At what height above the surface of earth acceleration due to gravity reduces by 1 % ?

Explanation

To find the height above the surface of the Earth at which the acceleration due to gravity reduces by 1%, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Formula for Gravity**:
The acceleration due to gravity at the surface of the Earth (G) is given by the formula:
\[ G = \frac{GM}{R^2} \]
where \( G \) is the universal gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

2. **Gravity at Height (h)**:
At a height \( h \) above the surface, the acceleration due to gravity \( g' \) can be expressed as:
\[ g' = \frac{GM}{(R + h)^2} \]

3. **Setting Up the Equation**:
According to the problem, we want to find the height \( h \) where \( g' \) is 1% less than \( G \). This can be expressed as:
\[ g' = 0.99G \]

4. **Substituting the Values**:
We can substitute the expressions for \( g' \) and \( G \) into the equation:
\[ \frac{GM}{(R + h)^2} = 0.99 \cdot \frac{GM}{R^2} \]

5. **Canceling GM**:
Since \( GM \) appears in both sides of the equation, we can cancel it out:
\[ \frac{1}{(R + h)^2} = 0.99 \cdot \frac{1}{R^2} \]

6. **Cross-Multiplying**:
Cross-multiplying gives us:
\[ R^2 = 0.99(R + h)^2 \]

7. **Expanding the Right Side**:
Expanding the right side:
\[ R^2 = 0.99(R^2 + 2Rh + h^2) \]

8. **Rearranging the Equation**:
Rearranging the equation:
\[ R^2 - 0.99R^2 = 0.99(2Rh + h^2) \]
\[ 0.01R^2 = 0.99(2Rh + h^2) \]

9. **Dividing by 0.99**:
Dividing both sides by 0.99:
\[ \frac{0.01R^2}{0.99} = 2Rh + h^2 \]

10. **Assuming h is much smaller than R**:
If \( h \) is much smaller than \( R \), we can ignore \( h^2 \) in comparison to \( 2Rh \):
\[ \frac{0.01R^2}{0.99} \approx 2Rh \]

11. **Solving for h**:
Rearranging gives:
\[ h \approx \frac{0.01R^2}{0.99 \cdot 2R} \]
\[ h \approx \frac{0.01R}{1.98} \]

12. **Substituting the Radius of Earth**:
The average radius of the Earth \( R \) is approximately 6400 km:
\[ h \approx \frac{0.01 \cdot 6400}{1.98} \approx \frac{64}{1.98} \approx 32.32 \text{ km} \]

### Final Answer:
The height above the surface of the Earth at which the acceleration due to gravity reduces by 1% is approximately **32 km**.