Skip to main content
\(\require{cancel}\)
Physics LibreTexts

21.6: A General Central Force

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

Radial:

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

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:

Contributor