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Gravitation Part 2

Section: Physics

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Question : 9 of 110

Marks: +1, -0

The escape velocity from a spherical planet

A

10\ \text{km/s}

. The escape velocity from another planet

B

whose density and radius are

10\%

of those of planet

A

\_\_\_\_

\text{m/s}

[22 Jan 2026 Shift 1]

1000
$200\sqrt{5}$
$1000\sqrt{2}$
$100\sqrt{10}$

Solution:

Concept:

Escape velocity

v_e

depends on a planet's radius

R

and density

\rho

as

v_e \propto R\sqrt{\rho}

.

Explanation:

The escape velocity formula is

v_e = \sqrt{\frac{2GM}{R}}

.

For a spherical planet, mass

M = \frac{4}{3}\pi R^3 \rho

.

Substituting gives

v_e = \sqrt{\frac{2G}{R} \cdot \frac{4}{3}\pi R^3 \rho} = R\sqrt{\frac{8}{3}\pi G \rho}

.

Thus

v_e \propto R\sqrt{\rho}

.

Given

v_{eA} = 10\ \text{km/s} = 10000\ \text{m/s}

.

Planet B has

R_B = 0.1 R_A

and

\rho_B = 0.1 \rho_A

.

Using proportionality:

\frac{v_{eB}}{v_{eA}} = \frac{R_B \sqrt{\rho_B}}{R_A \sqrt{\rho_A}} = \frac{0.1 R_A \sqrt{0.1 \rho_A}}{R_A \sqrt{\rho_A}} = 0.1 \sqrt{0.1} = \frac{1}{10\sqrt{10}}

.

Therefore

v_{eB} = 10000 \times \frac{1}{10\sqrt{10}} = \frac{1000}{\sqrt{10}} = 100\sqrt{10}\ \text{m/s}

.

Answer:

100\sqrt{10}\ \text{m/s}

, which corresponds to option (D).

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