If the change in the value of `g` at a height `h` above the surface of earth is the same as at a depth `d` below it (both `h` and `d` are much smaller than the radius of the earth), then

Options

A. h = d
B. 2h = d
C. h = 2d
D. `h^(2) = d`

Answer

To solve the problem, we need to find the relationship between the height \( h \) above the Earth's surface and the depth \( d \) below the Earth's surface, given that the change in the value of \( g \) at height \( h \) is the same as at depth \( d \).

### Step-by-Step Solution:

1. **Understanding the change in \( g \) at height \( h \)**:
The formula for the acceleration due to gravity at a height \( h \) above the Earth's surface is given by:
\[ g_h = \frac{g}{(1 + \frac{h}{R})^2} \]
where \( R \) is the radius of the Earth. For small \( h \) compared to \( R \), we can use the binomial expansion:
\[ g_h \approx g \left(1 - 2\frac{h}{R}\right) \]
Therefore, the change in \( g \) at height \( h \) is:
\[ \Delta g_h = g - g_h = g - g \left(1 - 2\frac{h}{R}\right) = 2\frac{h}{R}g \]

2. **Understanding the change in \( g \) at depth \( d \)**:
The formula for the acceleration due to gravity at a depth \( d \) below the Earth's surface is given by:
\[ g_d = g \left(1 - \frac{d}{R}\right) \]
Thus, the change in \( g \) at depth \( d \) is:
\[ \Delta g_d = g - g_d = g - g \left(1 - \frac{d}{R}\right) = \frac{d}{R}g \]

3. **Setting the changes equal**:
According to the problem, the changes in \( g \) at height \( h \) and at depth \( d \) are equal:
\[ \Delta g_h = \Delta g_d \]
Substituting the expressions we derived:
\[ 2\frac{h}{R}g = \frac{d}{R}g \]

4. **Cancelling common terms**:
Since \( g \) and \( R \) are non-zero, we can cancel them from both sides:
\[ 2h = d \]

5. **Final relationship**:
Rearranging gives us the relationship:
\[ d = 2h \]

### Conclusion:
The relationship between the height \( h \) above the Earth's surface and the depth \( d \) below the Earth's surface is:
\[ d = 2h \]