The escape velocity for a rocket from earth is 11.2 kms^-1 value on a planet where acceleration

step 1 Hello students in this question we have the escape speed on the surface of earth is 11 .2 kilome

The escape velocity for a rocket from earth is 11.2 kms^-1...

The escape velocity for a rocket from earth is $11.2 \mathrm{kms}^{-1}$ value on a planet where acceleration due to gravity is double that on earth and diameter of the planet is twice that of earth will be $=\ldots \ldots \ldots \ldots \mathrm{kms}^{-1}$ (A) $11.2$ (B) $22.4$ (C) $5.6$ (C) $53.6$

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Key Concepts

Escape Velocity

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

Gravitational Acceleration

Gravitational acceleration at the surface of a planet is defined by g = GM/R² and represents the gravitational pull experienced by an object on that surface. This parameter encapsulates both the mass of the planet and its radius, establishing a direct connection between the planet's gravitational field and the escape velocity.

Scaling of Planetary Parameters

Scaling planetary dimensions such as doubling of the planet's radius and simultaneous changes in the surface gravitational acceleration directly affect the escape velocity. Since escape velocity depends on both gravity and the radius (v? ? ?(gR)), altering these parameters by specific factors (such as doubling each) results in a predictable change in the escape velocity according to the scaling relationship.

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