AIIMS2017Physics-Gravitation

AIIMS 2017 Physics Escape Velocity MCQ Question

Type: MCQ-conceptual-Medium-Class 11

The escape velocity from the earth is about 11 kms1kms ^{-1}.The escape velocity from a planet having twice he radius and the same mean density as the earth is

A

22 kms1kms ^{-1}

B

22 kms1kms ^{-1}

C

22 kms1kms ^{-1}

D

22 kms1kms ^{-1}

Correct Answer

Option C

Detailed Explanation

The escape velocity VeV_e is derived from the gravitational potential energy and kinetic energy relationship, specifically given by the formula Ve=2GMRV_e = \sqrt{\frac{2GM}{R}}, where GG is the gravitational constant, MM is the mass of the celestial body, and RR is its radius. Option C, Ve=GMRV_e = \sqrt{\frac{GM}{R}}, is incorrect because it omits the factor of 2, which is essential for calculating the velocity needed to escape a gravitational field. Options A and D also misrepresent the relationship by introducing incorrect constants or factors that do not align with the fundamental principles of gravitational physics.

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