A satellite of mass $ M $ is launched vertically upwards with an initial speed $ u $ from the surface of the earth. After it reaches height $ R $ ( $ R= $ radius of the earth), it ejects a rocket of mass $ \frac{M}{10} $ so that subsequently the satellite moves in a circular orbit. The kinetic energy of the rocket is ( $ G $ is the gravitational constant; $ M_{e} $ is the mass of the earth):

Question

A satellite of mass $ M $ is launched vertically upwards with an initial speed $ u $ from the surface of the earth. After it reaches height $ R $ ( $ R= $ radius of the earth), it ejects a rocket of mass $ \frac{M}{10} $ so that subsequently the satellite moves in a circular orbit. The kinetic energy of the rocket is ( $ G $ is the gravitational constant; $ M_{e} $ is the mass of the earth):, $ \frac{M}{20}\left(u^{2}+\frac{113}{200} \frac{G M_{e}}{R}\right) $, $ 5 M\left(u^{2}-\frac{119}{200} \frac{G M_{e}}{R}\right) $, $ \frac{3 M}{8}\left(u+\sqrt{\frac{5 G M_{e}}{6 R}}\right)^{2} $, $ \frac{M}{20}\left(u-\sqrt{\frac{2 G M_{e}}{3 R}}\right)^{2} $

MARKS App · Accepted Answer

- GM e M R + 1 2 Mu 2 = - GM e M 2 R + 1 2 Mv 2 v = u 2 - G M e R V τ → Transverse velocity of rocket V R → Radial velocity of rocket V = G M e 2 R M 10 V T = 9 M 10 G M e 2 R M 10 V r = M u 2 - GM e R Kinetic energy = 1 2 M 10 V T 2 + V r 2 = M 20 81 GM e 2 R + 100 u 2 - 100 GM e R = M 20 100 u 2 - 119 G M e 2 R = 5 M u 2 - 119 GM e 200 R

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