A body is projected vertically upwards from the surface of earth with a velocity equal to one third of escape velocity. The maximum height attained by the body will be :
(Take radius of earth $=6400 \mathrm{~km}$ and $\mathrm{g}=10 \mathrm{~ms}^{-2}$ )
Solution
<p>Applying conservation of energy</p>
<p>$$ - {{G{M_e}m} \over {{R_e}}} + {1 \over 2}m{\left( {{1 \over 3}\sqrt {{{2G{m_e}} \over {{R_e}}}} } \right)^2} = - {{G{M_e}m} \over {{R_e} + h}}$$</p>
<p>$$ - {{G{M_e}m} \over {{R_e}}} + {{G{M_e}m} \over {9{R_e}}} = - {{G{M_e}m} \over {{R_e} + h}}$$</p>
<p>${8 \over {9{R_e}}} = {1 \over {{R_e} + h}}$</p>
<p>$\Rightarrow h = {{{R_e}} \over 8} = {{6400} \over 8} = 800$ km</p>
About this question
Subject: Physics · Chapter: Gravitation · Topic: Kepler's Laws
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