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JEE Mains · Physics · STD 11 - Gravitation
  • JEE Mains 2020
  • Hard
  • MCQ

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

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Answer & Solution

Correct Answer

(B) \(5 \mathrm{m}\left(\mathrm{u}^{2}-\frac{119}{200} \frac{\mathrm{GM}}{\mathrm{R}}\right)\)

Step-by-step Solution

Detailed explanation

Applying energy conservation

Applying energy conservation

\({\mathrm{K}_{1}+\mathrm{U}_{\mathrm{i}}=\mathrm{K}_{\mathrm{f}}+\mathrm{U}_{\mathrm{f}}}\)

\({\frac{1}{2} \mathrm{mu}^{2}+\left(-\frac{\mathrm{GMm}}{\mathrm{R}}\right)=\frac{1}{2} \mathrm{mv}^{2}-\frac{\mathrm{GMm}}{2 \mathrm{R}}}\)

\(\mathrm{v}=\sqrt{\mathrm{u}^{2}-\frac{\mathrm{GM}}{\mathrm{R}}}\)

By momentum conservation, we have

\(\frac{\mathrm{m}}{10} \mathrm{v}_{\mathrm{T}}=\frac{9 \mathrm{m}}{10} \sqrt{\frac{\mathrm{GM}}{2 \mathrm{R}}}\)

and \(\frac{\mathrm{m}}{10} \mathrm{v}_{\mathrm{r}}=\mathrm{mv}\)

\(\Rightarrow \frac{\mathrm{m}}{10} \mathrm{v}_{\mathrm{r}}=\mathrm{m} \sqrt{\mathrm{u}^{2}-\frac{\mathrm{GM}}{\mathrm{R}}}\)

Kinetic energy of rocket

\({=\frac{1}{2} \mathrm{m}\left(\mathrm{v}_{\mathrm{T}}^{2}+\mathrm{v}_{\mathrm{r}}^{2}\right)}\)

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

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

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

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