The acceleration due to gravity on a planet is same as that...

The acceleration due to gravity on a planet is same as that on earth and its radius is four times that of earth. What will be the value of escape velocity on that planet if it is $\mathrm{V}_{\mathrm{e}}$ on the earth (A) $\mathrm{V}_{\mathrm{e}}$ (B) $2 \mathrm{~V}_{\mathrm{e}}$ (C) $4 \mathrm{~V}_{\mathrm{e}}$ (D) $\mathrm{V}_{\mathrm{e}} / 2$

The acceleration due to gravity on a planet is same as that on earth and its radius is four times that of earth. What will be the value of escape velocity on that planet if it is $\mathrm{V}_{\mathrm{e}}$ on the earth
(A) $\mathrm{V}_{\mathrm{e}}$
(B) $2 \mathrm{~V}_{\mathrm{e}}$
(C) $4 \mathrm{~V}_{\mathrm{e}}$
(D) $\mathrm{V}_{\mathrm{e}} / 2$

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Key Concepts

Escape Velocity

Escape velocity is the minimum speed required for an object to break free from the gravitational pull of a celestial body without further propulsion. It is determined by the formula v? = ?(2GM/R), where G is the gravitational constant, M is the mass of the body, and R is its radius.

Gravitational Acceleration

Gravitational acceleration at the surface of a planet is given by g = GM/R², linking the mass of the planet and its radius. This relationship is crucial for understanding how mass and distance affect the strength of a planet's gravitational field.

Scaling Laws in Gravitational Systems

In gravitational systems, scaling laws help to understand how changes in physical parameters like radius affect other quantities such as escape velocity. By combining the escape velocity formula with the gravitational acceleration relation, one can derive that escape velocity can be expressed as v? = ?(2gR). Thus, if the gravitational acceleration remains the same while the radius increases by a factor (as in the example where the radius is four times that of Earth), the escape velocity scales with the square root of the radius, leading to an increase by a factor of ?4 = 2.

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