If the value of `g` at the surface of the earth is `9.8 m//sec^(2)`, then the value of `g` at a place `480` km above the surface of the earth will be (Radius of the earth is `6400` km)

Options

A. `9.8 m//s^(2)`
B. `7.2 m//s^(2)`
C. `8.5 m//s^(2)`
D. `4.2 m//s^(2)`

Answer

To find the value of `g` at a height of `480 km` above the Earth's surface, we can use the formula that relates the acceleration due to gravity at a height `h` above the surface of the Earth to the acceleration due to gravity at the surface.

### Step-by-Step Solution:

1. **Identify the known values:**
- Acceleration due to gravity at the Earth's surface, \( g = 9.8 \, \text{m/s}^2 \)
- Radius of the Earth, \( R = 6400 \, \text{km} = 6400 \times 10^3 \, \text{m} \)
- Height above the Earth's surface, \( h = 480 \, \text{km} = 480 \times 10^3 \, \text{m} \)

2. **Total distance from the center of the Earth at height \( h \):**
\[ r = R + h = 6400 \times 10^3 \, \text{m} + 480 \times 10^3 \, \text{m} = (6400 + 480) \times 10^3 \, \text{m} = 6880 \times 10^3 \, \text{m} \]

3. **Use the formula for acceleration due to gravity at height \( h \):**
The formula for the acceleration due to gravity at height \( h \) is given by:
\[ g' = g \left( \frac{R^2}{(R + h)^2} \right) \]

4. **Substituting the values into the formula:**
\[ g' = 9.8 \left( \frac{(6400 \times 10^3)^2}{(6880 \times 10^3)^2} \right) \]

5. **Calculating the squares:**
\[ g' = 9.8 \left( \frac{(6400^2)}{(6880^2)} \right) \times 10^{-6} \]

6. **Calculating the ratio:**
\[ \frac{6400^2}{6880^2} = \left( \frac{6400}{6880} \right)^2 \]
Calculate \( \frac{6400}{6880} \):
\[ \frac{6400}{6880} \approx 0.928 \]
Then square it:
\[ (0.928)^2 \approx 0.861 \]

7. **Final calculation of \( g' \):**
\[ g' \approx 9.8 \times 0.861 \approx 8.45 \, \text{m/s}^2 \]

8. **Rounding off:**
\[ g' \approx 8.5 \, \text{m/s}^2 \]

### Conclusion:
The value of \( g \) at a height of \( 480 \, \text{km} \) above the surface of the Earth is approximately \( 8.5 \, \text{m/s}^2 \).