The escape velocity of a body from the surface of the earth is 11.2 km/s. If a body is projected with a velocity twice the escape velocity, then the velocity of the body far away from the earth will be:
Show Hint
- 11.2 km/s
- $11.2 \sqrt{3}$ km/s
- $11.2 \sqrt{2}$ km/s
- 22.4 km/s
The Correct Option is B
Solution and Explanation
Velocity at infinity $v_{\infty} = \sqrt{v^{2} - v_{e}^{2}}$.
Step 2: Analysis
Given $v = 2v_{e}$. $v_{\infty} = \sqrt{(2v_{e})^{2} - v_{e}^{2}} = \sqrt{4v_{e}^{2} - v_{e}^{2}} = \sqrt{3v_{e}^{2}}$.
Step 3: Conclusion
$v_{\infty} = v_{e}\sqrt{3} = 11.2 \sqrt{3}$ km/s.
Final Answer: (B)
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