Acceleration due to gravity at earth's surface if 'g' `m//s^(2)`. Find the effective value of acceleration due to gravity at a height of 32 km from sea level : `(R_(e)=6400 km)`

Options

A. `0.5 g m//s^(2)`
B. `0.99 g m//s^(2)`
C. `1.01 g m//s^(2)`
D. `0.90 g m//s^(2)`

Answer

To solve the problem of finding the effective value of acceleration due to gravity at a height of 32 km from sea level, we can use the formula for gravitational acceleration at a height \( h \) above the Earth's surface:

\[ g' = g \left(1 - \frac{h}{R_e}\right) \]

Where:
- \( g' \) is the acceleration due to gravity at height \( h \),
- \( g \) is the acceleration due to gravity at the Earth's surface (approximately \( 9.81 \, m/s^2 \)),
- \( h \) is the height above the Earth's surface (32 km in this case),
- \( R_e \) is the radius of the Earth (6400 km).

### Step-by-step solution:

1. **Identify the known values**:
- \( g = 9.81 \, m/s^2 \)
- \( h = 32 \, km = 32000 \, m \)
- \( R_e = 6400 \, km = 6400000 \, m \)

2. **Convert height to the same unit as the radius**:
- Since \( R_e \) is given in kilometers, we convert \( h \) to kilometers:
\[ h = 32 \, km \]

3. **Substitute the values into the formula**:
- Using the formula:
\[ g' = g \left(1 - \frac{h}{R_e}\right) \]
- Substitute \( g \), \( h \), and \( R_e \):
\[ g' = 9.81 \left(1 - \frac{32}{6400}\right) \]

4. **Calculate the fraction**:
- Calculate \( \frac{32}{6400} \):
\[ \frac{32}{6400} = 0.005 \]

5. **Substitute back into the equation**:
- Now substitute back:
\[ g' = 9.81 \left(1 - 0.005\right) = 9.81 \times 0.995 \]

6. **Perform the multiplication**:
- Calculate \( 9.81 \times 0.995 \):
\[ g' \approx 9.81 \times 0.995 \approx 9.76 \, m/s^2 \]

7. **Final result**:
- The effective value of acceleration due to gravity at a height of 32 km from sea level is approximately:
\[ g' \approx 9.76 \, m/s^2 \]