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(C) Using the law of conservation of mechanical energy between the surface of the earth and the maximum height $h$:Total energy at surface = Total ene

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$2/3$

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(C) Using the law of conservation of mechanical energy between the surface of the earth and the maximum height $h$:Total energy at surface = Total energy at maximum height$-\frac{G M m}{R_E} + \frac{1}{2} m v^2 = -\frac{G M m}{R_E + h} + 0$Given $h = \frac{4}{5} R_E$,so $R_E + h = R_E + \frac{4}{5} R_E = \frac{9}{5} R_E$.Substituting $GM = g R_E^2$:$-\frac{g R_E^2 m}{R_E} + \frac{1}{2} m v^2 = -\frac{g R_E^2 m}{\frac{9}{5} R_E}$$-g R_E + \frac{v^2}{2} = -\frac{5}{9} g R_E$$\frac{v^2}{2} = g R_E - \frac{5}{9} g R_E = \frac{4}{9} g R_E$$v^2 = \frac{8}{9} g R_E$$v = \sqrt{\frac{8}{9} g R_E} = \frac{2}{3} \sqrt{2 g R_E}$Since escape velocity $v_E = \sqrt{2 g R_E}$,we have $v = \frac{2}{3} v_E$.Therefore,$\frac{v}{v_E} = \frac{2}{3}$.

An object is thrown directly away from the surface of the earth with an initial speed $v$. The object reaches up to a height of $\frac{4}{5} R_E$ from the earth's surface,where $R_E$ is the radius of the earth. If the escape velocity of the object is $v_E$,then the value of $\frac{v}{v_E}$ is:

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