JEE Mains · Physics · STD 11 - Gravitation
- JEE Mains 2019
- Hard
- MCQ
The energy required to take a satellite to a height \(‘h’\) above Earth surface (radius of Earth \(= 6.4 \times 10^3\,km\) ) is \(E_1\) and kinetic energy required for the satellite to be in a circular orbit at this height is \(E_2.\) The value of \(h\) for which \(E_1\) and \(E_2\) are equal is
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Answer & Solution
Correct Answer
(B) \(3.2\times 10^3\,km\)
Step-by-step Solution
Detailed explanation
\({E_1} =-\frac{{GMm}}{{R + h}} - \left( { - \frac{{GMm}}{R}} \right)\)
\({E_2} = \frac{1}{2}m{\left( {\sqrt {\frac{{GM}}{{R + h}}} } \right)^2} = \frac{{GMm}}{{2\left( {R + h} \right)}}\)
\({E_1} = {E_2}\,\,;\,\,h = \frac{R}{2}\)
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