Physics · Gravitation

JEE Advanced 2021 — Paper 2 — Question 8

The distance between two stars of masses $3 \mathrm{M}_{\mathrm{S}}$ and $6 \mathrm{M}_{\mathrm{S}}$ is $9 R$ . Here $R$ is the mean distance between the centers of the Earth and the Sun, and $\mathrm{M}_{\mathrm{S}}$ is the mass of the Sun. The two stars orbit around their common centre of mass in circular orbits with period nT , where T is the period of Earth's revolution around the Sun. The value of $n$ is $\qquad$

Answer: 9

Numerical answer — enter this value.

Step-by-step solution

Both will revolve about common center of mass

$\begin{matrix}x=\frac{3{{M}_{s}}}{6{{M}_{s}}+3{{M}_{s}}}\times 9R=3R\#\left( i \right) \\\frac{G6{{M}_{s}}\times 3{{M}_{s}}}{{{(9R)}^{2}}}=6{{M}_{s}}\left( {{\omega }^{2}}x \right)\#\left( ii \right) \\\end{matrix}$

Solving equation (i) and (ii) we get $\begin{array}{*{35}{r}}{} & {{\omega }^{2}}=\frac{\text{G}{{\text{M}}_{\text{s}}}}{81{{\text{R}}^{3}}} \\\end{array}$

For the motion of earth around Sun ${{\text{T}}^{2}}=\frac{4{{\pi }^{2}}{{\text{R}}^{3}}}{\text{G}{{\text{M}}_{\text{S}}}}$

From (iii) and (iv) ${{\text{T}}^{'2}}=81\text{ }\!\!~\!\!\text{ }{{\text{T}}^{2}}$ $\text{{T}'}=9\text{ }\!\!~\!\!\text{ T}$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2021
Paper: Paper 2
Subject: Physics
Chapter: Gravitation
Topic: Planetary Motion & Binary Star System (Kepler's Law)