A uniform solid sphere of radius R produces a gravitational acceleration `a_g` on its surface. At what two distances from the centre of the sphere the acceleration due to gravity is `a_g // 4` ?

Options

A. 4R, 0.50R
B. 2R, 0.25R
C. 3R, 0.33R
D. 2R, 0.50R

Explanation

To solve the problem of finding the distances from the center of a uniform solid sphere where the gravitational acceleration is \( \frac{a_g}{4} \), we will consider two scenarios: one at a height above the surface and another at a depth below the surface.

### Step-by-Step Solution:

1. **Understanding Gravitational Acceleration on the Surface:**
The gravitational acceleration \( a_g \) at the surface of a uniform solid sphere of radius \( R \) is given by:
\[ a_g = \frac{GM}{R^2} \]
where \( G \) is the gravitational constant and \( M \) is the mass of the sphere.

2. **Finding Height Above the Surface:**
To find the height \( h \) above the surface where the gravitational acceleration is \( \frac{a_g}{4} \):
\[ \frac{a_g}{4} = \frac{GM}{(R + h)^2} \]
Setting the two expressions for \( a_g \) equal:
\[ \frac{GM}{(R + h)^2} = \frac{1}{4} \cdot \frac{GM}{R^2} \]
Cancel \( GM \) from both sides:
\[ \frac{1}{(R + h)^2} = \frac{1}{4R^2} \]
Taking the reciprocal:
\[ (R + h)^2 = 4R^2 \]
Taking the square root:
\[ R + h = 2R \]
Thus:
\[ h = 2R - R = R \]
Therefore, the distance from the center of the sphere is:
\[ d_1 = R + h = R + R = 2R \]

3. **Finding Depth Below the Surface:**
Now, we consider the depth \( d \) below the surface where the gravitational acceleration is \( \frac{a_g}{4} \):
The formula for gravitational acceleration at depth \( d \) is:
\[ g_d = g \left(1 - \frac{d}{R}\right) \]
Setting this equal to \( \frac{a_g}{4} \):
\[ \frac{a_g}{4} = a_g \left(1 - \frac{d}{R}\right) \]
Dividing both sides by \( a_g \):
\[ \frac{1}{4} = 1 - \frac{d}{R} \]
Rearranging gives:
\[ \frac{d}{R} = 1 - \frac{1}{4} = \frac{3}{4} \]
Thus:
\[ d = \frac{3}{4}R \]
The distance from the center of the sphere is:
\[ d_2 = R - d = R - \frac{3}{4}R = \frac{1}{4}R \]

### Final Distances:
- The two distances from the center of the sphere where the gravitational acceleration is \( \frac{a_g}{4} \) are:
- \( d_1 = 2R \) (above the surface)
- \( d_2 = \frac{1}{4}R \) (below the surface)