Q. 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:

Solution:

$U_{surface} + E_1 = U_h$
KE of satelite is zero at earth surface & at height h $ - \frac{GM_{e}m}{R_{e}} + E_{1} = - \frac{GM_{e}m}{\left(R_{e}+h\right)} $
$ E_{1} =GM_{e}m \left( \frac{1}{R_{e}} - \frac{1}{R_{e} +h}\right) $
$E_{1} = \frac{GM_{e}m}{\left(R_{e} +h\right)} \times\frac{h}{R_{e}} $
Gravitational attraction $ F_{G} = ma_{C} = \frac{mv^{2}}{\left(R_{e} +h\right)} $
$E_{2 } \Rightarrow \frac{mv^{2}}{\left(R_{e} +h\right)} = \frac{GM_{e}m}{\left(R_{e} + h\right)^{2}} $
$ mv^{2} = \frac{GM_{e}m}{\left(R_{e} +h\right) } $
$E_{2} = \frac{mv^{2}}{2} = \frac{GM_{e}m}{2 \left(R_{e} + h\right)}$
$ E_{1} =E_{2} $
$ \frac{h}{R_{e}} = \frac{1}{2} \Rightarrow h = \frac{R_{e}}{2} = 3200 $ km

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