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. **Understand the Problem**:
We need to find the height \( h \) above the Earth's surface where the acceleration due to gravity \( g' \) is 99% of the acceleration due to gravity at the surface \( g \).

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

At a height \( h \), the acceleration due to gravity \( g' \) is given by:
\[ g' = \frac{GM}{(R + h)^2} \]

3. **Set Up the Equation**:
Since we want \( g' \) to be 99% of \( g \), we can write:
\[ g' = 0.99g \]
Substituting the expressions for \( g \) and \( g' \):
\[ \frac{GM}{(R + h)^2} = 0.99 \cdot \frac{GM}{R^2} \]

4. **Cancel Common Terms**:
The \( GM \) terms cancel out from both sides:
\[ \frac{1}{(R + h)^2} = 0.99 \cdot \frac{1}{R^2} \]

5. **Cross-Multiply**:
Cross-multiplying gives:
\[ R^2 = 0.99(R + h)^2 \]

6. **Expand the Equation**:
Expanding the right side:
\[ R^2 = 0.99(R^2 + 2Rh + h^2) \]
This simplifies to:
\[ R^2 = 0.99R^2 + 1.98Rh + 0.99h^2 \]

7. **Rearranging the Equation**:
Rearranging gives:
\[ R^2 - 0.99R^2 = 1.98Rh + 0.99h^2 \]
Simplifying further:
\[ 0.01R^2 = 1.98Rh + 0.99h^2 \]

8. **Neglecting \( h^2 \) for Small Heights**:
Since \( h \) is much smaller than \( R \), we can neglect the \( h^2 \) term:
\[ 0.01R^2 \approx 1.98Rh \]

9. **Solving for \( h \)**:
Rearranging gives:
\[ h \approx \frac{0.01R^2}{1.98R} = \frac{0.01R}{1.98} \]
Substituting \( R \approx 6400 \) km (or \( 6400000 \) m):
\[ h \approx \frac{0.01 \times 6400000}{1.98} \approx \frac{64000}{1.98} \approx 32323.23 \text{ m} \approx 32.32 \text{ km} \]

10. **Final Answer**:
Thus, the height above the surface of the Earth at which the acceleration due to gravity reduces by 1% is approximately:
\[ h \approx 32 \text{ km} \]