Question:

The escape velocity of a projectile on the earth's surface is 11.2 km/s. A body is projected out with thrice this speed. The speed of the body far away from the earth will be?

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When an object is projected with speed higher than the escape velocity, its final velocity at infinity will be directly related to the initial speed.
Updated On: Apr 15, 2025
  • 22.4 km/s
  • 31.7 km/s
  • 33.6 km/s
  • None of these
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The Correct Option is B

Solution and Explanation


The escape velocity \( v_e \) is the minimum speed needed for an object to escape the gravitational pull of the earth. If a body is projected with thrice this velocity, the speed of the body at infinity will be: \[ v_{\infty} = \sqrt{3} \times v_e \] Given that \( v_e = 11.2 \text{ km/s} \), then: \[ v_{\infty} = \sqrt{3} \times 11.2 = 31.7 \text{ km/s} \] Thus, the speed of the body far away from the earth is 31.7 km/s.
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