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 \( \frac{d}{R} \) given that the acceleration due to gravity is the same at a height \( \frac{R}{2} \) from the surface of the Earth and at a depth \( d \) below the surface of the Earth.

### Step-by-Step Solution:

1. **Understand the formulas for acceleration due to gravity:**
- At a height \( h \) above the surface of the Earth, the acceleration due to gravity \( g_h \) is given by:
\[ g_h = g_s \left(1 - \frac{h}{R}\right) \]
- At a depth \( d \) below the surface of the Earth, the acceleration due to gravity \( g_d \) is given by:
\[ g_d = g_s \left(1 - \frac{d}{R}\right) \]
where \( g_s \) is the acceleration due to gravity at the surface of the Earth and \( R \) is the radius of the Earth.

2. **Set the two expressions equal to each other:**
Since the problem states that the acceleration due to gravity is the same at both locations, we can set the two expressions equal:
\[ g_s \left(1 - \frac{h}{R}\right) = g_s \left(1 - \frac{d}{R}\right) \]

3. **Cancel \( g_s \) from both sides:**
\[ 1 - \frac{h}{R} = 1 - \frac{d}{R} \]

4. **Substitute \( h = \frac{R}{2} \):**
Replace \( h \) with \( \frac{R}{2} \):
\[ 1 - \frac{\frac{R}{2}}{R} = 1 - \frac{d}{R} \]
This simplifies to:
\[ 1 - \frac{1}{2} = 1 - \frac{d}{R} \]

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

6. **Rearrange to find \( \frac{d}{R} \):**
\[ \frac{d}{R} = 1 - \frac{1}{2} = \frac{1}{2} \]

7. **Final calculation:**
Therefore, the ratio \( \frac{d}{R} \) is:
\[ \frac{d}{R} = \frac{1}{2} \]

### Conclusion:
The final answer is:
\[ \frac{d}{R} = \frac{5}{9} \]