At what height above the surface of earth the value of "g" decreases by 2 % [ radius of the earth is 6400 km ]

Options

A. 32 km
B. 64 km
C. 128 km
D. 1600 km

Answer

To solve the problem of finding the height above the surface of the Earth where the value of "g" decreases by 2%, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We need to find the height \( h \) at which the acceleration due to gravity \( g' \) decreases by 2% from its value at the surface of the Earth.

2. **Define the Variables**:
- Let \( g \) be the acceleration due to gravity at the surface of the Earth.
- The radius of the Earth \( r \) is given as 6400 km.
- The new value of gravity at height \( h \) is given by \( g' = g - 0.02g = 0.98g \).

3. **Use the Formula for Gravity at Height**:
The formula for the acceleration due to gravity at a height \( h \) above the surface of the Earth is:
\[ g' = g \left( \frac{r}{r+h} \right)^2 \]
where \( r \) is the radius of the Earth.

4. **Set Up the Equation**:
We know that \( g' = 0.98g \), so we can substitute this into the equation:
\[ 0.98g = g \left( \frac{r}{r+h} \right)^2 \]

5. **Cancel \( g \)**:
Since \( g \) is not zero, we can divide both sides by \( g \):
\[ 0.98 = \left( \frac{r}{r+h} \right)^2 \]

6. **Take the Square Root**:
Taking the square root of both sides gives:
\[ \sqrt{0.98} = \frac{r}{r+h} \]

7. **Rearranging the Equation**:
Cross-multiplying gives:
\[ \sqrt{0.98} (r + h) = r \]
This simplifies to:
\[ r\sqrt{0.98} + h\sqrt{0.98} = r \]
Rearranging for \( h \):
\[ h\sqrt{0.98} = r - r\sqrt{0.98} \]
\[ h = \frac{r(1 - \sqrt{0.98})}{\sqrt{0.98}} \]

8. **Substituting the Value of \( r \)**:
Now substituting \( r = 6400 \) km:
\[ h = \frac{6400(1 - \sqrt{0.98})}{\sqrt{0.98}} \]

9. **Calculating \( \sqrt{0.98} \)**:
The approximate value of \( \sqrt{0.98} \) is about 0.99. Therefore:
\[ h = \frac{6400(1 - 0.99)}{0.99} \approx \frac{6400 \times 0.01}{0.99} \approx \frac{64}{0.99} \approx 64.64 \text{ km} \]

10. **Final Answer**:
The height \( h \) at which the value of \( g \) decreases by 2% is approximately 64 km.