Question

The escape velocity of a body on a planet ‘A’ is 12 kms^–1. The escape velocity of the body on another planet ‘B’, whose density is four times and radius is half of the planet ‘A’, is

• A. 12 kms^–1
• B. 24 kms^–1
• C. 36 kms^–1
• D. 6 kms^–1

Answer 1

The Correct Option is A

To find the escape velocity of a body on planet 'B', we start by understanding the given problem. The escape velocity \( v_e \) for any planet is given by the formula:

v_e = \sqrt{2gR}

where:

• g is the acceleration due to gravity on the surface of the planet.
• R is the radius of the planet.

Another expression for escape velocity in terms of the mass (M) and radius (R) is:

v_e = \sqrt{\frac{2GM}{R}}

The mass M can be expressed as M = \rho \frac{4}{3}\pi R^3, where \rho is the density of the planet.

Thus, the escape velocity becomes:

v_e = \sqrt{\frac{2G(\rho \frac{4}{3}\pi R^3)}{R}} or simplified to v_e = \sqrt{\frac{8}{3}\pi G \rho R^2}

For planet 'A', the escape velocity is given as 12 km/s, with density \rho_A and radius R_A. For planet 'B':

• Density \rho_B = 4\rho_A
• Radius R_B = \frac{1}{2}R_A

Substituting the conditions for planet 'B' into the escape velocity formula:

v_{eB} = \sqrt{\frac{8}{3}\pi G (4\rho_A) \left(\frac{1}{2}R_A\right)^2}

This simplifies to:

v_{eB} = \sqrt{\frac{8}{3}\pi G \rho_A R_A^2}

Upon further simplification, it matches the escape velocity formula of planet 'A':

v_{eB} = v_{eA}

Since v_{eA} = 12 km/s, the escape velocity on planet 'B' is also 12 km/s.

Therefore, the correct answer is:

12 km/s