Binary stars ,time period of orbit?

AI Thread Summary
In a binary star system, two equal mass stars orbit each other with a period τ. When their motion is halted, the stars begin to fall towards each other due to gravitational attraction. The initial distance between the stars is 2R, where R is derived from Kepler's law relating the period to the semimajor axis. To determine the time until collision, one can use Newton's laws and the equations of motion, although the integration can be complex. The discussion emphasizes the importance of understanding gravitational interactions and the relationship between orbital mechanics and time until collision.
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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.
 
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hi humanist rho! :smile:

start by finding the initial distance between them :wink:
 
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}%<br /> \right) }

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

t=\int_{2R}^{0}\frac{dr}{\sqrt{\frac{2}{m}\left( E-\frac{L^{2}}{2mr^{2}}+%<br /> \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)?
 

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