Question:

A body which is initially at rest at a height \( R \) above the surface of the Earth of radius \( R \), falls freely towards the Earth. The velocity on reaching the surface of the Earth is:

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For free fall from height \( h \), the velocity on reaching the surface can be found using energy conservation: \[ v = \sqrt{2 g h} \] or using gravitational potential.
Updated On: Mar 24, 2025
  • \( \sqrt{2gR} \)
  • \( \sqrt{gR} \)
  • \( \sqrt{\frac{3}{2} gR} \)
  • \( \sqrt{4gR} \)
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The Correct Option is B

Solution and Explanation

Step 1: {Apply conservation of energy}
The total energy remains constant: \[ {Increase in kinetic energy} = {Decrease in potential energy} \] Step 2: {Using gravitational potential energy}
\[ \frac{1}{2} mv^2 = mgR \left( \frac{1}{1 + \frac{h}{R}} \right) \] For \( h = R \): \[ mv^2 = mgR \] \[ v = \sqrt{gR} \] Thus, the correct answer is \( \sqrt{gR} \).
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