A man weighs `80 kg` on the surface of earth of radius `r`. At what height above the surface of earth his weight will be `40 kg`?

Options

A. `R/2`
B. `sqrt(2)R`
C. `(sqrt(2)-1)R`
D. `(sqrt(2)+1)R`

Explanation

To solve the problem, we need to determine the height above the surface of the Earth where a man weighing 80 kg on the surface will weigh 40 kg.

### Step-by-Step Solution:

1. **Understanding Weight and Mass**:
- The weight \( W \) of an object is given by the formula:
\[ W = m \cdot g \]
where \( m \) is the mass of the object 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 80 kg. Thus, we can express this as:
\[ W_1 = m \cdot g \]
where \( W_1 = 80 \, \text{kg} \).

3. **Weight at Height \( h \)**:
- At height \( h \), the weight of the man is 40 kg:
\[ W_2 = m \cdot g' \]
where \( W_2 = 40 \, \text{kg} \) and \( g' \) is the acceleration due to gravity at height \( h \).

4. **Relating the Weights**:
- Since the mass \( m \) remains constant, we can set up the ratio of the weights:
\[ \frac{W_1}{W_2} = \frac{g}{g'} \]
- Substituting the values:
\[ \frac{80}{40} = \frac{g}{g'} \implies 2 = \frac{g}{g'} \]
- This means:
\[ g' = \frac{g}{2} \]

5. **Expression for \( g' \)**:
- The acceleration due to gravity at a height \( h \) above the surface of the Earth is given by:
\[ g' = \frac{G \cdot M}{(r + h)^2} \]
where \( G \) is the gravitational constant and \( M \) is the mass of the Earth, and \( r \) is the radius of the Earth.

6. **Using the Relationship**:
- We know that:
\[ g = \frac{G \cdot M}{r^2} \]
- Therefore, substituting \( g \) into the equation for \( g' \):
\[ \frac{g}{2} = \frac{G \cdot M}{(r + h)^2} \]

7. **Setting Up the Equation**:
- From the above equations, we have:
\[ \frac{G \cdot M}{2} = \frac{G \cdot M}{(r + h)^2} \]
- Canceling \( G \cdot M \) from both sides gives:
\[ \frac{1}{2} = \frac{1}{(r + h)^2} \]

8. **Cross-Multiplying**:
- Cross-multiplying gives:
\[ (r + h)^2 = 2 \]

9. **Taking the Square Root**:
- Taking the square root of both sides:
\[ r + h = r \sqrt{2} \]

10. **Solving for \( h \)**:
- Rearranging gives:
\[ h = r \sqrt{2} - r = r (\sqrt{2} - 1) \]

### Final Answer:
The height \( h \) above the surface of the Earth at which the man's weight will be 40 kg is:
\[ h = r (\sqrt{2} - 1) \]