Q.8 If a projectile lifts off from the surface of the Earth with a speed of 11.2 km.s–1 , then it can escape
from the Earth’s gravitational field completely. This is called the escape velocity. If the radius of
the Earth were 2 times larger and the mass 8 times larger, then the escape velocity (in km.s–1 )
would be
(A) 5.6 (B) 11.2 (C) 22.4 (D) 44.8
Escape velocity from Earth’s surface is 11.2 km/s. For a modified Earth with radius doubled (2R) and mass increased 8 times (8M), the new escape velocity becomes 22.4 km/s .
Escape Velocity Formula
Escape velocity \( v_e \) is derived from conservation of energy, where kinetic energy at launch equals the absolute value of gravitational potential energy: \( \frac{1}{2} m v_e^2 = \frac{G M m}{R} \) [web:2]. This simplifies to \( v_e = \sqrt{\frac{2 G M}{R}} \), independent of the projectile’s mass m .
For original Earth: \( v_e = 11.2 \) km/s = \( \sqrt{\frac{2 G M}{R}} \) .
Modified Earth Calculation
New parameters: mass \( M’ = 8M \), radius \( R’ = 2R \). Substitute into formula: \( v_e’ = \sqrt{\frac{2 G (8M)}{2R}} = \sqrt{8 \cdot \frac{2 G M}{2R}} = \sqrt{4} \cdot v_e = 2 \times 11.2 = 22.4 \) km/s .
Since \( v_e \propto \sqrt{\frac{M}{R}} \), the factor is \( \sqrt{\frac{8M / 2R}{M / R}} = \sqrt{4} = 2 \) .
Option Analysis
| Option | Value | Analysis |
|---|---|---|
| (A) | 5.6 | Incorrect; implies halving velocity (factor 0.5), perhaps confusing with orbital velocity or inverse scaling. |
| (B) | 11.2 | Incorrect; assumes no change, ignoring mass/radius effects. |
| (C) | 22.4 | Correct; matches 2 times original due to net factor of 4 in \( \frac{M}{R} \) . |
| (D) | 44.8 | Incorrect; implies factor 4 (doubling again), maybe squaring the scaling. |
Exam Relevance
Common in CSIR NET Life Sciences (biophysics contexts) and JEE Physics. Variations test scaling: mass ×8 raises \( v_e \) by \( \sqrt{8} \approx 2.83 \), radius ×2 lowers by \( 1/\sqrt{2} \approx 0.707 \), product = 2 .
