Question

Escape velocity on Earth is ve = √(2gR). A planet of half the radius of Earth has the same density as Earth. The escape speed on that planet is ve / N. Find the value of N.

Answer 1

Correct Answer: 2

Concept:
Escape velocity of a planet is given by \[ v_e = \sqrt{\frac{2GM}{R}} \] where \(G\) is gravitational constant, \(M\) is mass of the planet and \(R\) is its radius. Also, \[ M = \rho \left(\frac{4}{3}\pi R^3\right) \] Thus escape velocity can also be written as \[ v_e = \sqrt{\frac{2G\left(\rho \frac{4}{3}\pi R^3\right)}{R}} \] \[ v_e \propto R \] when density \(\rho\) remains constant. Step 1: Relation between escape velocity and radius. Since density of the planet is same as that of earth: \[ v_e \propto R \] Step 2: Radius of the new planet. Given: \[ R' = \frac{R}{2} \] Step 3: Escape velocity of the planet. \[ v' = v_e \times \frac{R'}{R} \] \[ v' = v_e \times \frac{1}{2} \] \[ v' = \frac{v_e}{2} \] Step 4: Determine \(N\). Given \[ v' = \frac{v_e}{N} \] Thus \[ N = 2 \] \[ \boxed{N = 2} \]