Question

The energy required to take a satellite to a height ${}^{\prime }{h}^{\prime }$ 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. $1.28\times 10^{4}km$
• B. $6.4\times 10^{3}km$
• C. $3.2\times 10^{3}km$
• D. $1.6\times 10^{3}km$

Answer 1

Answer: (C)

$U_{surface}+E_1=U_h$
$KE$ of satelite is zero at earth surface & at height h $-\frac{G{M}_{e}m}{R_e}+E_1=-\frac{G{M}_{e}m}{\left(R_e+h\right)}$
$E_1=G{M}_{e}m\left(\frac{1}{R_e}-\frac{1}{R_e+h}\right)$
$E_1=\frac{G{M}_{e}m}{\left(R_e+h\right)}\times \frac{h}{R_e}$
Gravitational attraction $F_G=m{a}_C=\frac{m{v}^2}{\left(R_e+h\right)}$
$E_2\Rightarrow \frac{m{v}^2}{\left(R_e+h\right)}=\frac{G{M}_{e}m}{{\left(R_e+h\right)}^2}$
$m{v}^2=\frac{G{M}_{e}m}{\left(R_e+h\right)}$
$E_2=\frac{m{v}^2}{2}=\frac{G{M}_{e}m}{2\left(R_e+h\right)}$
$E_1=E_2$
$\frac{h}{R_e}=\frac{1}{2}\Rightarrow h=\frac{R_e}{2}=3200km$