Question:
Consider two satellites $S_{1}$ and $S_{2}$ with periods of revolution $1 \mathrm{hr}$. and $8 \mathrm{hr}$.
respectively revolving around a planet in circular orbits. The ratio of
angular velocity of satellite $S_{1}$ to the angular velocity of satellite $S_{2}$ is -
Correct Option: 1
Solution:
(1)
We know that $\omega=\frac{2 \pi}{T}$
given : Ratio of time period
$\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{1}{8}$
$\Rightarrow \omega \propto \frac{1}{T}$
$\Rightarrow \frac{\omega_{1}}{\omega_{2}}=\frac{T_{2}}{T_{1}}$
$\Rightarrow \frac{\omega_{1}}{\omega_{2}}=\frac{8}{1}$
$\Rightarrow \omega_{1}: \omega_{2}=8: 1$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.