Question

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:

• A. 1
• B. (2)\(^{\frac{1}{3}}\)
• C. (2)\(^{-\frac{1}{3}}\)
• D. 2

Answer 1

The Correct Option is B

The correct answer is option (B): \((2) ^{\frac{1}{3}}\)

Answer 2

Weight on a Different Planet Problem

Step 1: Relationship between Mass, Density, and Radius

Let \( M_e, R_e, \) and \( \rho_e \) be the mass, radius, and average density of Earth, respectively. Let \( M_p, R_p, \) and \( \rho_p \) be the mass, radius, and average density of the planet, respectively. Given that \( M_p = 2M_e \) and \( \rho_p = \rho_e \). Density is defined as mass divided by volume: \( \rho = \frac{M}{V} \). Assuming spherical shapes, \( V = \frac{4}{3} \pi R^3 \). Thus:

\( \rho_e = \frac{M_e}{\frac{4}{3}\pi R_e^3} \) and \( \rho_p = \frac{M_p}{\frac{4}{3}\pi R_p^3} \)

Since \( \rho_e = \rho_p \):

\( \frac{M_e}{R_e^3} = \frac{M_p}{R_p^3} \)

\( \frac{M_e}{R_e^3} = \frac{2M_e}{R_p^3} \)

\( R_p^3 = 2R_e^3 \implies R_p = 2^{1/3} R_e \)

Step 2: Relationship between Weight and Gravitational Acceleration

Weight (W) is given by \( W = mg \), where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity.

Step 3: Gravitational Acceleration

The acceleration due to gravity (g) 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. Let \( g_e \) and \( g_p \) be the acceleration due to gravity on Earth and the planet, respectively.

\( \frac{g_e}{g_p} = \frac{M_e / R_e^2}{M_p / R_p^2} = \frac{M_e}{R_e^2} \times \frac{R_p^2}{M_p} = \frac{M_e}{R_e^2} \times \frac{(2^{1/3} R_e)^2}{2M_e} = \frac{2^{2/3}}{2} = 2^{-1/3} \)

\( g_p = 2^{1/3} g_e \)

Step 4: Weight on the Planet

Since the mass of the object remains constant, the weight on the planet (\( W_p \)) is:

\( W_p = mg_p = m(2^{1/3} g_e) = 2^{1/3} (mg_e) = 2^{1/3} W \)

Conclusion:

The object will weigh \( \mathbf{2^{1/3}W} \) on the planet (Option 1).