Gravitation

JEE Main level, single correct, medium

167 of 223

Question 167 of 223

A satellite is launched into a circular orbit of radius ' R ' around the earth. A second satellite is launched into an orbit of radius 1.03 R . The time period of revolution of the second satellite is larger than the first one approximately by

Show answer and solution

Answer: (C) 4.5%4.5 \%



T1∝(R)3/2T_1 \propto(R)^{3 / 2}
and
T2∝(1.03R)3/2T_2 \propto(1.03 R)^{3 / 2}
⇒T2=(1.03R)3/2⋅T1≈1.045T1\Rightarrow \quad T_2=(1.03 R)^{3 / 2} \cdot T_1 \approx 1.045 T_1
So
T2T_2 will larger by 4.5%4.5 \% w.r.t. T1T_1.

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.