Question

The least energy required to launch a satellite of mass $100\,kg$ from the surface of a planet of mass $M$ and radius $3200\, km$ in a circular orbit at an altitude of $6400 \,km$ is:

• A. $\frac{ GM }{38400}$
• B. $\frac{ GM }{8860}$
• C. $\frac{ GM }{98600}$
• D. $\frac{ GM }{9580}$

Answer 1

Answer: (A)

given $R=3200\, km , R^1=6400\, kg , m =100 \,kg$
Gravitational potential energy $=\frac{-G M m}{r}$
and orbital velocity, $v_0=\sqrt{G M / R+h}$
$E_f =\frac{1}{2} m v_0^2-\frac{G M m}{3 R}=\frac{1}{2} m \frac{G M}{3 R}-\frac{G M m}{3 R}$
$=\frac{G M m}{3 R}\left(\frac{1}{2}-1\right)=\frac{-G M m}{6 R} $
$E_i =\frac{-G M m}{R}+K \text { also } E_i=E_f$
$ \Rightarrow K=\frac{5 G M m}{6 R} $
$\therefore k =\frac{5 G M \times 100}{6 \times 3200} \times 10^{-3}$
$=\frac{G M}{6 \times 32 \times 200}=\frac{G M}{38400}$