Editing Visual binary (section)

===Newton's generalisation===
Consider a binary star system. This consists of two objects, of mass <math>m_1</math> and <math>m_2</math>, orbiting around their centre of mass. <math>m_1</math> has position vector <math>r_1</math> and orbital velocity <math>v_1</math>, and <math>m_2</math> has position vector <math>r_2</math> and orbital velocity <math>v_2</math> relative to the centre of mass.  The separation between the two stars is denoted <math>r</math>, and is assumed to be constant. Since the gravitational force acts along a line joining the centers of both stars, we can assume the stars have an equivalent time period around their center of mass, and therefore a constant separation between each other.<ref>{{cite web
 | url = http://www.egglescliffe.org.uk/physics/gravitation/binary/binary.html
 | title = The Physics of Binary Stars
 | access-date = 2013-10-15
 | archive-date = 2013-10-15
 | archive-url = https://web.archive.org/web/20131015144315/http://www.egglescliffe.org.uk/physics/gravitation/binary/binary.html
 | url-status = dead
 }}</ref>

To arrive at Newton's version of Kepler's 3rd law we can start by considering [[Newton's 2nd law]] which states: "The net force acting on an object is proportional to the objects mass and resultant acceleration." 
:<math> F_{net} = \sum \, F_{i} = ma </math> 
where <math>F_{net}</math> is the net force acting on the object of mass <math>m</math>, and <math>a</math> is the acceleration of the object.<ref name="Astro">{{cite book
 | author = Bradley W. Carroll
 | author2 = Dale A. Ostlie
 | name-list-style = amp  
 | title = An Introduction to Modern Astrophysics 
 | date = 2013 
 | isbn = 978-1292022932 
 | publisher = Pearson
}}</ref>

Applying the definition of [[centripetal acceleration]] to Newton's second law gives a force of 
:<math> F = \frac{mv^2}{r} </math> <ref name="phy">{{cite book
 | author = Hugh D. Young
 | title = University Physics 
 | date = 2010
 | isbn = 978-0321501301 
 | publisher = [[Bertrams]] 
}}</ref> 
Then using the fact that the orbital velocity is given as 
:<math> v = \frac{2\pi r}{T} </math> <ref name=phy />  
we can state the force on each star as 
:<math> F_{1} = \frac{4\pi^2 m_{1}r_{1}}{T^2} </math> and <math> F_{2} = \frac{4\pi^2 m_{2}r_{2}}{T^2} </math>

If we apply [[Newton's 3rd law]]- "For every action there is an equal and opposite reaction"
:<math> F_{12} = -F_{21} </math> <ref name=Astro /> 
We can set the force on each star equal to each other.
:<math> \frac{4\pi^2 m_{1}r_{1}}{T^2} = \frac{4\pi^2 m_{2}r_{2}}{T^2} </math> 
This reduces to 
:<math> r_{1}m_{1} = r_{2}m_{2} </math> 
If we assume that the masses are not equal, then this equation tells us that the smaller mass remains farther from the centre of mass than does the larger mass.

The separation <math>r</math> of the two objects is 
:<math> r = r_{1}+r_{2} </math> 
Since <math>r_1</math> and <math>r_2</math> would form a line starting from opposite directions and joining at the centre of mass.

Now we can substitute this expression into one of the equations describing the force on the stars and rearrange for <math>r_1</math> to find an expression relating the position of one star to the masses of both and the separation between them. Equally, this could have been solved for <math>r_2</math>. We find that
:<math> r_{1} = \frac{m_{2}a}{(m_{1}+m_{2})} </math>

Substituting this equation into the equation for the force on one of the stars, setting it equal to Newton's Universal Law of Gravitation (namely, <math>F=Gm_{1}m_{2}/a^2</math>,<ref name=Astro /> and solving for the period squared yields the required result.
:<math> T^2 = \frac{4\pi^2 a^3}{G(m_{1}+m_{2})} </math><ref name=Astro /> 
This is Newton's version of Kepler's 3rd Law. Unless <math>G</math> is in non-standard units, this will not work if mass is measured in solar masses, the orbital period is measured in years, and the orbital semi-major axis is measured in astronomical units (e.g., use the Earth's orbital parameters).  It will work if [[SI units]], for instance, are used throughout.