Question

Imagine a new planet having the same density as that of the earth, but it is two times bigger than the earth in size. If the acceleration due to gravity on the surface of the new planet is g' and that on the surface of the earth is g, then

• A. \( g' = \frac{g}{4} \)
• B. \( g' = 8g \)
• C. \( g' = 2g \)
• D. \( g' = 4g \)

Hint:

When density is constant, \( g \propto R \). This is a quick mental shortcut. If radius doubles, g doubles. If mass is constant, \( g \propto \frac{1}{R^2} \). Always check what parameter is kept constant or related.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem compares the acceleration due to gravity on the surface of Earth to that of a hypothetical new planet. We are given that the new planet has the same density as Earth but twice its size (radius).
Step 2: Key Formula or Approach:
The acceleration due to gravity 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 its radius.
We can express mass \( M \) in terms of density (\( \rho \)) and volume (\( V \)) for a sphere: \( M = \rho \times V = \rho \times \frac{4}{3}\pi R^3 \).
Substituting \( M \) into the formula for \( g \): \[ g = \frac{G \left( \rho \frac{4}{3}\pi R^3 \right)}{R^2} = G \rho \frac{4}{3}\pi R \] So, \( g \propto \rho R \).
Step 3: Detailed Explanation:
Let the properties of Earth be \( g, R, \rho \).
For the new planet, let the properties be \( g', R', \rho' \).
Given:
- Same density: \( \rho' = \rho \)
- Two times bigger in size (radius): \( R' = 2R \)
Now, form the ratio: \[ \frac{g'}{g} = \frac{G \rho' \frac{4}{3}\pi R'}{G \rho \frac{4}{3}\pi R} = \frac{\rho' R'}{\rho R} \] Substitute the given relations: \[ \frac{g'}{g} = \frac{\rho (2R)}{\rho R} = 2 \] Therefore, \( g' = 2g \).
Step 4: Final Answer:
The acceleration due to gravity on the surface of the new planet is \( 2g \).