Question

An artificial satellite is moving in a circular orbit around the earth with a speed equal to $\frac{3}{8}$ times the magnitude of the escape velocity from the earth. If the satellite is stopped suddenly in its orbit and allowed to fall freely onto the earth, then the speed with which it hits the surface of the earth is $\sqrt{\frac{23}{n}}\text{km/s}$. Find $n$. (take g = 10 m/s and $R_e$ = 6400 Km)

Answer 1

Escape velocity $v_e=\sqrt{2g{R}_e}$
Given, speed of the satelliite$v=\frac{3}{8}\sqrt{2g{R}_e}=\sqrt{\frac{9g{R}_e}{32}}$ --- (i)
Orbital velocity at height ${}^{\prime }{h}^{\prime }=\sqrt{\frac{GM}{R_e+h}}$
$=\sqrt{\frac{GM.R_e^2}{R_e^2(R_e+h)}}=\sqrt{\frac{R_e^{2}g}{R_e+h}}$ ---(ii)

Equating (i) and (ii),
$\sqrt{\frac{9g{R}_e}{32}}=\sqrt{\frac{R_e^{2}g}{R_e+h}}$
$\Rightarrow 9R_e+9h=32R_e$
$\Rightarrow h=\frac{23}{9}{R}_e$

Now, total energy at height ${}^{\prime }{h}^{\prime }$ = total energy at the surface of the earth.
$0-\frac{G{M}_{e}m}{R_e}=\frac{1}{2}m{v}^2-\frac{G{M}_{e}m}{R_e+h}$
$\frac{1}{2}m{v}^2=\frac{G{M}_{e}m}{R_e}-\frac{G{M}_{e}m}{R_e+\frac{23}{9}{R}_e}$
$\frac{1}{2}m{v}^2=\frac{23}{32}\frac{G{M}_{e}m}{R_e}$
$\Rightarrow v=\sqrt{\frac{23}{16}\frac{G{M}_e}{R_e}}$
$\Rightarrow v=\sqrt{\frac{23}{16}g{R}_e}=\sqrt{\frac{23}{2}}\text{km/s}$
$(\because \sqrt{g{R}_e}=8\text{km/sec})$

$\therefore n=2$