Planet P has a mean uniform density that is twice that of Planet Q, and a radius that is h… — JAMB UTME

Planet $P$ has a mean uniform density that is twice that of Planet $Q$ , and a radius that is half that of Planet $Q$ . What is the ratio of the escape velocity at the surface of Planet $P$ to the escape velocity at the surface of Planet $Q$ ?

$\frac{1}{\sqrt{2}}$ Answer
B
$\frac{1}{2}$
C
$1$
D
$\sqrt{2}$

Answer

\frac{1}{\sqrt{2}}

The correct answer is derived by substituting the planet's mass in terms of radius and density (

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

) into the escape velocity equation

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

. Simplifying gives

v_e = R\sqrt{\frac{8}{3}\pi G \rho}

, which means escape velocity is proportional to

R\sqrt{\rho}

. Halving the radius and doubling the density yields a factor of

\frac{1}{2} \times \sqrt{2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}

Step-by-Step Solution

Express the mass of a uniform spherical planet in terms of its radius

R

and density

\rho

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

Mass equals volume multiplied by mean density.

Substitute mass into the escape velocity formula

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

v_e = \sqrt{\frac{2G \left(\frac{4}{3}\pi R^3 \rho\right)}{R}} = R\sqrt{\frac{8}{3}\pi G \rho}

This establishes the proportionality

v_e \propto R\sqrt{\rho}

Set up the ratio of escape velocity for Planet

P

to Planet

Q

using

R_P = \frac{1}{2}R_Q

and

\rho_P = 2\rho_Q

\frac{v_{e,P}}{v_{e,Q}} = \frac{R_P\sqrt{\rho_P}}{R_Q\sqrt{\rho_Q}} = \frac{\left(\frac{1}{2}R_Q\right)\sqrt{2\rho_Q}}{R_Q\sqrt{\rho_Q}} = \frac{1}{2}\sqrt{2} = \frac{1}{\sqrt{2}}

Evaluating the ratio of proportional quantities yields the final scale factor.

Key Concept

Dependence of Escape Velocity on Planet Density and Radius

Estimated Time:2m 0s

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