If a man weighs 90 kg on the surface of the earth, the height above the surface of the earth of radius, R where the weight is 30 kg is

Options

A. 0.73 R
B. `sqrt(3) R`
C. `R`
D. `2R`

Answer

To solve the problem, we need to find the height \( h \) above the surface of the Earth where a man who weighs 90 kg on the surface weighs only 30 kg.

### Step-by-Step Solution:

1. **Understanding Weight on Earth**:
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. On the surface of the Earth, the weight of the man is:
\[ W_1 = 90 \, \text{kg} \times g = 900 \, \text{N} \]

2. **Weight at Height \( h \)**:
At height \( h \), the weight of the man is given as 30 kg, which translates to:
\[ W_2 = 30 \, \text{kg} \times g' = 300 \, \text{N} \]
where \( g' \) is the acceleration due to gravity at height \( h \).

3. **Gravitational Force Relation**:
The gravitational force at a distance \( r \) (radius of the Earth) is:
\[ g = \frac{GM}{r^2} \]
At height \( h \), the distance from the center of the Earth becomes \( r + h \):
\[ g' = \frac{GM}{(r + h)^2} \]

4. **Setting Up the Ratio**:
We can set up the ratio of the gravitational forces at the two locations:
\[ \frac{g'}{g} = \frac{W_2}{W_1} = \frac{300}{900} = \frac{1}{3} \]
Therefore,
\[ \frac{GM/(r + h)^2}{GM/r^2} = \frac{1}{3} \]
Simplifying gives:
\[ \frac{r^2}{(r + h)^2} = \frac{1}{3} \]

5. **Cross Multiplying**:
Cross multiplying gives:
\[ 3r^2 = (r + h)^2 \]

6. **Expanding the Equation**:
Expanding the right side:
\[ 3r^2 = r^2 + 2rh + h^2 \]
Rearranging gives:
\[ 2r^2 - 2rh - h^2 = 0 \]

7. **Using the Quadratic Formula**:
This is a quadratic equation in terms of \( h \):
\[ h^2 + 2rh - 2r^2 = 0 \]
Using the quadratic formula \( h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = 2r, c = -2r^2 \):
\[ h = \frac{-2r \pm \sqrt{(2r)^2 - 4 \cdot 1 \cdot (-2r^2)}}{2 \cdot 1} \]
Simplifying gives:
\[ h = \frac{-2r \pm \sqrt{4r^2 + 8r^2}}{2} = \frac{-2r \pm \sqrt{12r^2}}{2} = \frac{-2r \pm 2\sqrt{3}r}{2} \]
Thus,
\[ h = (-1 + \sqrt{3})r \]

8. **Final Calculation**:
Since \( \sqrt{3} \approx 1.732 \):
\[ h \approx (-1 + 1.732)r \approx 0.732r \]

### Conclusion:
The height \( h \) above the surface of the Earth where the man weighs 30 kg is approximately \( 0.732R \).