The escape velocity of a body on the surface of the earth is 11.2 km / sec. If the mass of the

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The escape velocity of a body on the surface of the earth is...

The escape velocity of a body on the surface of the earth is $11.2 \mathrm{~km} / \mathrm{sec}$. If the mass of the earth is increases to twice its present value and the radius of the earth becomes half, the escape velocity becomes $=\ldots \ldots \ldots \ldots \mathrm{kms}^{-1}$ $(\Delta) 56$

Key Concepts

Dependence of Escape Velocity on Mass and Radius

The escape velocity is directly influenced by both the mass and the radius of the celestial body, as seen in the formula v? = ?(2GM/R). Changes in the mass and radius will affect the gravitational pull: increasing the mass leads to a stronger gravitational field, thereby increasing the escape velocity, while decreasing the radius increases the gravitational field strength at the surface, also resulting in a higher escape velocity. This relationship is key to analyzing how modifications to these parameters alter escape speed.

Escape Velocity

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

Gravitational Potential Energy

Gravitational potential energy represents the energy an object possesses due to its position in a gravitational field. In the context of escape velocity, the energy required to move an object from the surface of a celestial body to a point at infinity (where the gravitational influence is negligible) is balanced with its kinetic energy, making it a fundamental concept for understanding escape mechanics.

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