# 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 $\vec R = \dfrac{m_1\vec r_1 + m_2\vec r_2}{M}$ where $$M = m_1 + m_2$$. Now, using $$\vec r = \vec r_1 - \vec r_2$$, $\vec r_1 = \vec R + \dfrac{m_2}M \vec r\\\vec r_2 = \vec R - \dfrac{m_1}M \vec r$

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

Looking at Kinetic Energy, \begin{align*}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{align*} where $$\mu = \dfrac{m_1m_2}M$$, 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, $$T = \dfrac12 \mu\dot r^2$$.

#### Equation of Motion

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

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

Using $$\dot{\vec r} = \dot r \ \hat r + r\dot\theta\ \hat\theta$$, $T = \dfrac12 \mu\left(\dot r^2 + r^2\dot\theta^2\right)$

Thus, the total energy, \begin{align*} 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{align*} where $$U_{\text{eff}} = \dfrac{L^2}{2\mu r^2} + U(r)$$

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

This will be updated

Note by Kishore S Shenoy
8 months, 3 weeks ago

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