The weight of an object at earth’s surface is 700 g. What will be its weight at the surface of a planet whose radius is `1//2` and mass is `1//7` of that of the earth?

Options

A. 200g-wt
B. 400g-Wt
C. 50g-Wt
D. 300g-Wt

Answer

To find the weight of an object on the surface of a planet with a different radius and mass compared to Earth, we can follow these steps:

### Step 1: Understand the relationship between weight, mass, and gravity
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.

### Step 2: Identify the given values
- The weight of the object on Earth's surface is \( W_e = 700 \, \text{g} \).
- The mass of the planet \( M_p = \frac{1}{7} M_e \) (where \( M_e \) is the mass of Earth).
- The radius of the planet \( R_p = \frac{1}{2} R_e \) (where \( R_e \) is the radius of Earth).

### Step 3: Calculate the acceleration due to gravity on the new planet
The formula for acceleration due to gravity is:
\[ g = \frac{G \cdot M}{R^2} \]
For the new planet, we substitute the values:
\[ g_p = \frac{G \cdot M_p}{R_p^2} \]
Substituting \( M_p \) and \( R_p \):
\[ g_p = \frac{G \cdot \left(\frac{1}{7} M_e\right)}{\left(\frac{1}{2} R_e\right)^2} \]
This simplifies to:
\[ g_p = \frac{G \cdot \left(\frac{1}{7} M_e\right)}{\frac{1}{4} R_e^2} \]
\[ g_p = \frac{4G \cdot M_e}{7R_e^2} \]
Since \( g_e = \frac{G \cdot M_e}{R_e^2} \), we can express \( g_p \) in terms of \( g_e \):
\[ g_p = \frac{4}{7} g_e \]

### Step 4: Calculate the weight of the object on the new planet
Using the weight formula again:
\[ W_p = m \cdot g_p \]
Substituting \( g_p \):
\[ W_p = m \cdot \left(\frac{4}{7} g_e\right) \]
Since \( W_e = m \cdot g_e \) and \( W_e = 700 \, \text{g} \):
\[ W_p = \frac{4}{7} W_e \]
Substituting \( W_e \):
\[ W_p = \frac{4}{7} \cdot 700 \, \text{g} \]
Calculating:
\[ W_p = 400 \, \text{g} \]

### Final Answer
The weight of the object at the surface of the new planet is \( 400 \, \text{g} \).

---