Question

A solid sphere of mass M and radius R is placed inside a spherical shell of mass M and radius 4R such that their surfaces touch each other. The gravitational force due to the spherical shell and solid sphere on a body of unit mass placed at the centre of the spherical shell is (G = Universal gravitational constant)

• A. $\frac{25GM}{144R^2}$
• B. $\frac{GM}{9R^2}$
• C. $\frac{7GM}{144R^2}$
• D. Zero

Hint:

Key Shell Theorem points: 1. Field inside a shell is zero. 2. Field outside a spherical distribution is $\frac{GM}{r^2}$ (point mass approx).

Answer 1

The Correct Option is B

Step 1: Understanding Gravitational Fields:
We need to find the total gravitational field (force on unit mass) at the center of the spherical shell (let's call this point $O$). The total field is the vector sum of the field due to the shell ($\vec{E}_{shell}$) and the field due to the solid sphere ($\vec{E}_{sphere}$).
Step 2: Field due to the Shell:
According to the Shell Theorem, the gravitational field inside a uniform spherical shell is zero everywhere. \[ E_{shell} = 0 \] Thus, the shell contributes no force on the mass at its center.
Step 3: Field due to the Solid Sphere:
The solid sphere has radius $R$ and is placed inside the shell of radius $4R$ touching its surface.

• Let the center of the shell be $O$.
• Let the center of the solid sphere be $C$.
• Since the solid sphere (radius $R$) touches the inner surface of the shell (radius $4R$), the distance between their centers is $d = 4R - R = 3R$.

The test mass is at $O$. We need to calculate the field due to the solid sphere at point $O$. Since the distance $d = 3R$ is greater than the radius of the solid sphere ($R$), point $O$ is external to the solid sphere. For external points, a sphere behaves as if its entire mass is concentrated at its center.
Step 4: Calculation:
Field due to solid sphere at distance $d$: \[ E_{sphere} = \frac{GM}{d^2} \] Substituting $d = 3R$: \[ E_{sphere} = \frac{GM}{(3R)^2} = \frac{GM}{9R^2} \]
Step 5: Total Field:
\[ E_{total} = E_{shell} + E_{sphere} = 0 + \frac{GM}{9R^2} = \frac{GM}{9R^2} \]