Question

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

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

Answer 1

Answer: (A)

given $R=3200km,R^1=6400kg,m=100kg$
Gravitational potential energy $=\frac{-GMm}{r}$
and orbital velocity, $v_0=\sqrt{GM/R+h}$
$E_f=\frac{1}{2}m{v}_0^2-\frac{GMm}{3R}=\frac{1}{2}m\frac{GM}{3R}-\frac{GMm}{3R}$
$=\frac{GMm}{3R}\left(\frac{1}{2}-1\right)=\frac{-GMm}{6R}$
$E_i=\frac{-GMm}{R}+K\text{ also }{E}_i=E_f$
$\Rightarrow K=\frac{5GMm}{6R}$
$\therefore k=\frac{5GM\times 100}{6\times 3200}\times 10^{-3}$
$=\frac{GM}{6\times 32\times 200}=\frac{GM}{38400}$