What is the weight of a 70 kg body on the surface of a planet mass is `1/7 th` that of earth and radius is is half of earth ?

Options

A. 20 kgf
B. 40 kgf
C. 70 kgf
D. 140 kgf

Answer

To find the weight of a 70 kg body on the surface of a planet whose mass is \( \frac{1}{7} \) that of Earth and radius is half of Earth, we can use the formula for gravitational force:

\[ F = \frac{G \cdot M \cdot m}{R^2} \]

Where:
- \( F \) is the weight of the body,
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the planet,
- \( m \) is the mass of the body,
- \( R \) is the radius of the planet.

### Step 1: Determine the mass and radius of the planet
Given:
- Mass of the planet \( M = \frac{1}{7} M_E \) (where \( M_E \) is the mass of Earth)
- Radius of the planet \( R = \frac{1}{2} R_E \) (where \( R_E \) is the radius of Earth)

### Step 2: Substitute the values into the weight formula
The weight of the body on the planet can be expressed as:

\[ F = \frac{G \cdot \left(\frac{1}{7} M_E\right) \cdot m}{\left(\frac{1}{2} R_E\right)^2} \]

### Step 3: Simplify the equation
Substituting \( R \) into the equation gives:

\[ F = \frac{G \cdot \left(\frac{1}{7} M_E\right) \cdot m}{\frac{1}{4} R_E^2} \]

This can be simplified to:

\[ F = \frac{G \cdot M_E \cdot m}{R_E^2} \cdot \frac{4}{7} \]

### Step 4: Recognize the weight on Earth
The weight of the body on the surface of the Earth is given by:

\[ F_E = \frac{G \cdot M_E \cdot m}{R_E^2} \]

### Step 5: Relate the two weights
Thus, we can express the weight on the new planet as:

\[ F = \frac{4}{7} F_E \]

### Step 6: Calculate the weight on Earth
Given that the mass of the body \( m = 70 \, \text{kg} \), the weight on Earth \( F_E \) can be calculated as:

\[ F_E = m \cdot g = 70 \, \text{kg} \cdot g \]

Assuming \( g \approx 9.8 \, \text{m/s}^2 \), we can find \( F_E \):

\[ F_E \approx 70 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 \approx 686 \, \text{N} \]

### Step 7: Calculate the weight on the new planet
Now substituting back into our equation for \( F \):

\[ F = \frac{4}{7} \cdot 686 \, \text{N} \approx 392 \, \text{N} \]

### Step 8: Convert to kgf
To convert Newtons to kgf (kilogram-force), we use the relation \( 1 \, \text{kgf} \approx 9.8 \, \text{N} \):

\[ F \approx \frac{392 \, \text{N}}{9.8 \, \text{N/kgf}} \approx 40 \, \text{kgf} \]

### Final Answer
Thus, the weight of the 70 kg body on the surface of the planet is approximately **40 kgf**.

---