Given mass of the moon is (1718) of the mass of the earth...

Given mass of the moon is $(1718)$ of the mass of the earth and corresponding radius is $(1 / 4)$ of the earth, If escape velocity on the earth surface is $11.2 \mathrm{kms}^{-1}$ the value of same on the surface of moon is $=\ldots \ldots \ldots \ldots \mathrm{kms}^{-1}$. (A) $0.14$ (B) $0.5$ (C) $2.5$ (D) 5

Key Concepts

Dependence on Mass and Radius

The escape velocity depends directly on the square root of the mass and inversely on the square root of the radius of a celestial body. This means that for two bodies, the ratio of their escape velocities can be determined by taking the square root of the ratio of their masses divided by the ratio of their radii. This proportional relationship helps in comparing different bodies’ gravitational pulls, regardless of their absolute sizes and masses.

Use of Ratios in Gravitational Comparisons

When comparing escape velocities of different celestial bodies, one can use known ratios of their masses and radii to derive the escape velocity of one body from that of another. This method involves canceling common factors and applying the square root function to the resulting ratio. Such scaling techniques are essential in astrophysics as they simplify calculations and provide insights into how differences in physical properties affect gravitational behavior.

Escape Velocity Formula

Escape velocity is the minimum speed needed for an object to break free from the gravitational attraction of a celestial body without further propulsion. It is given by the formula v? = ?(2GM/R), where G is the gravitational constant, M is the mass of the body, and R is its radius. This formula is a fundamental concept in astrophysics and classical mechanics, illustrating the interplay between mass and distance in gravitational fields.

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