CBSE · Class 11 · Physics · GravitationThe value of escape velocity from the surface of the Earth is approximately:
Step-by-Step Solution
Using $v_e = \sqrt{2gR}$ where $g = 9.8\text{ m/s}^2$ and $R = 6.4 \times 10^6\text{ m}$, we get $v_e \approx 11.2\text{ km/s}$.
Detailed Options Breakdown
Option : $11.2\text{ km/s}$ (Correct Answer)
Correct choice. Refer to the step-by-step verified solution guidelines above for details.
Option 1: $9.8\text{ km/s}$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Gravitation.
Option 2: $8.0\text{ km/s}$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Gravitation.
Option 3: $42.0\text{ km/s}$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Gravitation.
💡 Study Guide: This question tests core syllabus concepts from Gravitation. For formulas, key summaries, and mock exam reference guides, read the full Gravitation Revision Notes.