Motion in Gravitational Fields

I would like to share some insight on how different trajectories can form in a central field. I'll use laws of Newtonian gravitation and I would limit myself to examples in gravitation.

Two Body Central Force

In a two body system, we normally assume the heavier to be fixed. I will not. Also, I will proceed by taking the motion of the reduced mass of the system.

Going down to basic equations, coordinate of C.M. becomes R=m1r1+m2r2M\vec R = \dfrac{m_1\vec r_1 + m_2\vec r_2}{M} where M=m1+m2M = m_1 + m_2. Now, using r=r1r2\vec r = \vec r_1 - \vec r_2, r1=R+m2Mrr2=Rm1Mr\vec r_1 = \vec R + \dfrac{m_2}M \vec r\\\vec r_2 = \vec R - \dfrac{m_1}M \vec r

Note that the momentum P=MR˙=0\vec P = M\dot {\vec R} = 0 implies that the centre of mass is stationary (R=0\vec R = 0)

Looking at Kinetic Energy, T=12(m1r˙12+m2r˙22)=12[m1(R˙+m2Mr˙)2+m1(R˙m1Mr˙)2]=12MR˙2+12μr˙2\begin{aligned}T &= \dfrac12\left(m_1\dot r_1^2+m_2\dot r_2^2\right)\\ &=\dfrac12\left[m_1\left(\dot R + \dfrac{m_2}M \dot r\right)^2+m_1\left(\dot R - \dfrac{m_1}M \dot r\right)^2\right]\\ &=\dfrac12 M\dot R^2+\dfrac12 \mu\dot r^2 \end{aligned} where μ=m1m2M\mu = \dfrac{m_1m_2}M, the reduced mass of the system. It's time for us to shift to the centre of mass frame of reference.

The Centre of Mass reference frame

From now on, all standard variables will be referred on the C.M. frame.

In C.M. frame, for example, our kinetic energy becomes, T=12μr˙2T = \dfrac12 \mu\dot r^2.

Equation of Motion

Let our central force and it's potential be F=f(r) r^U=U(r)\vec F = -f(r)\ \hat r\\U = U(r)

The angular momentum of the system (w.r.t. CM) will become, L=(m1r12+m2r22)ω=μr2θ˙\vec L = \left(m_1r_1^2+m_2r_2^2\right)\omega = \mu r^2 \dot \theta

Using r˙=r˙ r^+rθ˙ θ^\dot{\vec r} = \dot r \ \hat r + r\dot\theta\ \hat\theta, T=12μ(r˙2+r2θ˙2)T = \dfrac12 \mu\left(\dot r^2 + r^2\dot\theta^2\right)

Thus, the total energy, E=T+U=12μ(r˙2+r2θ˙2)+U(r)=12μr˙2+L22μr2+U(r)E=12μr˙2+Ueff\begin{aligned} E &= T+U\\ &= \dfrac12 \mu\left(\dot r^2 + r^2\dot\theta^2\right) + U(r)\\ &= \dfrac12 \mu\dot r^2 + \dfrac{L^2}{2\mu r^2} + U(r)\\ E&= \dfrac12 \mu\dot r^2 + U_{\text{eff}}\end{aligned} where Ueff=L22μr2+U(r)U_{\text{eff}} = \dfrac{L^2}{2\mu r^2} + U(r)

Note: I've written rotational kinetic energy in terms of LL is because θ\theta term is messy and LL is constant.

This won't be updated. I'm tired, sorry.

Note by Kishore S. Shenoy
3 years ago

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Great Work

Ram Mohith - 3 years ago

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Okay, this is dead. I have no idea what I intended to say after this. Good day!

Kishore S. Shenoy - 2 months, 3 weeks ago

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