At what depth below the surface of earth, value of accelaration due to gravity is same as the value at height `h =R1, where R is the radius of earth.

Explanation

To solve the problem of finding the depth below the surface of the Earth where the acceleration due to gravity is the same as the value at a height equal to the radius of the Earth, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**: We need to find the depth \( d \) below the Earth's surface where the acceleration due to gravity \( g' \) is equal to the acceleration due to gravity at a height \( h \) equal to the radius of the Earth \( R \).

2. **Acceleration Due to Gravity Inside the Earth**: The formula for the acceleration due to gravity at a depth \( d \) inside the Earth is given by:
\[ g' = g \left(1 - \frac{d}{R}\right) \]
where \( g \) is the acceleration due to gravity at the surface of the Earth.

3. **Acceleration Due to Gravity at Height**: The formula for the acceleration due to gravity at a height \( h \) above the Earth is:
\[ g' = \frac{g}{(1 + \frac{h}{R})^2} \]
Since we are interested in the case where \( h = R \), we substitute \( h \) with \( R \):
\[ g' = \frac{g}{(1 + 1)^2} = \frac{g}{2^2} = \frac{g}{4} \]

4. **Setting the Two Expressions Equal**: We need to set the two expressions for \( g' \) equal to each other:
\[ g \left(1 - \frac{d}{R}\right) = \frac{g}{4} \]

5. **Cancelling \( g \)**: Since \( g \) is a common factor on both sides, we can cancel it (assuming \( g \neq 0 \)):
\[ 1 - \frac{d}{R} = \frac{1}{4} \]

6. **Solving for \( d \)**: Rearranging the equation gives:
\[ \frac{d}{R} = 1 - \frac{1}{4} = \frac{3}{4} \]
Therefore, multiplying both sides by \( R \):
\[ d = \frac{3}{4} R \]

### Final Answer:
The depth \( d \) below the surface of the Earth at which the acceleration due to gravity is the same as that at a height equal to the radius of the Earth is:
\[ d = \frac{3}{4} R \]