Question

There is a planet which is 8 times massive and 27 times denser than the earth. If \( g' \) and \( g \) are the accelerations due to gravity on the surfaces of the planet and the earth respectively, then

• A. \( g' = 8g \)
• B. \( g' = 27g \)
• C. \( g' = \dfrac{9}{4}g \)
• D. \( g' = 18g \)

Hint:

Use the derived formula \( g \propto M^{1/3} \rho^{2/3} \) when both mass and density change together. It's especially helpful when radius is not directly given.

Answer 1

The Correct Option is D

Step 1: Use the relation for gravity on a planet: \[ g = \dfrac{GM}{R^2}, \quad \text{where } M = \text{mass}, R = \text{radius} \] Step 2: Use the density formula: \[ \rho = \dfrac{M}{\dfrac{4}{3}\pi R^3} \Rightarrow R^3 = \dfrac{M}{\rho} \Rightarrow R = \left( \dfrac{M}{\rho} \right)^{1/3} \] Step 3: Substitute into gravity expression: \[ g = \dfrac{GM}{\left( \dfrac{M}{\rho} \right)^{2/3}} = G M^{1/3} \rho^{2/3} \Rightarrow g' \propto M^{1/3} \rho^{2/3} \] Step 4: Plug in values: \[ M = 8M_e, \quad \rho = 27\rho_e \Rightarrow g' = g \cdot 8^{1/3} \cdot 27^{2/3} = g \cdot 2 \cdot 9 = \boxed{18g} \]