# Binary stars ,time period of orbit?

• humanist rho
In summary: When motion is stopped, L =0The center of force would be the same as the center of mass. However, it makes a big difference in terms of the solution of the problem. In center of mass frame of reference, the dynamics will be simplified because the center of mass will always be in the middle of the two objects.
humanist rho

## Homework Statement

A binary system consists of two stars of equal mass m orbiting each other in a circular orbit under the influence of gravitational forces. The period of the orbit is τ . At t = 0, the motion is stopped and the stars are allowed to fall towards each other. After what time t, expressed in terms of τ , do they collide?

## The Attempt at a Solution

I don't know where to start.

Last edited:
hi humanist rho!

start by finding the initial distance between them

humanist rho said:

## Homework Statement

A binary system consists of two stars of equal mass m orbiting each other in a circular orbit under the influence of gravitational forces. The period of the orbit is τ . At t = 0, the motion is stopped and the stars are allowed to fall towards each other. After what time t, expressed in terms of τ , do they collide?

## The Attempt at a Solution

I don't know where to start.
firstly find τ. then
find distance between them them. make a function of force(then find acceleration) in terms of distance between them and then you need to do integration(2 times) ...
I have not tried it but i think this will work.

humanist rho said:

## Homework Statement

A binary system consists of two stars of equal mass m orbiting each other in a circular orbit under the influence of gravitational forces. The period of the orbit is τ . At t = 0, the motion is stopped and the stars are allowed to fall towards each other. After what time t, expressed in terms of τ , do they collide?

## The Attempt at a Solution

I don't know where to start.

Newton's form of Kepler's Law regarding the relationship of period to semimajor axis of an orbit can help.

Consider the period of an elliptical orbit whose minor axis is gradually shrunk towards zero and whose aphelion remains at distance r. The "orbit" eventually resembles a straight line drop and bounce, becoming exact as the minor axis → 0.

I didn't get it yet. :(

$\tau ^{2}=\frac{16\pi ^{2}R^{3}}{Gm}$, by Kepler's law.

then?

humanist rho said:
I didn't get it yet. :(

$\tau ^{2}=\frac{16\pi ^{2}R^{3}}{Gm}$, by Kepler's law.

then?

From that you have R in terms of ##\tau##. So how far apart are the two stars when they begin their fall towards each other?

The next task is to find the time for two objects of mass M, starting from rest at that separation, to collide due to gravitational attraction. There are various ways to approach this part of the problem, including writing and solving the corresponding differential equation of motion, or making clever use of Newton's form of Kepler's Law as I mentioned in my post above.

Sorry, i can't understand even now.

They'll be at a distance 2R apart.(i think)

and the equation of motion is,

$\ddot{r}=\frac{L^{2}}{mr^{3}}-\frac{Gm}{2r^{2}}$

how can i solve this?

I tried in terms of energy also.

$E=\frac{1}{2}m\dot{r}^{2}+\frac{L^{2}}{2mr^{2}}-\frac{Gm^{2}}{r}$

$\dot{r}=\sqrt{\frac{2}{m}\left( E-\frac{L^{2}}{2mr^{2}}+\frac{Gm^{2}}{r}% \right) }$

$\frac{dr}{\sqrt{\frac{2}{m}\left( E-\frac{L^{2}}{2mr^{2}}+\frac{Gm^{2}}{r}% \right) }}=dt$

$t=\int_{2R}^{0}\frac{dr}{\sqrt{\frac{2}{m}\left( E-\frac{L^{2}}{2mr^{2}}+% \frac{Gm^{2}}{r}\right) }}$

When motion is stopped, L =0

then,

$t=\int_{2R}^{0}\frac{dr}{\sqrt{\frac{2}{m}\left( E+\frac{Gm^{2}}{r}\right)}}$

Now i don't know to proceed.

And about the third method, by using Newtons laws cleverly, i have no idea. But it sounds interesting.

You shouldn't need to include terms for angular momentum in your differential equations if the bodies are falling along a straight line towards each other -- angular momentum will always be zero. Even so, solving the equation can be a bit tricky.

The "clever" method relies on the fact that the period of an elliptical orbit depends only on the length of its major axis and the gravitational parameter ##\mu## of the system. It's Newton's version of Kepler's period law:
$$P = \frac{2\pi}{\sqrt{\mu}}a^{3/2}$$
If you imagine the two bodies following elliptical orbits with aphelions at their starting locations, then you can make the minor axes as small as you like without changing the period. As the minor axes approach zero length, the ellipses degenerate into straight lines -- the bodies fall straight towards each other. The "orbits" would then be falling straight in and bouncing straight back out from the collision point. It is in all ways equivalent to the case in the given problem. Just determine the appropriate length for the semimajor axis a, the gravitational parameter ##\mu##, and the portion of the full period T that corresponds to the infall.

When we take this problem in center of mass frame of reference, where will be center of force, which acts on the reduced mass, located? What difference does it make to the solution of the problem (with respect to the integration limits)?

## 1. What are binary stars?

Binary stars are a system of two stars that are gravitationally bound to each other and orbit around a common center of mass. These stars are often close enough to appear as a single point of light in the night sky.

## 2. How do binary stars form?

Binary stars can form in two ways: either through the fragmentation of a single cloud of gas and dust, or through the capture of one star by another star's gravitational pull. The exact mechanism of formation depends on the specific conditions of the stellar environment.

## 3. What is the time period of orbit for binary stars?

The time period of orbit, also known as the orbital period, is the time it takes for the two stars in a binary system to complete one full orbit around each other. The length of this period varies depending on the distance between the stars and their masses, but it can range from a few days to thousands of years.

## 4. How is the time period of orbit calculated?

The time period of orbit can be calculated using Kepler's third law of planetary motion, which states that the square of the orbital period is proportional to the cube of the semi-major axis of the orbit. This law can be applied to binary stars, with the semi-major axis being the average distance between the two stars.

## 5. Why is studying binary stars important?

Studying binary stars can provide valuable information about the formation and evolution of stars, as well as the properties and behavior of different types of stars. Binary star systems can also serve as laboratories for testing theories and models of stellar dynamics and interactions. In addition, binary stars are often used as standard candles for measuring astronomical distances, making them important tools in the study of the universe.

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