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 find the height above the Earth's surface at which a person's weight becomes one-fourth of their weight on the surface of the Earth, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Weight**:
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 of the Earth.

2. **Weight at Height \( h \)**:
At a height \( h \) above the Earth's surface, the acceleration due to gravity \( g' \) can be expressed as:
\[ g' = \frac{GM}{(R + h)^2} \]
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 \( h \)**:
The weight of the person at height \( h \) becomes:
\[ W' = mg' = m \cdot \frac{GM}{(R + h)^2} \]

4. **Setting Up the Equation**:
According to the problem, the weight at height \( h \) is one-fourth of the weight on the surface:
\[ W' = \frac{W}{4} \]
Substituting the expressions for \( W \) and \( W' \):
\[ m \cdot \frac{GM}{(R + h)^2} = \frac{mg}{4} \]

5. **Cancelling \( m \) and Substituting \( g \)**:
Since \( g = \frac{GM}{R^2} \), we can substitute this into the equation:
\[ \frac{GM}{(R + h)^2} = \frac{1}{4} \cdot \frac{GM}{R^2} \]
Cancelling \( GM \) from both sides (assuming \( GM \neq 0 \)):
\[ \frac{1}{(R + h)^2} = \frac{1}{4R^2} \]

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

7. **Expanding the Right Side**:
Expanding the right side:
\[ 4R^2 = R^2 + 2Rh + h^2 \]

8. **Rearranging the Equation**:
Rearranging gives:
\[ 4R^2 - R^2 = 2Rh + h^2 \]
\[ 3R^2 = 2Rh + h^2 \]

9. **Rearranging to Form a Quadratic Equation**:
Rearranging this into standard quadratic form:
\[ h^2 + 2Rh - 3R^2 = 0 \]

10. **Using the Quadratic Formula**:
Using 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. **Finding the Positive Solution**:
The two possible solutions are:
\[ h = \frac{2R}{2} = R \quad \text{(positive solution)} \]
\[ h = \frac{-6R}{2} = -3R \quad \text{(not physically meaningful)} \]

Thus, the height \( h \) above the Earth's surface at which the weight of a person becomes one-fourth of their weight on the surface of the Earth is:
\[ \boxed{R} \]