At what depth below the surface of earth the acceleration due to gravity becomes one-third of its value at the surface of earth ?

Answer

To find the depth below the surface of the Earth where the acceleration due to gravity becomes one-third of its value at the surface, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the acceleration due to gravity at the surface**:
The acceleration due to gravity at the surface of the Earth is given by the formula:
\[ g = \frac{GM}{R^2} \]
where \( G \) is the universal gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

2. **Acceleration due to gravity at depth \( d \)**:
When we go to a depth \( d \) below the surface, the acceleration due to gravity \( g' \) can be expressed as:
\[ g' = g \left(1 - \frac{d}{R}\right) \]
This formula comes from the fact that only the mass of the sphere of radius \( R - d \) contributes to the gravitational force at that depth.

3. **Set up the equation for \( g' \)**:
We want to find the depth \( d \) where the acceleration due to gravity \( g' \) is one-third of \( g \):
\[ g' = \frac{g}{3} \]
Substituting this into our equation gives:
\[ \frac{g}{3} = g \left(1 - \frac{d}{R}\right) \]

4. **Simplify the equation**:
Dividing both sides by \( g \) (assuming \( g \neq 0 \)):
\[ \frac{1}{3} = 1 - \frac{d}{R} \]

5. **Rearranging the equation**:
Rearranging gives:
\[ \frac{d}{R} = 1 - \frac{1}{3} = \frac{2}{3} \]

6. **Solve for \( d \)**:
Thus, we can express \( d \) as:
\[ d = \frac{2}{3} R \]

7. **Substituting the radius of the Earth**:
The average radius of the Earth \( R \) is approximately \( 6400 \) km. Therefore:
\[ d = \frac{2}{3} \times 6400 \text{ km} = \frac{12800}{3} \text{ km} \approx 4266.67 \text{ km} \]

### Final Answer:
The depth below the surface of the Earth where the acceleration due to gravity becomes one-third of its value at the surface is approximately **4266.67 km**.