At what height from the surface of earth, the acceleration due to gravity decreases by 2% ? [Radius of earth = 6,400 km.]

Answer

To find the height from the surface of the Earth where the acceleration due to gravity decreases by 2%, we can follow these steps:

### Step-by-Step Solution

1. **Understand the relationship between gravity at height and at the surface**:
The acceleration due to gravity at a height \( h \) above the Earth's surface is given by:
\[ g' = \frac{GM}{(R + h)^2} \]
where \( g' \) is the gravity at height \( h \), \( G \) is the universal gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

2. **Express the decrease in gravity**:
We know that the acceleration due to gravity decreases by 2%. Therefore, we can express this as:
\[ g' = g - 0.02g = 0.98g \]
where \( g \) is the acceleration due to gravity at the surface of the Earth, given by:
\[ g = \frac{GM}{R^2} \]

3. **Set up the equation**:
We can substitute \( g' \) into the equation:
\[ 0.98g = \frac{GM}{(R + h)^2} \]
Substituting \( g = \frac{GM}{R^2} \) into the equation gives:
\[ 0.98 \left(\frac{GM}{R^2}\right) = \frac{GM}{(R + h)^2} \]

4. **Cancel \( GM \) from both sides**:
Since \( GM \) is common on both sides, we can cancel it out:
\[ 0.98 \cdot \frac{1}{R^2} = \frac{1}{(R + h)^2} \]

5. **Cross-multiply**:
Cross-multiplying gives:
\[ 0.98(R + h)^2 = R^2 \]

6. **Expand and rearrange**:
Expanding the left side:
\[ 0.98(R^2 + 2Rh + h^2) = R^2 \]
Rearranging gives:
\[ 0.98R^2 + 1.96Rh + 0.98h^2 = R^2 \]
Simplifying this results in:
\[ 1.96Rh + 0.98h^2 = R^2 - 0.98R^2 \]
\[ 1.96Rh + 0.98h^2 = 0.02R^2 \]

7. **Assume \( h \) is small compared to \( R \)**:
If \( h \) is small compared to \( R \), we can neglect \( h^2 \):
\[ 1.96Rh \approx 0.02R^2 \]

8. **Solve for \( h \)**:
Rearranging gives:
\[ h \approx \frac{0.02R^2}{1.96R} = \frac{0.02R}{1.96} \]
Substituting \( R = 6400 \, \text{km} \):
\[ h \approx \frac{0.02 \times 6400}{1.96} \approx \frac{128}{1.96} \approx 65.31 \, \text{km} \]

### Final Answer
The height from the surface of the Earth where the acceleration due to gravity decreases by 2% is approximately **65.31 km**.