Physics · Gravitation

JEE Main 2024 — 29 January, Shift 2 — Question 43

A planet takes 200 days to complete one revolution around the Sun. If the distance of the planet from Sun is reduced to one fourth of the original distance, how many days will it take to complete one revolution?

Option A:
25
Correct
Option B:
50
Option C:
100
Option D:
20

Answer: A

Step-by-step solution

$\mathrm{T}^{2} \propto \mathrm{r}^{3}$

$\frac{\mathrm{T}_{1}^{2}}{\mathrm{r}_{1}^{3}}=\frac{\mathrm{T}_{2}^{2}}{\mathrm{r}_{2}^{3}}$

$\frac{(200)^{2}}{\mathrm{r}^{3}}=\frac{\mathrm{T}_{2}^{2}}{\left(\frac{\mathrm{r}}{4}\right)^{3}}$

$\implies \frac{200 \times 200}{4 \times 4 \times 4}=\mathrm{T}_{2}^{2}$

$\mathrm{T}_{2}=\frac{200}{4 \times 2}$

$\mathrm{T}_{2}=25$ days

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 29 January, Shift 2
Subject: Physics
Chapter: Gravitation
Topic: Planetary Motion & Binary Star System (Kepler's Law)