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 with a mass that is \( \frac{1}{7} \) that of Earth and a radius that is half of Earth's radius, we can follow these steps:

### Step 1: Understand the formula for gravitational acceleration
The formula for acceleration due to gravity \( g' \) on the surface of a planet is given by:
\[ g' = \frac{G \cdot M}{R^2} \]
where:
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the planet,
- \( R \) is the radius of the planet.

### Step 2: Define the variables for Earth
Let:
- \( M_e \) = mass of Earth,
- \( R_e \) = radius of Earth.

### Step 3: Define the variables for the planet
According to the problem:
- The mass of the planet \( M_p = \frac{1}{7} M_e \),
- The radius of the planet \( R_p = \frac{1}{2} R_e \).

### Step 4: Substitute the values into the formula for \( g' \)
Substituting the values for the mass and radius of the planet into the formula for gravitational acceleration:
\[ g' = \frac{G \cdot \left(\frac{1}{7} M_e\right)}{\left(\frac{1}{2} R_e\right)^2} \]

### Step 5: Simplify the expression
Calculating the denominator:
\[ \left(\frac{1}{2} R_e\right)^2 = \frac{1}{4} R_e^2 \]
Thus, we can rewrite \( g' \):
\[ g' = \frac{G \cdot \left(\frac{1}{7} M_e\right)}{\frac{1}{4} R_e^2} = \frac{4G \cdot M_e}{7R_e^2} \]

### Step 6: Relate \( g' \) to \( g \) (acceleration due to gravity on Earth)
We know that the acceleration due to gravity on Earth \( g \) is given by:
\[ g = \frac{G \cdot M_e}{R_e^2} \]
Thus, we can express \( g' \) in terms of \( g \):
\[ g' = \frac{4}{7} g \]

### Step 7: Calculate the weight of the body on the planet
The weight \( W' \) of the body on the planet is given by:
\[ W' = m \cdot g' \]
Substituting the values:
\[ W' = 70 \, \text{kg} \cdot \left(\frac{4}{7} g\right) \]

### Step 8: Simplify the expression for weight
\[ W' = 70 \cdot \frac{4}{7} g = 40g \]

### Step 9: Convert weight to kgf
Since \( g \) is approximately \( 9.8 \, \text{m/s}^2 \), the weight in kgf (kilogram-force) is:
\[ W' = 40 \, \text{kgf} \]

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