Gravitation

JEE Main level, single correct, medium

174 of 223

Question 174 of 223

The time period of a satellite, revolving above earth's surface at a height equal to RR will be (Given g=π2ms−2,R=g={\pi }^{2}m{s}^{-2},R= radius of earth)

Show answer and solution

Answer: (C) 32R\sqrt{32R}

The velocity of the satellite revolving round the Earth at a height rr from the centre of the Earth is given by
v=(GMr)...(1)v=\sqrt{\left(\dfrac{GM}{r}\right)}...\left(1\right)
The formula to calculate the time period in the circular orbit is given by
T=2πrv...(2)T=\dfrac{2\pi r}{v}...\left(2\right)
In this case: r=R+R=2Rr=R+R=2R
Substitute the expression for the velocity from equation (1) into equation (2) and simplify to obtain the required time period.
T=2πrGMr=2π×2RgR2=4πR2πR=42R=32R\begin{matrix}T & = & \dfrac{2\pi r}{\sqrt{\dfrac{GM}{r}}} \\ & = & \dfrac{2\pi \times 2R}{\sqrt{\dfrac{gR}{2}}} \\ & = & \dfrac{4\pi R\sqrt{2}}{\pi \sqrt{R}} \\ & = & 4\sqrt{2R} \\ & = & \sqrt{32R}\end{matrix}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Gravitation and orbits simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Gravitation (Motion of satellite), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.