The minimum energy required to launch a satel | vedclass

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Question

(D) The total energy of the satellite at the surface of the Earth is $E_i = K_i + U_i = K_i - \frac{G M m}{R}$.At an altitude of $2R$ from the surface

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$\frac{5 G m M}{6 R}$

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(D) The total energy of the satellite at the surface of the Earth is $E_i = K_i + U_i = K_i - \frac{G M m}{R}$.At an altitude of $2R$ from the surface,the distance from the center of the Earth is $r = R + 2R = 3R$.The orbital velocity $v$ at distance $r = 3R$ is given by $v = \sqrt{\frac{G M}{3R}}$.The total energy at the orbit is $E_f = K_f + U_f = \frac{1}{2} m v^2 - \frac{G M m}{3R} = \frac{1}{2} m \left(\frac{G M}{3R}\right) - \frac{G M m}{3R} = \frac{G M m}{6R} - \frac{G M m}{3R} = -\frac{G M m}{6R}$.By the law of conservation of energy,$E_i = E_f$:$K_i - \frac{G M m}{R} = -\frac{G M m}{6R}$.$K_i = \frac{G M m}{R} - \frac{G M m}{6R} = \frac{5 G M m}{6R}$.

The minimum energy required to launch a satellite of mass $m$ from the surface of the Earth of mass $M$ and radius $R$ into a circular orbit at an altitude of $2R$ from the surface of the Earth is:

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