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

Options

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

Answer

To solve the problem, we need to analyze the change in the acceleration due to gravity (g) at a height \( h \) above the Earth's surface and at a depth \( x \) below the Earth's surface. We are given that these changes are equal.

### Step-by-Step Solution:

1. **Define the variables**:
- Let \( g \) be the acceleration due to gravity at the surface of the Earth.
- Let \( g_h \) be the acceleration due to gravity at height \( h \).
- Let \( g_x \) be the acceleration due to gravity at depth \( x \).

2. **Acceleration due to gravity at height \( h \)**:
- The formula for the acceleration due to gravity at height \( h \) is given by:
\[ g_h = \frac{g R^2}{(R + h)^2} \]
- For \( h \) much smaller than the radius of the Earth \( R \), we can use the binomial approximation:
\[ g_h \approx g \left(1 - \frac{2h}{R}\right) \]
- Therefore, the change in \( g \) at height \( h \) is:
\[ g - g_h \approx g - g \left(1 - \frac{2h}{R}\right) = \frac{2gh}{R} \]

3. **Acceleration due to gravity at depth \( x \)**:
- The formula for the acceleration due to gravity at depth \( x \) is:
\[ g_x = g \left(1 - \frac{x}{R}\right) \]
- The change in \( g \) at depth \( x \) is:
\[ g - g_x = g - g \left(1 - \frac{x}{R}\right) = \frac{gx}{R} \]

4. **Set the changes equal**:
- According to the problem, the change in \( g \) at height \( h \) is equal to the change in \( g \) at depth \( x \):
\[ \frac{2gh}{R} = \frac{gx}{R} \]

5. **Cancel \( g \) and \( R \)**:
- Since \( g \) and \( R \) are constants and non-zero, we can cancel them from both sides:
\[ 2h = x \]

6. **Final Result**:
- Rearranging gives us:
\[ x = 2h \]
- This means that the depth \( x \) is twice the height \( h \).

### Conclusion:
The required relationship is \( x = 2h \).