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7.Gravitation
hard
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
A
$1.6\times 10^3\,km$
B
$3.2\times 10^3\,km$
C
$6.4\times 10^3\,km$
D
$1.28\times 10^4\,km$
(JEE MAIN-2019)
Solution
${E_1} =-\frac{{GMm}}{{R + h}} – \left( { – \frac{{GMm}}{R}} \right)$
${E_2} = \frac{1}{2}m{\left( {\sqrt {\frac{{GM}}{{R + h}}} } \right)^2} = \frac{{GMm}}{{2\left( {R + h} \right)}}$
${E_1} = {E_2}\,\,;\,\,h = \frac{R}{2}$
Standard 11
Physics
