A satellite of massM is… - JEE Main 2020 Physics PYQ

JEE Main 2020 — Physics Mechanics

2020mcqhard

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):

Official previous-year question

Held on 7 Jan 2020 · Verified 6 Jul 2026.

Options

A
$\frac{M}{20}({u}^{2}+\frac{113}{200}\frac{G{M}_{e}}{R})$
B
$5M({u}^{2}-\frac{119}{200}\frac{G{M}_{e}}{R})$
C
$\frac{3M}{8}{(u+\sqrt{\frac{5G{M}_{e}}{6R}})}^{2}$
D
$\frac{M}{20}{(u-\sqrt{\frac{2G{M}_{e}}{3R}})}^{2}$

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Solution

$\frac{-{\mathrm{GM}}_{e}M}{R}+\frac{1}{2}{\mathrm{Mu}}^{2}=\frac{-{\mathrm{GM}}_{e}M}{2R}+\frac{1}{2}{\mathrm{Mv}}^{2}$

JEE Main 2020 Physics, Gravitation — solution figure 1

$v=\sqrt{{u}^{2}-\frac{G{M}_{e}}{R}}$ JEE Main 2020 Physics, Gravitation — solution figure 2

${V}_{\tau }\rightarrow$ Transverse velocity of rocket

${V}_{R}\rightarrow$ Radial velocity of rocket

$V=\sqrt{\frac{G{M}_{e}}{2R}}$

$\frac{M}{10}{V}_{T}=\frac{9M}{10}\sqrt{\frac{G{M}_{e}}{2R}}$

$\frac{M}{10}{V}_{r}=M\sqrt{{u}^{2}-\frac{{\mathrm{GM}}_{e}}{R}}$

Kinetic energy $=\frac{1}{2}\frac{M}{10}({V}_{T}^{2}+{V}_{r}^{2})=\frac{M}{20}(81\frac{{\mathrm{GM}}_{e}}{2R}+100{u}^{2}-100\frac{{\mathrm{GM}}_{e}}{R})$

$=\frac{M}{20}(100{u}^{2}-\frac{119G{M}_{e}}{2R})$

$=5M({u}^{2}-\frac{119{\mathrm{GM}}_{e}}{200R})$

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Options

$\frac{M}{20}({u}^{2}+\frac{113}{200}\frac{G{M}_{e}}{R})$

$5M({u}^{2}-\frac{119}{200}\frac{G{M}_{e}}{R})$

$\frac{3M}{8}{(u+\sqrt{\frac{5G{M}_{e}}{6R}})}^{2}$

$\frac{M}{20}{(u-\sqrt{\frac{2G{M}_{e}}{3R}})}^{2}$

Solution

$\frac{-{\mathrm{GM}}_{e}M}{R}+\frac{1}{2}{\mathrm{Mu}}^{2}=\frac{-{\mathrm{GM}}_{e}M}{2R}+\frac{1}{2}{\mathrm{Mv}}^{2}$

JEE Main 2020 Physics, Gravitation — solution figure 1

$v=\sqrt{{u}^{2}-\frac{G{M}_{e}}{R}}$ JEE Main 2020 Physics, Gravitation — solution figure 2

${V}_{\tau }\rightarrow$ Transverse velocity of rocket

${V}_{R}\rightarrow$ Radial velocity of rocket

$V=\sqrt{\frac{G{M}_{e}}{2R}}$

$\frac{M}{10}{V}_{T}=\frac{9M}{10}\sqrt{\frac{G{M}_{e}}{2R}}$

$\frac{M}{10}{V}_{r}=M\sqrt{{u}^{2}-\frac{{\mathrm{GM}}_{e}}{R}}$

Kinetic energy $=\frac{1}{2}\frac{M}{10}({V}_{T}^{2}+{V}_{r}^{2})=\frac{M}{20}(81\frac{{\mathrm{GM}}_{e}}{2R}+100{u}^{2}-100\frac{{\mathrm{GM}}_{e}}{R})$

$=\frac{M}{20}(100{u}^{2}-\frac{119G{M}_{e}}{2R})$

$=5M({u}^{2}-\frac{119{\mathrm{GM}}_{e}}{200R})$

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Options

Did you get this right?

Solution

Your note

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More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

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Keep practicing Mechanics