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# 21.6: A General Central Force

[ "article:topic", "authorname:tatumj" ]

Let us suppose that we have a particle that is moving under the influence of a central force $$F(r)$$. The equations of motion are

$m( \ddot{r} -r \dot{ \theta }^2) = F(r) \tag{21.6.2}\label{eq:21.6.2}$

Transverse:

$r \ddot{\theta} + 2 \dot{r} \dot{\theta} = 0. \tag{21.6.3}\label{eq:21.6.3}$

These can also be written

$\ddot{r} - \dot{ \theta}^2 = a(r) \tag{21.6.4}\label{eq:21.6.4}$

$r^2 \dot{ \theta} = h. \tag{21.6.5}\label{eq:21.6.5}$

Here $$a$$ is the radial force per unit mass (i.e. the radial acceleration) and $$h$$ is the (constant) angular momentum per unit mass. [If you are unsure of why Equations $$\ref{eq:21.6.3}$$  and  $$\ref{eq:21.6.5}$$ are the same, differentiate equation $$\ref{eq:21.6.5}$$  with respect to time.]

These are two simultaneous equations in $$r, \theta, t$$ . In principle, if we could eliminate $$t$$ between them, we would obtain a relation between $$r$$ and $$\theta$$, which would tell us the shape of the path pursued by the particle. In Chapter 9 of my Celestial Mechanics notes we do this for the gravitational case, and we find that the path is an ellipse of the form $$r = \frac{l}{1 + e \cos \theta }$$. Or perhaps we could eliminate $$r$$ and hence find out how the angle $$\theta$$ changes with time. Or again we might be able to eliminate $$\theta$$ and hence get a relation telling us how $$r$$ varies with the time. Yet again we might be told the shape of the path $$r(\theta)$$, and asked to find the force law $$F(r)$$. Or again, rather than the force, we might be given the form of the potential energy $$V(r)$$, which is related to the force by $$F = - dV/dr$$. The potential $$\Phi$$  is the potential energy per unit mass, and $$-d \Phi /dr$$ is the radial force per unit mass - i.e. it is the radial acceleration $$a(r)$$ of the orbiting particle. The angular momentum of the particle, which is constant, is $$L - mr^2 \dot{ \theta}$$, and the angular momentum per unit mass is $$h = r^2 \dot{\theta}$$, which is twice the rate at which the radius vector sweeps out area.

We might also remember that, if we are given the potential energy $$V$$ or the potential $$\Phi$$ in an inertial frame, we might also want to work in a co-rotating frame, making use of the equivalent potential energy $$V' = V + \frac{L^2}{2mr^2}$$ or the equivalent potential $$\Omega ' = \Omega + \frac{h^2}{2^2}$$.

One last thing to bear in mind before starting any problems of this class. It turns out that, very often, a change of variable $$u = 1/r$$ turns out to be useful. Conservation of angular momentum then takes the form $$\dot{\theta} / u ^2 = h$$  Also

$\dot{r} = \frac{dr}{du}\frac{du}{dt}= \frac{dr}{d\theta}\frac{dr}{dt}\frac{du}{d\theta} = - \frac{\dot{\theta}}{u^2}\frac{du}{d\theta} = - h \frac{du}{d\theta} \tag{21.6.6}\label{eq:21.6.6}$

and

$\ddot{r} = \frac{d}{dt}\left(-h \frac{du}{d\theta}\right) = -h \frac{d}{dt}\frac{du}{d\theta}= - h \frac{d\theta}{dt}\frac{d}{d\theta}\frac{du}{d\theta}= -h.hu^2.\frac{d^2u}{d\theta^2}= -h^2u^2\frac{d^2u}{d\theta^2}. \tag{21.6.7}\label{eq:21.6.7}$

Equations 21.5.4 and $$\ref{eq:21.6.5}$$ now become

$h^2u^2\frac{d^2u}{d\theta^2}+h^2u^3 = -a(r). \tag{21.6.8}\label{eq:21.6.8}$

and

$\dot{\theta} = hu^2. \tag{21.6.9}\label{eq:21.6.9}$

We can now easily eliminate the time which was one of our aims:

$h^2u^2 \frac{d^2u}{d \theta^2} + h^2u^3 = -a(r). \tag{21.6.10 }\label{eq:21.6.10}$

[As ever, check the dimensions.] This equation, which does not contain the time, when integrated will give us the equation to the path.

With these remarks in mind, let us try a few problems. For example: