Question

The minimum energy required to launch a $m$ kg satellite from the earth's surface in a circular orbit at an altitude $2R$, where $R$ is the radius of earth is

• A. $\frac{5}{3}mgR$
• B. $\frac{4}{3}mgR$
• C. $\frac{5}{6}mgR$
• D. $\frac{5}{4}mgR$

Answer 1

Answer: (C)

$\frac{5}{6}mgR$

$\text{Energy at earth's surface}=\text{potential energy+kinetic energy}=-\frac{GMm}{R}+K.E$
$\text{Energy at earth's surface}=-\frac{GMm}{R}+0=-\frac{GMm}{R}$
$\text{Energy at altitude of 2R}=E=P.E+K.E.$
$E=-\frac{GMm}{3R}+\frac{1}{2}m(\frac{GM}{3R}{)}^{\frac{1}{2}\times 2}$
$[As,orbitalvelocity=(\frac{GM}{R}{)}^{\frac{1}{2}}]$
$\text{By taking the difference of energies, we get}$
$E=-\frac{GMm}{6R}-(-\frac{GMm}{R})$
$E=\frac{5GMm}{6R}=\frac{5}{6}mgR$ $As,GM=g{R}^2$