At what height above the earth's surface is the acceleration due to gravity 1% less than its value at the surface ? [R = 6400 km]

Answer

To find the height above the Earth's surface where the acceleration due to gravity is 1% less than its value at the surface, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship of gravity at height (h)**:
The acceleration due to gravity at a height \( h \) above the Earth's surface is given by the formula:
\[ g' = g \left(1 - \frac{2h}{R}\right) \]
where \( g' \) is the acceleration due to gravity at height \( h \), \( g \) is the acceleration due to gravity at the surface, and \( R \) is the radius of the Earth.

2. **Set up the equation for 1% less gravity**:
We know that \( g' \) is 1% less than \( g \):
\[ g' = g \left(1 - 0.01\right) = 0.99g \]

3. **Equate the two expressions for \( g' \)**:
Now we can set the two expressions for \( g' \) equal to each other:
\[ g \left(1 - \frac{2h}{R}\right) = 0.99g \]

4. **Cancel \( g \) from both sides**:
Since \( g \) is not zero, we can divide both sides by \( g \):
\[ 1 - \frac{2h}{R} = 0.99 \]

5. **Rearrange the equation**:
Rearranging gives:
\[ \frac{2h}{R} = 1 - 0.99 = 0.01 \]

6. **Solve for \( h \)**:
Multiply both sides by \( R \) and then divide by 2:
\[ 2h = 0.01R \quad \Rightarrow \quad h = \frac{0.01R}{2} \]

7. **Substitute the value of \( R \)**:
Given \( R = 6400 \) km, substitute this value into the equation:
\[ h = \frac{0.01 \times 6400 \text{ km}}{2} \]

8. **Calculate \( h \)**:
\[ h = \frac{64 \text{ km}}{2} = 32 \text{ km} \]

Thus, the height above the Earth's surface where the acceleration due to gravity is 1% less than its value at the surface is **32 km**.