The height above the earth`s surface at which the weight of a person becomes 1/4th of his weight on the surface of earth is (R is the radius of earth)

Options

A. 2 R
B. R
C. 3 R
D. 5 R

Answer

To solve the problem, we need to find the height \( H \) above the Earth's surface where the weight of a person becomes one-fourth of their weight on the surface of the Earth.

### Step-by-Step Solution:

1. **Understand the Weight on the Surface of the Earth**:
The weight of a person on the surface of the Earth is given by:
\[ W = mg \]
where \( m \) is the mass of the person and \( g \) is the acceleration due to gravity at the surface.

2. **Express Weight in Terms of Gravitational Force**:
The gravitational force acting on the person at the surface can also be expressed using Newton's law of gravitation:
\[ W = \frac{GMm}{R^2} \]
where \( G \) is the universal gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

3. **Weight at Height \( H \)**:
At a height \( H \) above the Earth's surface, the weight \( W' \) of the person is given by:
\[ W' = \frac{GMm}{(R + H)^2} \]

4. **Set Up the Equation for One-Fourth Weight**:
According to the problem, the weight at height \( H \) is one-fourth of the weight at the surface:
\[ W' = \frac{1}{4} W \]
Substituting the expressions for \( W \) and \( W' \):
\[ \frac{GMm}{(R + H)^2} = \frac{1}{4} \cdot \frac{GMm}{R^2} \]

5. **Cancel Common Terms**:
Since \( GMm \) appears on both sides, we can cancel it out:
\[ \frac{1}{(R + H)^2} = \frac{1}{4R^2} \]

6. **Cross-Multiply**:
Cross-multiplying gives:
\[ 4R^2 = (R + H)^2 \]

7. **Expand the Right Side**:
Expanding the right side:
\[ 4R^2 = R^2 + 2RH + H^2 \]

8. **Rearranging the Equation**:
Rearranging the equation gives:
\[ 3R^2 = 2RH + H^2 \]

9. **Rearranging Further**:
Rearranging it into a standard quadratic form:
\[ H^2 + 2RH - 3R^2 = 0 \]

10. **Using the Quadratic Formula**:
Applying the quadratic formula \( H = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = 2R, c = -3R^2 \):
\[ H = \frac{-2R \pm \sqrt{(2R)^2 - 4 \cdot 1 \cdot (-3R^2)}}{2 \cdot 1} \]
\[ H = \frac{-2R \pm \sqrt{4R^2 + 12R^2}}{2} \]
\[ H = \frac{-2R \pm \sqrt{16R^2}}{2} \]
\[ H = \frac{-2R \pm 4R}{2} \]

11. **Calculating the Two Possible Values**:
This gives us two solutions:
\[ H = \frac{2R}{2} = R \quad \text{(taking the positive root)} \]
\[ H = \frac{-6R}{2} = -3R \quad \text{(not physically meaningful)} \]

12. **Final Answer**:
Therefore, the height \( H \) above the Earth's surface where the weight becomes one-fourth of the weight on the surface is:
\[ H = R \]