Question #163614

How can we change two body system into one.


1
Expert's answer
2021-02-15T00:54:46-0500

Suppose we have a two body system (body 1, body 2) with a potential depending only on their relative distance. The total energy is :

E=m1v122+m2v222+V(r1r2)E = \frac{m_1 v_1^2}{2} + \frac{m_2 v_2^2}{2} + V(r_1-r_2)

and thus we introduce the relative position r=r1r2r=r_1-r_2, position of a center of mass R=m1r1+m2r2m1+m2R=\frac{m_1 r_1 + m_2 r_2}{m_1+m_2} and we reformulate the positions r1,r2r_1, r_2 with these new variables :

r1=R+m2m1+m2rr_1 = R+\frac{m_2}{m_1+m_2} r

r2=Rm1m1+m2rr_2 = R - \frac{m_1}{m_1+m_2}r

And now the velocities :

v1=vMC+m2m1+m2vv_1 = v_{MC} + \frac{m_2}{m_1+m_2} v

v2=vMCm1m1+m2vv_2 = v_{MC} - \frac{m_1}{m_1+m_2} v

Now by reformulating the energy in these new variables we get :

E=(m1+m2)vMC22+12m1m2m1+m2v2+V(r)E =\frac{(m_1+m_2)v_{MC}^2}{2} + \frac{1}{2} \frac{m_1m_2}{m_1+m_2} v^2+V(r)

We see that the potential energy is independent from R and thus vMCv_{MC} is conserved (which is coherent with the physical explanation - the system is isolated and thus the total momentum is conserved). We can place ourselves in a inertial frame of reference where vMC=0v_{MC}=0. Now by introducing μ=m1m2m1+m2\mu=\frac{m_1 m_2}{m_1 + m_2} - reduced mass, we get an energy of a one body in a central field :

E=μv22+V(r)E = \frac{\mu v^2}{2} +V(r)

And now we can solve the problem completely in this formalism.


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