Question

The escape velocity of an object from a planet is \(16\,\text{km s}^{-1}\). If the escape velocity of the object from another planet having twice the density and three times the radius of the planet is \(V\sqrt{2}\,\text{m s}^{-1}\), then the value of \(V\) is

• A. \(12\)
• B. \(48\)
• C. \(18\)
• D. \(36\)

Hint:

For a spherical planet, \[ v_e=\sqrt{\frac{2GM}{R}} \] and since \[ M=\frac{4}{3}\pi R^3\rho, \] we get \[ v_e\propto R\sqrt{\rho}. \]

Answer 1

The Correct Option is B

Step 1: Express escape velocity in terms of density and radius.
v_e = √(2GM/R), M = (4/3)πR³ρ → v_e = √(8πGρR²/3) ∝ R√ρ.

Step 2: Scale to the new planet.
R' = 3R, ρ' = 2ρ → v_e'/v_e = (3R√(2ρ))/(R√ρ) = 3√2.

Step 3: Compute new escape velocity.
v_e' = 3√2 × 16 = 48√2 km/s → V = 48.

Step 4: Final Answer:
48.