A satellite of mass m is launched vertically upwards with an initial speed U from the surface of the Earth. After it reaches height 2 R (R = radius of Earth) it ejects a rocket of mass `(m)/(5)` So, that subsequently the satellite moves in a circular orbit. The kinetic energy of rocket is (M is mass of Earth)

Options

A. `(5K)/2`
B. `(66K)/5`
C. `(33K)/(10)`
D. 10K

Answer

C

Explanation

`K=(1)/(2)m u^(2)" "therefore (1)/(2)m u^(2)-(GMm)/(R)=(1)/(2) mv^(2)-(GMm)/(2R)`
`implies K-K=(1)/(2)mv^(2)-(K)/(2)-(K)/(2)" "implies mv=(m)/(5)v_(1)`
`implies v_(1)=5v" also, (4m)/(5)V_(S)=(m)/(5)V_(2)`
`implies V_(2)=4V_(S)" "therefore K_("rocket")=(1)/(2)((m)/(5))(V_(1)^(2)+V_(2)^(2))`
`=(m)/(10)(25V^(2)+16V_(S)^(2))=(5)/(2)mv^(2)+(8)/(5)mv_(S)^(2)=(5)/(2)K+(4K)/(5)=(33)/(10)K`