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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$\sqrt{3}v_{es}$

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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 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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