Question

A communication satellite of $500kg$ revolves around the earth in a circular orbit of radius $4.0\times {10}^{7}m$ in the equatorial plane of the earth from west to east. The magnitude of angular momentum of the satellite is

• A. $\sim 0.13\times {10}^{14}kg{m}^2{s}^{-1}$
• B. $\sim 1.3\times {10}^{14}kg{m}^2{s}^{-1}$
• C. $\sim 0.58\times {10}^{14}kg{m}^2{s}^{-1}$
• D. $\sim 2.58\times {10}^{14}kg{m}^2{s}^{-1}$

Answer 1

The correct option is C $\sim 0.58\times {10}^{14}kg{m}^2{s}^{-1}$

As the satellite is moving in equatorial plane with orbital radius $4\times {10}^{7}m$.
$\therefore$ Satellite is geostationary satellite.
Hence, the time taken by satellite to complete its one revolution, $T=24h=86400s$
Velocity of satellite, $v=\frac{2\pi r}{T}$
Angular momentum, $L=mvr$
$L=m\left(\frac{2\pi r}{T}\right)r=\frac{2\pi m}{T}{r}^2$
$\therefore L=\frac{2\times 3.14\times 500}{86400}\times (4\times {10}^7{)}^2$
$=0.58\times {10}^{14}kg{m}^2{s}^{-1}$.