Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
A geostationary satellite is revolving around the earth. If radius of the earth is ' $R$ ' and the angular speed about its own axis is ' $\omega$ ' then the radius of the orbit of the geostationary satellite is ( $\mathrm{g}=$ acceleration due to gravity)
PhysicsGravitationMHT CETMHT CET 2022 (07 Aug Shift 2)
Options:
  • A $\left[\frac{\mathrm{R}^2 \omega^2}{\mathrm{~g}}\right]^{1 / 3}$
  • B $\left[\frac{\mathrm{Rg}}{\omega^2}\right]^{1 / 3}$
  • C $\left[\frac{\mathrm{R}^2 \mathrm{~g}}{\omega}\right]^{1 / 3}$
  • D $\left[\frac{\mathrm{R}^2 g}{\omega^2}\right]^{1 / 3}$
Solution:
1628 Upvotes Verified Answer
The correct answer is: $\left[\frac{\mathrm{R}^2 g}{\omega^2}\right]^{1 / 3}$
Considering force balance:
Centrifugal force $=$ Gravitation force,
$\begin{aligned} & \therefore m \omega^2 r=\frac{G M m}{r^2} \\ & \Rightarrow r=\left\{\frac{G M}{\omega^2}\right\}^{1 / 3}\end{aligned}$
We know, $g=\frac{G M}{R^2}$
$\Rightarrow \mathrm{r}=\left(\frac{\mathrm{R}^2 \mathrm{~g}}{\omega^2}\right)^{1 / 3}$

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.