If the radius of the earth is 6400 km, the height above the surface of the earth where the value of acceleration due to gravity will be `4%` of its value on the surface of the earth is

Options

A. 6400 km
B. 64 km
C. 57600 km
D. 25600 km

Explanation

To solve the problem, we need to find the height \( h \) above the surface of the Earth where the acceleration due to gravity \( g' \) is 4% of its value on the surface of the Earth \( g \).

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

2. **Acceleration due to gravity on the surface**: The acceleration due to gravity at the surface of the Earth is:
\[ g = \frac{GM}{R^2} \]

3. **Setting up the equation**: We want \( g' \) to be 4% of \( g \):
\[ g' = 0.04g \]
Substituting the expressions for \( g' \) and \( g \):
\[ \frac{GM}{(R + h)^2} = 0.04 \cdot \frac{GM}{R^2} \]

4. **Canceling \( GM \)**: Since \( GM \) appears on both sides, we can cancel it:
\[ \frac{1}{(R + h)^2} = 0.04 \cdot \frac{1}{R^2} \]

5. **Cross-multiplying**: This gives us:
\[ R^2 = 0.04(R + h)^2 \]

6. **Expanding the equation**: Expanding the right side:
\[ R^2 = 0.04(R^2 + 2Rh + h^2) \]
Simplifying this:
\[ R^2 = 0.04R^2 + 0.08Rh + 0.04h^2 \]

7. **Rearranging the equation**: Bringing all terms to one side:
\[ R^2 - 0.04R^2 - 0.08Rh - 0.04h^2 = 0 \]
This simplifies to:
\[ 0.96R^2 - 0.08Rh - 0.04h^2 = 0 \]

8. **Using the quadratic formula**: This is a quadratic equation in \( h \):
\[ 0.04h^2 + 0.08Rh - 0.96R^2 = 0 \]
Using the quadratic formula \( h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 0.04 \), \( b = 0.08R \), and \( c = -0.96R^2 \).

9. **Calculating the discriminant**:
\[ b^2 - 4ac = (0.08R)^2 - 4(0.04)(-0.96R^2) \]
\[ = 0.0064R^2 + 0.1536R^2 = 0.16R^2 \]

10. **Finding \( h \)**:
\[ h = \frac{-0.08R \pm \sqrt{0.16R^2}}{2 \times 0.04} \]
\[ = \frac{-0.08R \pm 0.4R}{0.08} \]
This gives two solutions:
\[ h = \frac{0.32R}{0.08} = 4R \quad \text{(taking the positive root)} \]

11. **Substituting the radius of the Earth**:
Given \( R = 6400 \) km:
\[ h = 4 \times 6400 \text{ km} = 25600 \text{ km} \]

Thus, the height above the surface of the Earth where the value of acceleration due to gravity will be 4% of its value on the surface of the Earth is **25600 km**.