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(B) The total energy of the satellite at the surface of the planet is $E_i = K_i + U_i = K_i - \frac{GmM}{R}$,where $K_i$ is the kinetic energy provid

Vedclass · Accepted Answer

$\frac{5GmM}{6R}$

Vedclass · Answer

(B) The total energy of the satellite at the surface of the planet is $E_i = K_i + U_i = K_i - \frac{GmM}{R}$,where $K_i$ is the kinetic energy provided at the surface.In the final circular orbit at altitude $h = 2R$,the distance from the center is $r = R + h = 3R$.The orbital velocity $v_0$ is given by $v_0 = \sqrt{\frac{GM}{3R}}$.The total energy in the orbit is $E_f = K_f + U_f = \frac{1}{2}mv_0^2 - \frac{GmM}{3R} = \frac{1}{2}m\left(\frac{GM}{3R}\right) - \frac{GmM}{3R} = \frac{GmM}{6R} - \frac{2GmM}{6R} = -\frac{GmM}{6R}$.By the law of conservation of energy,$E_i = E_f$:$K_i - \frac{GmM}{R} = -\frac{GmM}{6R}$.$K_i = \frac{GmM}{R} - \frac{GmM}{6R} = \frac{5GmM}{6R}$.

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

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