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Gravitation

JEE Main level, single correct, medium

146 of 223

Question 146 of 223

The energy required to take a satellite to a height $h$ above the Earth surface (radius of Earth $=6.4\times {10}^{3}km$ ) is ${E}_{1}$ , and the 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

Show answer and solution

Answer: (C) $3.2\times {10}^{3}km$

{U}_{surface}+{E}_{1}={U}_{h}

KE

of satellite is zero at earth surface and at height

h

-\dfrac{G{M}_{e}m}{{R}_{e}}+{E}_{1}=-\dfrac{G{M}_{e}m}{\left({R}_{e}+h\right)}

{E}_{1}=G{M}_{e}m\left(\dfrac{1}{{R}_{e}}-\dfrac{1}{{R}_{e}+h}\right)

{E}_{1}=\dfrac{G{M}_{e}m}{\left({R}_{e}+h\right)}\times \dfrac{h}{{R}_{e}}

Gravitational attraction

{F}_{G}=m{a}_{C}=\dfrac{m{v}^{2}}{\left({R}_{e}+h\right)}

\Rightarrow \dfrac{m{v}^{2}}{\left({R}_{e}+h\right)}=\dfrac{G{M}_{e}m}{{\left({R}_{e}+h\right)}^{2}}

m{v}^{2}=\dfrac{G{M}_{e}m}{\left({R}_{e}+h\right)}

{E}_{2}=\dfrac{m{v}^{2}}{2}=\dfrac{G{M}_{e}m}{2\left({R}_{e}+h\right)}

{E}_{1}={E}_{2}

\dfrac{h}{{R}_{e}}=\dfrac{1}{2}\Rightarrow h=\dfrac{{R}_{e}}{2}=3200km

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