Question

Escape velocity from a planet of radius \(R\) and density \(\rho\) is given as \(10\,\text{km s}^{-1}\). Find the escape velocity from a planet of radius \(\dfrac{R}{10}\) and density \(\dfrac{\rho}{10}\).

• A. \(10\sqrt{100}\,\text{m s}^{-1}\)
• B. \(110\sqrt{10}\,\text{m s}^{-1}\)
• C. \(100\sqrt{10}\,\text{m s}^{-1}\)
• D. \(90\sqrt{10}\,\text{m s}^{-1}\)

Hint:

Remember:
\(v_e \propto R\sqrt{\rho}\)
For uniform density planets, scaling laws simplify calculations

Answer 1

The Correct Option is C

Concept:
Escape velocity from a planet is: \[ v_e=\sqrt{\frac{2GM}{R}} \] For a planet of uniform density \(\rho\), \[ M=\frac{4}{3}\pi R^3\rho \] Hence, \[ v_e \propto R\sqrt{\rho} \]
Step 1: Write the proportionality relation. \[ \frac{v_{e2}}{v_{e1}} = \frac{R_2\sqrt{\rho_2}}{R_1\sqrt{\rho_1}} \]
Step 2: Substitute the given values. \[ R_2=\frac{R}{10},\quad \rho_2=\frac{\rho}{10} \] \[ \frac{v_{e2}}{10\,\text{km s}^{-1}} = \frac{\frac{R}{10}\sqrt{\frac{\rho}{10}}}{R\sqrt{\rho}} = \frac{1}{10\sqrt{10}} \]
Step 3: Find the new escape velocity. \[ v_{e2}=\frac{10}{10\sqrt{10}}\,\text{km s}^{-1} = \frac{1}{\sqrt{10}}\,\text{km s}^{-1} \] Convert to m/s: \[ v_{e2}=\frac{1000}{\sqrt{10}}\,\text{m s}^{-1} =100\sqrt{10}\,\text{m s}^{-1} \] \[ \boxed{v_{e2}=100\sqrt{10}\,\text{m s}^{-1}} \]