QJEEM20Phcdd2: JEE Main 2020 Physics - Class 11 JEEM20Phcdd2 Solution: A satellite of mass m is launched vertically upwards with an…

Using energy conservation

K_i + U_i = K_f + U_f

$\Rightarrow$ ${1 \over 2}m{u^2} - {{GmM} \over R}$ = ${1 \over 2}m{v^2} - {{GmM} \over {2R}}$

$\Rightarrow$ v = $\sqrt {{u^2} - {{GM} \over R}}$

After ejecting a rocket of mass ${m \over {10}}$ the remaining part of mass ${{9m} \over {10}}$ will rotate the earth with orbital velocity v₀.

$\therefore$ v₀ = $\sqrt {{{GM} \over {2R}}}$

Applying momentum conservation along radial direction,

Before firing rocket momentum of satelite in radial direction = mv

And after firing rocket momentum of satelite in radial direction = 0 and momentum of rocket in radial direction = ${m \over {10}}{v_2}$

$\therefore$ mv = ${m \over {10}}{v_2}$

$\Rightarrow$ v₂ = 10v

Now applying momentum conservation along tangential direction we get,

0 = ${m \over {10}}{v_1}$ - ${{9m} \over {10}}{v_0}$

$\Rightarrow$ ${{9m} \over {10}}{v_0}$ = ${m \over {10}}{v_1}$

$\Rightarrow$ v₁ = 9v₀

$\therefore$ Total Kinetic Energy of rocket

= ${1 \over 2}{m \over {10}}\left( {v_1^2 + v_2^2} \right)$

= ${1 \over 2}{m \over {10}}\left( {81v_0^2 + 100{v^2}} \right)$

= ${m \over {20}}\left( {81\left( {{{GM} \over {2R}}} \right) + 100\left( {{u^2} - {{GM} \over R}} \right)} \right)$

= ${m \over {20}}\left( {100{u^2} + {{81GM} \over {2R}} - {{100GM} \over R}} \right)$

= ${m \over {20}}\left( {100{u^2} - {{119GM} \over {2R}}} \right)$

= $5m\left( {{u^2} - {{119GM} \over {200R}}} \right)$