The acceleration due to gravity at a height `(1//20)^(th)` the radius of the earth above earth s surface is `9m//s^(2)` Find out its approximate value at a point at an equal distance below the surface of the earth .

Options

A. (a)8.5
B. (b)9.5
C. (c)9.8
D. (d)11.5

Answer

To solve the problem, we need to find the acceleration due to gravity at a point below the Earth's surface, given that the acceleration due to gravity at a height of \( \frac{1}{20} \) of the Earth's radius above the surface is \( 9 \, \text{m/s}^2 \).

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Acceleration due to gravity at height \( h = \frac{R}{20} \) (where \( R \) is the radius of the Earth) is \( g' = 9 \, \text{m/s}^2 \).

2. **Use the Formula for Acceleration due to Gravity at Height:**
The formula for acceleration due to gravity at a height \( h \) above the Earth's surface is given by:
\[ g' = g \left( \frac{R}{R + h} \right)^2 \]
where \( g \) is the acceleration due to gravity at the Earth's surface (approximately \( 10 \, \text{m/s}^2 \)).

3. **Substitute the Values:**
Here, \( h = \frac{R}{20} \):
\[ g' = g \left( \frac{R}{R + \frac{R}{20}} \right)^2 \]
Simplifying this:
\[ g' = g \left( \frac{R}{R \left(1 + \frac{1}{20}\right)} \right)^2 = g \left( \frac{1}{1 + \frac{1}{20}} \right)^2 = g \left( \frac{1}{\frac{21}{20}} \right)^2 = g \left( \frac{20}{21} \right)^2 \]

4. **Set Up the Equation:**
We know \( g' = 9 \, \text{m/s}^2 \):
\[ 9 = g \left( \frac{20}{21} \right)^2 \]
Solving for \( g \):
\[ g = 9 \left( \frac{21}{20} \right)^2 \]

5. **Calculate \( g \):**
\[ g = 9 \times \frac{441}{400} = \frac{3969}{400} \approx 9.92 \, \text{m/s}^2 \]

6. **Use the Formula for Acceleration due to Gravity Below the Surface:**
The formula for acceleration due to gravity at a depth \( d \) below the Earth's surface is:
\[ g'' = g \left( 1 - \frac{d}{R} \right) \]
Here, \( d = \frac{R}{20} \):
\[ g'' = g \left( 1 - \frac{1}{20} \right) = g \left( \frac{19}{20} \right) \]

7. **Substitute \( g \) into the Formula:**
\[ g'' = 9.92 \left( \frac{19}{20} \right) = 9.92 \times 0.95 \approx 9.42 \, \text{m/s}^2 \]

8. **Final Answer:**
The approximate value of acceleration due to gravity at a point at an equal distance below the Earth's surface is \( \approx 9.42 \, \text{m/s}^2 \).