Question 8: Gravitation

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, then its velocity on reaching the surface of the Earth is

Accepted Answer

To determine the velocity of a body falling from rest at a height $ R $ above the Earth's surface, we must equate the increase in kinetic energy to the decrease in potential energy.
Given:
Initial height $ h = R $
Radius of Earth $ R $
Acceleration due to gravity $ g $
The change in potential energy when the body falls to the Earth's surface can be given by:
$ \Delta U = \frac{mgh}{1 + \frac{h}{R}} $
Substituting $ h = R $, we get:
$ \Delta U = \frac{mgR}{1 + \frac{R}{R}} = \frac{mgR}{2} $
By the conservation of energy principle, the increase in kinetic energy equals the decrease in potential energy:
$ \frac{1}{2} mv^2 = \frac{mgR}{2} $
Solving for $ v $:
$ mv^2 = mgR $
$ v^2 = gR $
$ v = \sqrt{gR} $
Thus, the velocity of the body when it reaches the Earth's surface is $ v = \sqrt{gR} $.

Answer

$ \sqrt{(2 g R)} $

Answer

$ \sqrt{(g R)} $

Answer

$ \sqrt{\frac{3}{2} g R} $

Answer

$ \sqrt{(4 g R)} $

Gravitation

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, then its velocity on reaching the surface of the Earth is

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