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

Answer: 3200

$K.E$. of satellite is zero at earth surface and at height $h$ from energy conservation
$U_{\text{surface}}+E_1=U_h$
$-\frac{G{M}_{e}m}{R_e}+E_1=-\frac{G{M}_{e}m}{\left(Re+h\right)}$
$\Rightarrow E_1=G{M}_{e}m\left(\frac{1}{R_e}-\frac{1}{R_e+h}\right)$
$\Rightarrow E_1=\frac{G{M}_{e}m}{\left(R_e+h\right)}\times \frac{h}{R_e}$
Gravitational attraction
$F_G=m{a}_c=\frac{m{v}^2}{\left(R_e+h\right)}=\frac{G{M}_{e}m}{{\left(R_e+h\right)}^2}$
$m{v}^2=\frac{G{M}_{e}m}{\left(R_e+h\right)}$
$E_2=\frac{m{v}^2}{2}=\frac{G{M}_{e}m}{2\left(R_e+h\right)}$
$E_1=E_2$
Cleary,$\frac{h}{R_e}=\frac{1}{2}$
$\Rightarrow h=\frac{R_e}{2}=3200$km