radius \[ \frac{R}{2}. \] Since mass is proportional to volume, \[ \frac{M_{\text{cavity}}}{M} = \frac{\left(\frac{R}{2}\right)^3}{R^3}. \] \[ M_{\text{cavity}}=\frac{M}{8}. \]

Step 3: Write the force due to the complete sphere.
The point mass \(m\) is at distance \(d\) from the centre of the original sphere.
Therefore, the force due to the complete sphere is \[ F_0=\frac{GMm}{d^2}. \]

Step 4: Write the distances from the cavity centres.
The centres of the two cavities are at distances \[ \frac{R}{2} \] on either side of the centre of the original sphere.
Therefore, distances of the point mass from the two cavity centres are \[ d+\frac{R}{2} \] and \[ d-\frac{R}{2}. \]

Step 5: Subtract the forces due to removed masses.
Force due to the left removed sphere is \[ F_1=\frac{G\left(\frac{M}{8}\right)m}{\left(d+\frac{R}{2}\right)^2}. \] Force due to the right removed sphere is \[ F_2=\frac{G\left(\frac{M}{8}\right)m}{\left(d-\frac{R}{2}\right)^2}. \] Hence, \[ F=\frac{GMm}{d^2} - \frac{GMm}{8\left(d+\frac{R}{2}\right)^2} - \frac{GMm}{8\left(d-\frac{R}{2}\right)^2}. \]

Step 6: Take \(\frac{GMm}{d^2}\) common.
\[ F=\frac{GMm}{d^2} \left[ 1-\frac{1}{8}\frac{d^2}{\left(d+\frac{R}{2}\right)^2} -\frac{1}{8}\frac{d^2}{\left(d-\frac{R}{2}\right)^2} \right]. \] Now, \[ \frac{d^2}{\left(d+\frac{R}{2}\right)^2} = \frac{1}{\left(1+\frac{R}{2d}\right)^2}, \] and \[ \frac{d^2}{\left(d-\frac{R}{2}\right)^2} = \frac{1}{\left(1-\frac{R}{2d}\right)^2}. \] Therefore, \[ F= \frac{GMm}{d^2} \left[ 1-\frac{1}{8}\frac{1}{\left(1+\frac{R}{2d}\right)^2} -\frac{1}{8}\frac{1}{\left(1-\frac{R}{2d}\right)^2} \right]. \]

Step 7: Final conclusion.
Hence, the gravitational force is \[ \boxed{ \frac{GMm}{d^2} \left[ 1-\frac{1}{8}\frac{1}{\left(1+\frac{R}{2d}\right)^2} -\frac{1}{8}\frac{1}{\left(1-\frac{R}{2d}\right)^2} \right] } \] Therefore, the correct option is \[ \boxed{(1)} \]