Question

The energy required to take a satellite to a height $h$ above Earth surface (radius of Earth $=6.4\times {10}^3\text{km}$) 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. $6.4\times {10}^3\text{km}$
• B. $3.2\times {10}^3\text{km}$
• C. $1.6\times {10}^3\text{km}$
• D. $28\times {10}^4\text{km}$

Answer 1

The correct option is B $3.2\times {10}^3\text{km}$

$K.E.$ of satellite is zero at height $h$

From energy conservation

$U_{surface}+E_1=U_h$
$-\frac{G{M}_{e}m}{R_e}+E_1=-\frac{G{M}_{e}m}{R_e+h}$
$\Rightarrow E_1=G{M}_{e}m(\frac{1}{R_e}-\frac{1}{R_e+h})\Rightarrow E_1=\frac{G{M}_{e}m}{(R_e+h)}\times \frac{h}{R_e}$

Using Gravitational attraction,

$F_G=m{a}_c=\frac{m{v}^2}{(R_e+h)}=\frac{G{M}_{e}m}{(R_e+h{)}^2}$
$m{v}^2=\frac{G{M}_{e}m}{R_e+h}$
$E_2=\frac{m{v}^2}{2}=\frac{G{M}_{e}m}{2(R_e+h)}$

Given that
$E_1=E_2$
$\frac{G{M}_{e}m}{(R_e+h)}\times \frac{h}{R_e}=\frac{G{M}_{e}m}{2(R_e+h)}$
$\frac{h}{R_e}=\frac{1}{2}\Rightarrow h=\frac{R_e}{2}=3200\text{km}$