The escape velocity from Earth is v_{es}. A b | vedclass

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(C) According to the law of conservation of energy,the total energy of the body at the surface of the Earth must be equal to the total energy in inter

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$\sqrt{3}v_{es}$

Vedclass · Answer

(C) According to the law of conservation of energy,the total energy of the body at the surface of the Earth must be equal to the total energy in interplanetary space.Let $v$ be the projection velocity and $v'$ be the velocity in interplanetary space.Total energy at surface = $\frac{1}{2}mv^2 - \frac{GMm}{R} = \frac{1}{2}mv^2 - \frac{1}{2}mv_{es}^2$.Total energy in interplanetary space = $\frac{1}{2}m(v')^2$.Equating the two: $\frac{1}{2}m(v')^2 = \frac{1}{2}mv^2 - \frac{1}{2}mv_{es}^2$.$v' = \sqrt{v^2 - v_{es}^2}$.Given $v = 2v_{es}$,we have $v' = \sqrt{(2v_{es})^2 - v_{es}^2} = \sqrt{4v_{es}^2 - v_{es}^2} = \sqrt{3v_{es}^2}$.Therefore,$v' = \sqrt{3}v_{es}$.

The escape velocity from Earth is $v_{es}$. $A$ body is projected with a velocity of $2v_{es}$. With what constant velocity will it move in the interplanetary space?

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