A body is released from a height equal to the radius (r) of the earth. The velocity of the body when it strikes the surface of the earth will be:
A body is released from a height equal to the radius (r) of the earth. The velocity of the body when it strikes the surface of the earth will be:
- \(\sqrt{gR}\)
- \(\frac{\sqrt{gR}}{2}\)
- \(\sqrt{2gR}\)
- \(\sqrt{4gR}\)
The Correct Option is A
Solution and Explanation
Solution:
Using the principle of energy conservation:
\( K_1 + U_1 = K_2 + U_2 \)
Where:
- \( K_1 \) is the initial kinetic energy,
- \( U_1 \) is the initial potential energy,
- \( K_2 \) is the final kinetic energy,
- \( U_2 \) is the final potential energy.
At the initial height, potential energy is:
\( U_1 = -\frac{GMm}{2R} \)
At the final height, the body strikes the surface, and the potential energy is:
\( U_2 = -\frac{GMm}{R} \)
The body is released, meaning its initial velocity is zero. Therefore, \( K_1 = 0 \).
Using conservation of energy:
\( 0 + \left(-\frac{GMm}{2R}\right) = \frac{1}{2}mv^2 + \left(-\frac{GMm}{R}\right) \)
Simplifying for velocity \( v \):
\( -\frac{GMm}{2R} = \frac{1}{2}mv^2 - \frac{GMm}{R} \)
\( \frac{GMm}{R} - \frac{GMm}{2R} = \frac{1}{2}mv^2 \)
\( \frac{GMm}{2R} = \frac{1}{2}mv^2 \)
\( \frac{GM}{R} = v^2 \)
\( v = \sqrt{\frac{GM}{R}} \)
Substitute \( g = \frac{GM}{R^2} \):
\( GM = gR^2 \)
\( v = \sqrt{\frac{gR^2}{R}} \)
\( v = \sqrt{gR} \)
Final Answer:
The velocity with which the body strikes the surface is \( v = \sqrt{gR} \).
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