Question

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 $h$ (in km) for which $E_{1}$ and $E_{2}$ are equal, is:

Answer 1

$K.E$. of satellite is zero at earth surface and at height $h$ from energy conservation
$U_{\text{surface}} +E_{1} =U_{h}$
$ -\frac{GM_{e}m}{R_{e}} +E_{1} = - \frac{GM_{e}m}{\left(Re +h\right)}$
$\Rightarrow E_{1} =GM_{e} m\left(\frac{1}{R_{e}} -\frac{1}{R_{e} +h}\right)$
$\Rightarrow E_{1}=\frac{GM_{e} m}{\left(R_{e} +h\right)} \times\frac{h}{R_{e}}$
Gravitational attraction
$F_{G} = ma_{c} = \frac{mv^{2}}{\left(R_{e} +h\right)} = \frac{GM_{e} m}{\left(R_{e} +h\right)^{2}}$
$mv^{2} = \frac{GM_{e} m}{\left(R_{e} +h\right)}$
$E_{2} =\frac{mv^{2}}{2} =\frac{GM_{e} m}{2\left(R_{e} +h\right)}$
$E_{1} = E_{2}$
Cleary,$\frac{h}{R_{e}} = \frac{1}{2}$
$\Rightarrow h = \frac{R_{e}}{2} = 3200$km