Value of g on the surface of earth is `9.8 m//s^(2)`. Find its
(a) at height `h = R` from the surface ,
(b) at depth `d = (R)/(2)` from the surface . (`R = ` radius of earth)

Explanation

To solve the problem, we will calculate the value of gravitational acceleration \( g \) at a height \( h = R \) from the surface of the Earth and at a depth \( d = \frac{R}{2} \) from the surface, where \( R \) is the radius of the Earth.

### Given:
- \( g = 9.8 \, \text{m/s}^2 \) (gravitational acceleration at the surface of the Earth)
- \( R \) = radius of the Earth

### (a) Finding \( g \) at height \( h = R \)

1. **Use the formula for gravitational acceleration at height \( h \)**:
\[ g' = \frac{g}{(1 + \frac{h}{R})^2} \]

2. **Substituting \( h = R \)**:
\[ g' = \frac{g}{(1 + \frac{R}{R})^2} = \frac{g}{(1 + 1)^2} = \frac{g}{2^2} = \frac{g}{4} \]

3. **Calculating \( g' \)**:
\[ g' = \frac{9.8 \, \text{m/s}^2}{4} = 2.45 \, \text{m/s}^2 \]

### (b) Finding \( g \) at depth \( d = \frac{R}{2} \)

1. **Use the formula for gravitational acceleration at depth \( d \)**:
\[ g' = g \left(1 - \frac{d}{R}\right) \]

2. **Substituting \( d = \frac{R}{2} \)**:
\[ g' = g \left(1 - \frac{\frac{R}{2}}{R}\right) = g \left(1 - \frac{1}{2}\right) = g \left(\frac{1}{2}\right) \]

3. **Calculating \( g' \)**:
\[ g' = 9.8 \, \text{m/s}^2 \times \frac{1}{2} = 4.9 \, \text{m/s}^2 \]

### Final Answers:
- (a) \( g \) at height \( h = R \) is \( 2.45 \, \text{m/s}^2 \)
- (b) \( g \) at depth \( d = \frac{R}{2} \) is \( 4.9 \, \text{m/s}^2 \)