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, we need to find out the weight of an object on a planet that has double the mass of Earth and the same average density as Earth.

### Step-by-Step Solution:

1. **Understanding Weight**:
The weight \( W \) of an object on Earth is given by the formula:
\[ W = mg \]
where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity on Earth.

2. **Acceleration due to Gravity**:
The acceleration due to gravity \( g \) on the surface of a planet is given by:
\[ g = \frac{GM}{R^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.

3. **Mass of the Planet**:
According to the problem, the mass of the planet \( M_p \) is double the mass of Earth \( M_e \):
\[ M_p = 2M_e \]

4. **Density Relation**:
The density \( \rho \) of a planet is given by:
\[ \rho = \frac{M}{V} \]
where \( V \) is the volume of the planet. Since the density of the planet is equal to that of Earth, we have:
\[ \rho_p = \rho_e \]

5. **Volume of the Planet**:
The volume of a sphere is given by:
\[ V = \frac{4}{3}\pi R^3 \]
Therefore, for Earth:
\[ V_e = \frac{4}{3}\pi R_e^3 \]
and for the planet:
\[ V_p = \frac{4}{3}\pi R_p^3 \]

6. **Setting Up the Density Equation**:
Since the densities are equal:
\[ \frac{M_e}{V_e} = \frac{M_p}{V_p} \]
Substituting the values:
\[ \frac{M_e}{\frac{4}{3}\pi R_e^3} = \frac{2M_e}{\frac{4}{3}\pi R_p^3} \]
Simplifying this gives:
\[ R_p^3 = 2R_e^3 \implies R_p = R_e \cdot 2^{1/3} \]

7. **Calculating Gravity on the New Planet**:
Now we can calculate the acceleration due to gravity \( g_p \) on the new planet:
\[ g_p = \frac{G M_p}{R_p^2} \]
Substituting \( M_p \) and \( R_p \):
\[ g_p = \frac{G (2M_e)}{(R_e \cdot 2^{1/3})^2} = \frac{2G M_e}{R_e^2 \cdot 2^{2/3}} = \frac{2^{1/3} G M_e}{R_e^2} = 2^{1/3} g_e \]

8. **Weight on the New Planet**:
The weight \( W_p \) of the object on the new planet is:
\[ W_p = m g_p = m (2^{1/3} g_e) = 2^{1/3} W \]

### Final Answer:
Thus, the weight of the object on the new planet will be:
\[ W_p = 2^{1/3} W \]