A planet has double the mass of the earth. Its average density is equal to that of the earth. An object weighing W on earth will weigh on that planet:

Options

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

Answer

To solve the problem, we need to determine 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 the given data**:
- Let the mass of the Earth be \( M_e \).
- The mass of the planet \( M_p = 2M_e \).
- The average density of the planet \( \rho_p = \rho_e \) (density of Earth).

2. **Using the formula for density**:
- Density is defined as mass divided by volume:
\[ \rho = \frac{M}{V} \]
- The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
- For Earth:
\[ \rho_e = \frac{M_e}{\frac{4}{3} \pi r_e^3} \]
- For the planet:
\[ \rho_p = \frac{M_p}{\frac{4}{3} \pi r_p^3} \]
- Since \( \rho_p = \rho_e \), we can set the two equations equal:
\[ \frac{2M_e}{\frac{4}{3} \pi r_p^3} = \frac{M_e}{\frac{4}{3} \pi r_e^3} \]

3. **Simplifying the equation**:
- Cancel out \( \frac{4}{3} \pi \) and \( M_e \) (assuming \( M_e \neq 0 \)):
\[ \frac{2}{r_p^3} = \frac{1}{r_e^3} \]
- Rearranging gives:
\[ r_p^3 = 2r_e^3 \]
- Taking the cube root:
\[ r_p = 2^{1/3} r_e \]

4. **Calculating the weight on the planet**:
- The weight of an object is given by:
\[ W = mg \]
- The gravitational acceleration \( g \) on the surface of a planet is given by:
\[ g = \frac{GM}{r^2} \]
- For Earth:
\[ g_e = \frac{GM_e}{r_e^2} \]
- For the planet:
\[ g_p = \frac{G(2M_e)}{(2^{1/3} r_e)^2} = \frac{2GM_e}{(2^{2/3} r_e^2)} = \frac{2^{1/3} GM_e}{r_e^2} \]
- Thus, we can express the weight on the planet \( W_p \):
\[ W_p = mg_p = m \cdot \frac{2^{1/3} GM_e}{r_e^2} \]

5. **Relating the weights**:
- Since \( W = mg_e = m \cdot \frac{GM_e}{r_e^2} \):
- We can relate \( W_p \) to \( W \):
\[ W_p = 2^{1/3} W \]

### Final Answer:
An object weighing \( W \) on Earth will weigh \( 2^{1/3} W \) on that planet.