A body is projected vertically upwards from the surface of earth with a velocity sufficient enough to carry it to infinity. The time taken by it to reach height h is ___________ s.
Solution
${1 \over 2}m{v^2} - {{GMm} \over r} = 0 \Rightarrow v = \sqrt {{{2GM} \over r}}$<br><br>${{dr} \over {dt}} = \sqrt {{{2GM} \over r}}$<br><br>$$ \Rightarrow \int\limits_{{R_e}}^{({R_e} + h)} {\sqrt r dr = \int\limits_0^t {\sqrt {2GM} dt} } $$<br><br>$$ \Rightarrow {2 \over 3}\left[ {{{({R_e} + h)}^{3/2}} - R_e^{3/2}} \right] = (t)\sqrt {2GM} $$<br><br>$$ \Rightarrow t = {1 \over 3}\sqrt {{{2{R_e}} \over g}} \left[ {{{\left( {1 + {h \over {{R_e}}}} \right)}^{{3 \over 2}}} - 1} \right]$$
About this question
Subject: Physics · Chapter: Gravitation · Topic: Kepler's Laws
This question is part of PrepWiser's free JEE Main question bank. 129 more solved questions on Gravitation are available — start with the harder ones if your accuracy is >70%.