A man weighs 100kg on the surface of the earth of radius R. At what height above the surface of the earth, he will weigh 50 kg?

Options

A. 0.41R
B. 0.51R
C. 0.31R
D. 0.61R

Explanation

To solve the problem of determining the height above the Earth's surface where a man weighing 100 kg on the surface will weigh 50 kg, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Weight and Gravitational Force**:
The weight of an object is given by the formula:
\[ W = mg \]
where \( W \) is the weight, \( m \) is the mass, and \( g \) is the acceleration due to gravity.

2. **Weight on the Surface of the Earth**:
On the surface of the Earth, the man weighs 100 kg. Therefore, we can express this as:
\[ W = 100g \]

3. **Weight at Height \( h \)**:
At a height \( h \) above the surface of the Earth, the weight of the man becomes 50 kg. Thus, we can express this as:
\[ W' = 50g' \]
where \( g' \) is the acceleration due to gravity at height \( h \).

4. **Relation Between \( g' \) and \( g \)**:
The acceleration due to gravity at a height \( h \) above the Earth's surface is given by:
\[ g' = g \left( \frac{R}{R + h} \right)^2 \]
where \( R \) is the radius of the Earth.

5. **Setting Up the Equation**:
Since we know the weights at both locations, we can set up the ratio:
\[ \frac{W'}{W} = \frac{g'}{g} \]
Substituting the known weights:
\[ \frac{50g'}{100g} = \frac{g'}{g} \]
This simplifies to:
\[ \frac{1}{2} = \left( \frac{R}{R + h} \right)^2 \]

6. **Solving for \( R + h \)**:
Taking the square root of both sides gives:
\[ \frac{R}{R + h} = \frac{1}{\sqrt{2}} \]
Cross-multiplying yields:
\[ R = \frac{R + h}{\sqrt{2}} \]
Rearranging gives:
\[ R\sqrt{2} = R + h \]
Thus:
\[ h = R\sqrt{2} - R = R(\sqrt{2} - 1) \]

7. **Calculating the Height**:
Using the approximate value of \( \sqrt{2} \approx 1.414 \):
\[ h \approx R(1.414 - 1) \approx R(0.414) \]

### Final Result:
The height \( h \) above the surface of the Earth where the man weighs 50 kg is approximately:
\[ h \approx 0.414 R \]