Acceleration due to gravity is same when an object is at height `R/2` from surface of earth and at depth 'd' below surface of earth. Find `d/R`

Answer

To solve the problem, we need to find the ratio \( \frac{d}{R} \) where \( d \) is the depth below the surface of the Earth and \( R \) is the radius of the Earth. We know that the acceleration due to gravity is the same at a height \( \frac{R}{2} \) above the surface and at a depth \( d \) below the surface.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- We need to find the ratio \( \frac{d}{R} \) where the gravitational acceleration at height \( \frac{R}{2} \) is equal to the gravitational acceleration at depth \( d \).

2. **Gravitational Acceleration at Height \( \frac{R}{2} \):**
- The formula for gravitational acceleration at a height \( h \) above the Earth's surface is given by:
\[ g_h = g \left( \frac{R}{R + h} \right)^2 \]
- Here, \( h = \frac{R}{2} \), so substituting this value:
\[ g_h = g \left( \frac{R}{R + \frac{R}{2}} \right)^2 = g \left( \frac{R}{\frac{3R}{2}} \right)^2 = g \left( \frac{2}{3} \right)^2 = g \cdot \frac{4}{9} \]

3. **Gravitational Acceleration at Depth \( d \):**
- The formula for gravitational acceleration at a depth \( d \) below the Earth's surface is given by:
\[ g_d = g \left( 1 - \frac{d}{R} \right) \]

4. **Setting the Two Accelerations Equal:**
- Since the problem states that the gravitational acceleration at height \( \frac{R}{2} \) is equal to that at depth \( d \):
\[ g \cdot \frac{4}{9} = g \left( 1 - \frac{d}{R} \right) \]

5. **Cancelling \( g \) from Both Sides:**
- We can cancel \( g \) (assuming \( g \neq 0 \)):
\[ \frac{4}{9} = 1 - \frac{d}{R} \]

6. **Solving for \( \frac{d}{R} \):**
- Rearranging the equation gives:
\[ \frac{d}{R} = 1 - \frac{4}{9} = \frac{5}{9} \]

7. **Final Result:**
- Therefore, the ratio \( \frac{d}{R} \) is:
\[ \frac{d}{R} = \frac{5}{9} \]