Two satellites are launched at a distance R f | vedclass

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(D) For a circular orbit at distance $R$,the orbital velocity is $v_0 = \sqrt{\frac{GM}{R}}$.For the second satellite,the initial total energy $E$ is

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$\frac{1}{7}R$

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(D) For a circular orbit at distance $R$,the orbital velocity is $v_0 = \sqrt{\frac{GM}{R}}$.For the second satellite,the initial total energy $E$ is given by $E = KE + PE = \frac{1}{2}m(\frac{v_0}{2})^2 - \frac{GMm}{R}$.Substituting $GM = v_0^2 R$,we get $E = \frac{1}{8}mv_0^2 - mv_0^2 = -\frac{7}{8}mv_0^2$.By conservation of angular momentum at the launch point $(R)$ and the point of minimum distance $(R_1)$,where velocity is $v_1$: $m(\frac{v_0}{2})R = mv_1 R_1$,so $v_1 = \frac{v_0 R}{2 R_1}$.By conservation of energy: $E = \frac{1}{2}mv_1^2 - \frac{GMm}{R_1}$.$-\frac{7}{8}mv_0^2 = \frac{1}{2}m(\frac{v_0 R}{2 R_1})^2 - \frac{mv_0^2 R}{R_1}$.Dividing by $mv_0^2$: $-\frac{7}{8} = \frac{R^2}{8 R_1^2} - \frac{R}{R_1}$.Let $x = \frac{R}{R_1}$,then $-\frac{7}{8} = \frac{1}{8}x^2 - x$.$x^2 - 8x + 7 = 0$.$(x-7)(x-1) = 0$.Since $R_1 < R$,$x = 7$,so $R_1 = \frac{R}{7}$.

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