At what depth from earth's surface does the acceleration due to gravity becomes `1/4` times that of its value at surface ?

Answer

To find the depth from the Earth's surface at which the acceleration due to gravity becomes \( \frac{1}{4} \) times its value at the surface, we can follow these steps:

### Step 1: Understand the relationship between gravity at depth and at the surface
The acceleration due to gravity at a depth \( d \) below the Earth's surface is given by the formula:
\[ g_d = g \left(1 - \frac{d}{R}\right) \]
where:
- \( g_d \) is the acceleration due to gravity at depth \( d \),
- \( g \) is the acceleration due to gravity at the surface,
- \( R \) is the radius of the Earth.

### Step 2: Set up the equation
We want to find the depth \( d \) where \( g_d = \frac{1}{4} g \). Therefore, we can set up the equation:
\[ g \left(1 - \frac{d}{R}\right) = \frac{1}{4} g \]

### Step 3: Simplify the equation
We can cancel \( g \) from both sides (assuming \( g \neq 0 \)):
\[ 1 - \frac{d}{R} = \frac{1}{4} \]

### Step 4: Solve for \( \frac{d}{R} \)
Rearranging the equation gives:
\[ \frac{d}{R} = 1 - \frac{1}{4} \]
\[ \frac{d}{R} = \frac{3}{4} \]

### Step 5: Solve for \( d \)
Now, we can find \( d \) by multiplying both sides by \( R \):
\[ d = \frac{3}{4} R \]

### Step 6: Conclusion
Thus, the depth from the Earth's surface at which the acceleration due to gravity becomes \( \frac{1}{4} \) times that at the surface is:
\[ d = \frac{3}{4} R \]