Physics · Gravitation

JEE Main 2025 — 24 January, Morning Shift — Question 60

A satellite is launched into a circular orbit of radius ' R ' around the earth. A second statellite 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 :-

Option A:
$3 \%$
Option B:
$4.5 \%$
Correct
Option C:
$9 \%$
Option D:
$2.5 \%$

Answer: B

Step-by-step solution

$\quad \mathrm{T}^{2}=\mathrm{KR}^{3}$

$\frac{2 \Delta \mathrm{~T}}{\mathrm{~T}}=\frac{3 \Delta \mathrm{R}}{\mathrm{R}}$

$\frac{2 \Delta \mathrm{~T}}{\mathrm{~T}}=\frac{3 \times 0.03 \mathrm{R}}{\mathrm{R}}$

$\frac{\Delta \mathrm{T}}{\mathrm{T}}=\frac{3 \times 0.03}{2} \times 100=4.5 \%$

Answer key and solution verified before publishing.

Practise Gravitation

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2025
Paper: 24 January, Morning Shift
Subject: Physics
Chapter: Gravitation
Topic: Motion of Satellites and Escape Speed