Question

A planet has mass \( 81 M_E \) and radius \( 9 R_E \), where \( M_E \) and \( R_E \) are the mass and radius of Earth respectively. The escape velocity from the planet is:

• A. Equal to Earth's escape velocity
• B. 3 times Earth's escape velocity
• C. \( \frac{1}{3} \) times Earth's escape velocity
• D. 9 times Earth's escape velocity

Hint:

Escape velocity scales as \( v_e \propto \sqrt{\frac{M}{R}} \). If mass is multiplied by \( k_1 \) and radius by \( k_2 \), the escape velocity changes by a factor of \( \sqrt{\frac{k_1}{k_2}} \). Here, \( \sqrt{\frac{81}{9}} = \sqrt{9} = 3 \).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to find the escape velocity of a planet given its mass and radius in terms of Earth's mass and radius.

Step 2: Key Formula or Approach:
The escape velocity \( v_e \) from the surface of a spherical body of mass \( M \) and radius \( R \) is given by:
\[ v_e = \sqrt{\frac{2GM}{R}} \] where \( G \) is the universal gravitational constant.

Step 3: Detailed Explanation:
Let the escape velocity of Earth be:
\[ v_E = \sqrt{\frac{2GM_E}{R_E}} \] For the given planet, mass \( M_P = 81 M_E \) and radius \( R_P = 9 R_E \).
The escape velocity from this planet is:
\[ v_P = \sqrt{\frac{2GM_P}{R_P}} \] Substitute the values of \( M_P \) and \( R_P \):
\[ v_P = \sqrt{\frac{2G(81M_E)}{9R_E}} \] \[ v_P = \sqrt{9 \times \frac{2GM_E}{R_E}} \] \[ v_P = 3 \sqrt{\frac{2GM_E}{R_E}} = 3 v_E \] Thus, the escape velocity from the planet is 3 times the escape velocity of Earth.

Step 4: Final Answer:
(B) 3 times Earth's escape velocity