Question

The radius of a planet is twice the radius of the earth. Both have almost equal average mass densities. If $V_P$ and $V_E$ are escape velocities of the planet and the earth respectively, then

• A. $V_E = 1.5\ V_P$
• B. $V_P = 1.5\ V_E$
• C. $V_P = 2\ V_E$
• D. $V_E = 3\ V_P$

Hint:

Escape velocity can be conceptualized in two ways: $v \propto \sqrt{\frac{M}{R}}$ when working with absolute mass, and $v \propto R\sqrt{\rho}$ when working with density. Under constant density conditions, escape velocity is strictly proportional to the radius. Doubling the radius automatically doubles the escape velocity.

Answer 1

The Correct Option is C

Step 1: State the comparison.
A planet has radius twice Earth's and the same average density. We compare escape velocities $V_P$ and $V_E$.
Step 2: Write escape velocity.
$v = \sqrt{\dfrac{2GM}{R}}$, and mass of a uniform sphere is $M = \dfrac{4}{3}\pi R^3\rho$.
Step 3: Substitute the mass.
$v = \sqrt{\dfrac{2G}{R}\cdot\dfrac{4}{3}\pi R^3\rho} = R\sqrt{\dfrac{8\pi G\rho}{3}}$.
Step 4: Identify the proportionality.
With $\rho$ the same for both bodies, the square-root factor is constant, so $v \propto R$.
Step 5: Form the ratio.
$\dfrac{V_P}{V_E} = \dfrac{R_P}{R_E}$.
Step 6: Insert the radius relation.
$R_P = 2R_E$ gives $\dfrac{V_P}{V_E} = 2$, so $V_P = 2V_E$, which is option (3).
\[ \boxed{V_P = 2V_E} \]