If the radius of the earth is 6400 km, the height above the surface of the earth, where the value of acceleration due to gravity will be 1% of its value from the surface of the earth is

Options

A. 6400 km
B. 64 km
C. 711 km
D. 57600 km

Answer

To solve the problem of finding the height above the surface of the Earth where the acceleration due to gravity is 1% of its value at the surface, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the given values**:
- Radius of the Earth, \( R = 6400 \) km = \( 6400 \times 10^3 \) m (since we need to work in meters).

2. **Formula for acceleration due to gravity**:
- The acceleration due to gravity at the surface of the Earth is given by:
\[ g = \frac{GM}{R^2} \]
- Where \( G \) is the universal gravitational constant and \( M \) is the mass of the Earth.

3. **Acceleration due to gravity at height \( h \)**:
- The acceleration due to gravity at a height \( h \) above the surface of the Earth is given by:
\[ g_h = \frac{GM}{(R + h)^2} \]

4. **Set up the equation for 1% of surface gravity**:
- We want \( g_h = 0.01g \). Therefore, we can write:
\[ \frac{GM}{(R + h)^2} = 0.01 \cdot \frac{GM}{R^2} \]

5. **Cancel \( GM \) from both sides**:
- This simplifies to:
\[ \frac{1}{(R + h)^2} = \frac{0.01}{R^2} \]

6. **Cross-multiply to eliminate the fraction**:
- We get:
\[ R^2 = 0.01(R + h)^2 \]

7. **Expand the right side**:
- This gives us:
\[ R^2 = 0.01(R^2 + 2Rh + h^2) \]

8. **Rearrange the equation**:
- Bringing all terms to one side:
\[ R^2 - 0.01R^2 - 0.02Rh - 0.01h^2 = 0 \]
- This simplifies to:
\[ 0.99R^2 - 0.02Rh - 0.01h^2 = 0 \]

9. **Assume \( h \) is much smaller than \( R \)**:
- For practical purposes, we can neglect \( h^2 \) and rewrite the equation as:
\[ 0.99R^2 - 0.02Rh = 0 \]

10. **Solve for \( h \)**:
- Rearranging gives:
\[ h = \frac{0.99R^2}{0.02R} = \frac{0.99R}{0.02} = 49.5R \]

11. **Substituting the value of \( R \)**:
- Now substituting \( R = 6400 \) km:
\[ h = 49.5 \times 6400 \text{ km} = 316800 \text{ km} \]

### Final Answer:
The height above the surface of the Earth where the value of acceleration due to gravity will be 1% of its value from the surface of the Earth is approximately **316800 km**.