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JEE Main 20207 JanuaryMorning Shift
Exam
JEE Main
Year
2020
Date
7 January
Shift
Morning Shift
Subject
Physics
Chapter
Gravitation

JEE Main 2020 · 7 January · Morning Shift · Physics · Gravitation

A satellite of mass M is launched vertically upwards with an...

A satellite of massMM is launched vertically upwards with an initial speed uu from the surface of the earth. After it reaches height RR ( R=R = radius of the earth), it ejects a rocket of mass M10\dfrac{M}{10} so that subsequently the satellite moves in a circular orbit. The kinetic energy of the rocket is ( GG is the gravitational constant; Me{M}_{e} is the mass of the earth):A satellite of mass M is launched vertically upwards with an initial speed u from the surface of the earth. After itA satellite of mass M is launched vertically upwards with an initial speed u from the surface of the earth. After it

Options

  1. AM20(u2+113200GMeR)\dfrac{M}{20}\left({u}^{2} + \dfrac{113}{200}\dfrac{G{M}_{e}}{R}\right)
  2. B5M(u2−119200GMeR)5M\left({u}^{2} - \dfrac{119}{200}\dfrac{G{M}_{e}}{R}\right)
  3. C\dfrac{3M}{8}{\left(u + \sqrt{\dfrac{5G{M}_{e}}{6R}}\right)}^{2}
  4. D\dfrac{M}{20}{\left(u - \sqrt{\dfrac{2G{M}_{e}}{3R}}\right)}^{2}

Correct answer

B. 5M(u2−119200GMeR)5M\left({u}^{2} - \dfrac{119}{200}\dfrac{G{M}_{e}}{R}\right)

Step-by-step solution

−GMeMR+12Mu2=−GMeM2R+12Mv2\dfrac{- {GM}_{e}M}{R} + \dfrac{1}{2}{Mu}^{2} = \dfrac{- {GM}_{e}M}{2R} + \dfrac{1}{2}{Mv}^{2}

A satellite of mass M is launched vertically upwards with an initial speed u from the surface of the earth. After it

v=u2−GMeRv = \sqrt{{u}^{2} - \dfrac{G{M}_{e}}{R}}

A satellite of mass M is launched vertically upwards with an initial speed u from the surface of the earth. After it

Vτ→{V}_{τ} \to Transverse velocity of rocket
VR→{V}_{R} \to Radial velocity of rocket

V=GMe2RV = \sqrt{\dfrac{G{M}_{e}}{2R}}

M10VT=9M10GMe2R\dfrac{M}{10}{V}_{T} = \dfrac{9M}{10}\sqrt{\dfrac{G{M}_{e}}{2R}}
M10Vr=Mu2−GMeR\dfrac{M}{10}{V}_{r} = M\sqrt{{u}^{2} - \dfrac{{GM}_{e}}{R}}

Kinetic energy =12M10(VT2+Vr2)=M20(81GMe2R+100u2−100GMeR)= \dfrac{1}{2}\dfrac{M}{10}\left({V}_{T}^{2} + {V}_{r}^{2}\right) = \dfrac{M}{20}\left(81\dfrac{{GM}_{e}}{2R} + 100{u}^{2} - 100\dfrac{{GM}_{e}}{R}\right)

=M20(100u2−119GMe2R)= \dfrac{M}{20}\left(100{u}^{2} - \dfrac{119G{M}_{e}}{2R}\right)
=5M(u2−119GMe200R)= 5M\left({u}^{2} - \dfrac{119{GM}_{e}}{200R}\right)

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