If g on the surface of the earth is `9.8m//s^(2)`, its value at a height of 6400 km is (Radius of the earth =6400 km)

Options

A. `4.9ms^(2)`
B. `9.8ms^(2)`
C. `2.45 ms^(-2)`
D. `19.6ms^(-2)`

Explanation

To find the value of gravitational acceleration (g') at a height (h) of 6400 km above the Earth's surface, we can use the formula that relates the gravitational acceleration at height to that on the surface of the Earth:

\[ g' = g \left( \frac{R^2}{(R + h)^2} \right) \]

Where:
- \( g' \) = gravitational acceleration at height \( h \)
- \( g \) = gravitational acceleration on the surface of the Earth (9.8 m/s²)
- \( R \) = radius of the Earth (6400 km)
- \( h \) = height above the Earth's surface (6400 km)

### Step-by-Step Solution:

1. **Convert units**: Since the radius of the Earth and height are given in kilometers, we first convert them to meters for consistency:
- \( R = 6400 \, \text{km} = 6400 \times 1000 \, \text{m} = 6.4 \times 10^6 \, \text{m} \)
- \( h = 6400 \, \text{km} = 6400 \times 1000 \, \text{m} = 6.4 \times 10^6 \, \text{m} \)

2. **Substitute values into the formula**:
- \( g' = 9.8 \left( \frac{(6.4 \times 10^6)^2}{(6.4 \times 10^6 + 6.4 \times 10^6)^2} \right) \)

3. **Calculate \( R + h \)**:
- \( R + h = 6.4 \times 10^6 + 6.4 \times 10^6 = 12.8 \times 10^6 \, \text{m} \)

4. **Calculate \( (R + h)^2 \)**:
- \( (R + h)^2 = (12.8 \times 10^6)^2 = 1.6384 \times 10^{14} \, \text{m}^2 \)

5. **Calculate \( R^2 \)**:
- \( R^2 = (6.4 \times 10^6)^2 = 4.096 \times 10^{13} \, \text{m}^2 \)

6. **Substitute \( R^2 \) and \( (R + h)^2 \) back into the formula**:
- \( g' = 9.8 \left( \frac{4.096 \times 10^{13}}{1.6384 \times 10^{14}} \right) \)

7. **Calculate the fraction**:
- \( \frac{4.096 \times 10^{13}}{1.6384 \times 10^{14}} = \frac{4.096}{16.384} = \frac{1}{4} \)

8. **Calculate \( g' \)**:
- \( g' = 9.8 \times \frac{1}{4} = 2.45 \, \text{m/s}^2 \)

### Final Result:
The value of gravitational acceleration at a height of 6400 km is \( g' = 2.45 \, \text{m/s}^2 \).

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