Question

A uniform solid sphere of radius \(R\) produces a gravitational acceleration of \(a_0\) on its surface. The distance of the point from the centre of the sphere where the gravitational acceleration becomes \[ \frac{a_0}{4} \] is

• A. \(4R\)
• B. \(\dfrac{3R}{2}\)
• C. \(2R\)
• D. \(3R\)

Hint:

Outside a spherical body, gravitational acceleration varies inversely as the square of the distance: \[ g\propto \frac{1}{r^2} \]

Answer 1

The Correct Option is C

Step 1: Write the surface gravity.
For a uniform sphere of mass $M$ and radius $R$, the gravitational acceleration on its surface is \[ a_0 = \frac{GM}{R^2} \] We want the distance from the centre where the value drops to $\dfrac{a_0}{4}$.
Step 2: Decide inside or outside.
Since the acceleration is decreasing below the surface value, we are looking at a point outside the sphere, where the whole mass acts as if concentrated at the centre.
Step 3: Write the field at distance $r$ outside.
\[ a = \frac{GM}{r^2} \]
Step 4: Impose the given condition.
We need $a = \dfrac{a_0}{4}$, so \[ \frac{GM}{r^2} = \frac{1}{4}\cdot\frac{GM}{R^2} \]
Step 5: Cancel and simplify.
The common factor $GM$ cancels: \[ \frac{1}{r^2} = \frac{1}{4R^2} \] so \[ r^2 = 4R^2 \]
Step 6: Take the positive root.
\[ r = 2R \] Therefore the gravitational acceleration falls to a quarter at a distance $2R$ from the centre, matching option (3). \[ \boxed{2R} \]