Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
2313 Upvotes
Verified Answer
The correct answer is:
$0.58 \times 10^{14} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}$
Given that, mass of satellite, $m=500 \mathrm{~kg}$
Radius of orbit, $r=4 \times 10^7 \mathrm{~m}$
Now, the satellite is revolving in equatorial plane so, its time period will be equal to time period of rotation of earth i.e., $T=24 \mathrm{~h}$.
$\therefore$ Angular velocity of satellite, $\omega=\frac{2 \pi}{T}$
Now, angular momentum, $L=m \omega r^2=m\left(\frac{2 \pi}{T}\right) r^2$
$$
\begin{aligned}
& =500 \times\left(\frac{2 \pi}{24 \times 3600}\right) \times\left(4 \times 10^7\right)^2 \\
& =0.58 \times 10^{14} \mathrm{~kg} \cdot \mathrm{m}^2 \mathrm{~s}^{-1}
\end{aligned}
$$
Radius of orbit, $r=4 \times 10^7 \mathrm{~m}$
Now, the satellite is revolving in equatorial plane so, its time period will be equal to time period of rotation of earth i.e., $T=24 \mathrm{~h}$.
$\therefore$ Angular velocity of satellite, $\omega=\frac{2 \pi}{T}$
Now, angular momentum, $L=m \omega r^2=m\left(\frac{2 \pi}{T}\right) r^2$
$$
\begin{aligned}
& =500 \times\left(\frac{2 \pi}{24 \times 3600}\right) \times\left(4 \times 10^7\right)^2 \\
& =0.58 \times 10^{14} \mathrm{~kg} \cdot \mathrm{m}^2 \mathrm{~s}^{-1}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.