JEE Mains · Physics · STD 11 - Gravitation
- JEE Mains 2020
- Hard
- Numerical
A satellite is moving in a low nearly circular orbit around the earth. Its radius is roughly equal to that of the earth's radius \(R _{ e }\). By firing rockets attached to it, its speed is instantaneously increased in the direction of its motion so that is become \(\sqrt{\frac{3}{2}}\) times larger. Due to this the farthest distance from the centre of the earth that the satellite reaches is \(R\), value of \(R\) is\(....R_e\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(3\)
Step-by-step Solution
Detailed explanation
\(V _{0}=\sqrt{\frac{ GM }{ R _{ e }}}\)
\(\frac{- GMm }{ R _{ e }}+\frac{1}{2} mv ^{2}=\frac{- GMm }{ R _{ max }}+\frac{1}{2} mv ^{\prime 2}\)\(...(i)\)
\(VR _{ e }= V ^{\prime} R _{\max }\)\(...(ii)\)
Solving \((i)\) \(\&\) \((ii)\)
\(R _{ max }=3 R _{e}\)
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