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A communication satellite of $500 \mathrm{~kg}$ revolves around the earth in a circular orbit of radius $4.0 \times 10^7 \mathrm{~m}$ in the equatorial plane of the earth from west to east. The magnitude of angular momentum of the satellite is
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$\sim 0.58\times 10^{14} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}$
As the satellite is moving in equatorial plane with orbital radius $4 \times 10^7 \mathrm{~m}$.
$\therefore \quad$ Satellite is geostationary satellite.
Hence, the time taken by satellite to complete its one revolution, $T=24 \mathrm{~h}=86400 \mathrm{~s}$
Velocity of satellite, $v=\frac{2 \pi r}{T}$
Angular momentum, $L=m v r$
$\begin{aligned} & L=m\left(\frac{2 \pi r}{T}\right) r=\frac{2 \pi m}{T} r^2 \\ & \therefore \quad L=\frac{2 \times 3.14 \times 500}{86400} \times\left(4 \times 10^7\right)^2 \\ & =0.58 \times 10^{14} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1} \\ & \end{aligned}$
$\therefore \quad$ Satellite is geostationary satellite.
Hence, the time taken by satellite to complete its one revolution, $T=24 \mathrm{~h}=86400 \mathrm{~s}$
Velocity of satellite, $v=\frac{2 \pi r}{T}$
Angular momentum, $L=m v r$
$\begin{aligned} & L=m\left(\frac{2 \pi r}{T}\right) r=\frac{2 \pi m}{T} r^2 \\ & \therefore \quad L=\frac{2 \times 3.14 \times 500}{86400} \times\left(4 \times 10^7\right)^2 \\ & =0.58 \times 10^{14} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1} \\ & \end{aligned}$
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