Consider a planet in some solar system which has a mass double the mass of the earth and density equal to the average density of the earth. An object weighing W on the earth will weigh

Options

A. W
B. 2W
C. `W/2`
D. `2^(1/3)W at the planet

Answer

To solve the problem step by step, we need to analyze the given conditions about the planet and the object weighing \( W \) on Earth.

### Step 1: Understand the given data
- The mass of the planet is double the mass of the Earth:
\[ M_{\text{planet}} = 2M_{\text{Earth}} \]
- The density of the planet is equal to the average density of the Earth:
\[ \rho_{\text{planet}} = \rho_{\text{Earth}} \]

### Step 2: Relate mass, density, and volume
The relationship between mass, density, and volume is given by:
\[ M = \rho V \]
For a sphere, the volume \( V \) is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Thus, we can express the mass of the planet in terms of its radius:
\[ M_{\text{planet}} = \rho_{\text{planet}} \cdot \frac{4}{3} \pi r_{\text{planet}}^3 \]
Since the density of the planet is equal to the density of the Earth, we can write:
\[ 2M_{\text{Earth}} = \rho_{\text{Earth}} \cdot \frac{4}{3} \pi r_{\text{planet}}^3 \]

### Step 3: Express the mass of the Earth
The mass of the Earth can also be expressed in terms of its radius:
\[ M_{\text{Earth}} = \rho_{\text{Earth}} \cdot \frac{4}{3} \pi r_{\text{Earth}}^3 \]

### Step 4: Set up the equation
Substituting the expression for \( M_{\text{Earth}} \) into the equation for \( M_{\text{planet}} \):
\[ 2 \left( \rho_{\text{Earth}} \cdot \frac{4}{3} \pi r_{\text{Earth}}^3 \right) = \rho_{\text{Earth}} \cdot \frac{4}{3} \pi r_{\text{planet}}^3 \]
We can cancel \( \rho_{\text{Earth}} \cdot \frac{4}{3} \pi \) from both sides:
\[ 2 r_{\text{Earth}}^3 = r_{\text{planet}}^3 \]

### Step 5: Solve for the radius of the planet
Taking the cube root of both sides gives:
\[ r_{\text{planet}} = r_{\text{Earth}} \cdot 2^{1/3} \]

### Step 6: Calculate the acceleration due to gravity on the planet
The formula for acceleration due to gravity \( g \) is:
\[ g = \frac{GM}{r^2} \]
For the planet, we have:
\[ g_{\text{planet}} = \frac{G \cdot 2M_{\text{Earth}}}{(r_{\text{planet}})^2} \]
Substituting \( r_{\text{planet}} = r_{\text{Earth}} \cdot 2^{1/3} \):
\[ g_{\text{planet}} = \frac{G \cdot 2M_{\text{Earth}}}{(r_{\text{Earth}} \cdot 2^{1/3})^2} \]
This simplifies to:
\[ g_{\text{planet}} = \frac{2GM_{\text{Earth}}}{r_{\text{Earth}}^2 \cdot 2^{2/3}} = 2^{1/3} \cdot g_{\text{Earth}} \]

### Step 7: Calculate the weight of the object on the planet
The weight of the object on the planet can be expressed as:
\[ W_{\text{planet}} = m \cdot g_{\text{planet}} = m \cdot (2^{1/3} \cdot g_{\text{Earth}}) \]
Since the weight \( W \) on Earth is given by \( W = m \cdot g_{\text{Earth}} \):
\[ W_{\text{planet}} = 2^{1/3} \cdot W \]

### Final Answer
Thus, the weight of the object on the planet is:
\[ W_{\text{planet}} = 2^{1/3} W \]