# Two-body problem: reduced mass velocity to individual velocity

I asked a question on finding the initial velocity of two objects of equal mass in order to follow a keplerian orbit with a given eccentricity but I am still having some trouble.

I have seen the reduced mass approach where you consider a reduced mass orbiting around the centre of mass of the original configuration. And they use the vic-viva equation to obtain the velocity of the reduced mass....but how do I find the initial velocity of each of my original masses? (The two objects are placed on the x axis so the velocities will be in the y axis)

If you have found the necessary relative velocity $$\mathbf v=\mathbf{v}_1-\mathbf{v}_2$$, then the velocities of the two individual masses are

$$\mathbf{v}_1=\frac{m_2}{m_1+m_2}\mathbf v$$

and

$$\mathbf{v}_2=-\frac{m_1}{m_1+m_2}\mathbf v.$$

Note that these satisfy

$$m_1\mathbf{v}_1+m_2\mathbf{v}_2=0,$$

which says that the center of mass isn’t moving.

These equations follow from those here.

Thank you, @MarcoCiafa, for noticing a confusing typo in the last equation.

• Just to clarify: the relative velocity V in your equation is the velocity of the reduced mass? So if we consider two masses of 0.5 placed at +- 10 units on the x axis (on a keplerian orbit of e=0.9). What is my initial velocity of each body? – Warrenmovic Feb 29 '20 at 9:08
• – G. Smith Feb 29 '20 at 23:47