The escape velocity from the earth is about 11 kms^-1. The escape velocity from a planet having

step 1 Here the correct answer will be that is option D and now I will give away the explanation So the

The escape velocity from the earth is about 11 kms^-1. The...

The escape velocity from the earth is about $11 \mathrm{kms}^{-1}$. The escape velocity from a planet having twice the radius and the same mean density as the earth is $=\ldots \ldots \ldots \ldots \mathrm{kms}^{-1}$.
(A) 22
(B) 11
(C) $5.5$
(D) $15.5$

Key Concepts

Scaling Laws in Gravitational Systems

Scaling laws describe how physical quantities change with respect to the dimensions of a system. In gravitational systems, if the size (radius) of a planet increases while its density remains the same, the mass increases in proportion to the volume (or the cube of the radius), which in turn affects the escape velocity. In these cases, the escape velocity scales in proportion to the radius because the mass-to-radius ratio changes predictably.

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 depends on both the mass and radius of the body, typically calculated using the formula v = ?(2GM/R), where G is the gravitational constant, M is the mass, and R is the radius.

Mean Density

Mean density is the average density of a celestial object and is defined as the mass of the object divided by its volume. When two bodies have the same mean density, any changes in their radii are directly related to changes in their masses. This relationship is crucial in determining how the escape velocity will vary when the radius is altered while maintaining the same density.

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