First slide

Gravitational force and laws

Question

An isolated triple star system consists of two identical stars, each of mass m and a fixed star of mass M. They revolve around the central star in the same circular orbit of radius r. The two orbiting stars are always at opposite ends of a diameter of the orbit. The time period of revolution of each star around the fixed star is equal to:

Moderate

A

$\frac{4 {πr}^{3 / 2}}{\sqrt{G (4 M + m)}}$
B

$\frac{2 {πr}^{3 / 2}}{\sqrt{GM}}$
C

$\frac{2 {πr}^{3 / 2}}{\sqrt{G (M + m)}}$
D

$\frac{4 {πr}^{3 / 2}}{\sqrt{G (M + m)}}$

Solution

Net gravitational force on any orbiting star provides necessary centripetal force.
$\begin{array}{l} So \frac{GMm}{r^{2}} + \frac{{Gm}^{2}}{4 r^{2}} = \frac{{mV}^{2}}{r} \\ \Rightarrow V = \sqrt{\frac{G (4 M + m)}{4 r}} \\ T = \frac{2 πr}{V} = \frac{4 {πr}^{3 / 2}}{\sqrt{G (4 M + m)}} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App