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# 21.2: Motion Under a Central Force

I consider the two-dimensional motion of a particle of mass $$m$$ under the influence of a conservative central force $$F(r)$$, which can be either attractive or repulsive, but depends only on the radial coordinate $$r$$. Recalling the formula $$\ddot{r} - r \dot{ \theta }^2$$ \)for acceleration in polar coordinates (the second term being the centripetal acceleration), we see that the equation of motion is

$m \ddot{r} - m r \dot{ \theta }^2 = F(r). \tag{21.2.1}\label{eq:21.2.1}$

This describes, in polar coordinates, two-dimensional motion in a plane. But since there are no transverse forces, the angular momentum $$m^2 \dot{ \theta }^2$$ is constant and equal to $$L$$, say. Thus we can write Equation $$\ref{eq:21.2.1}$$ as

$m \ddot{r} = F(r) + \frac{L^2}{mr^3}. \tag{21.2.2}\label{eq:21.2.2}$

This has reduced it to a one-dimensional equation; that is, we are describing, relative to a co-rotating frame, how the distance of the particle from the centre of attraction (or repulsion) varies with time. In this co-rotating frame it is as if the particle were subject not only to the force $$F(r)$$, but also to an additional force $$\frac{L^2}{mr^3}$$. In other words the total force on the particle (referred to the co-rotating frame) is

$F'(r) = F(r) + \frac{L^2}{mr^3}. \tag{21.2.3}\label{eq:21.2.3}$

Now $$F(r)$$, being a conservative force, can be written as minus the derivative of a potential energy function, $$F = - \frac{dV}{dr}$$. Likewise, $$\frac{L^2}{m^3}$$ is minus the derivative of $$\frac{L^2}{2mr^2}$$. Thus, in the co-rotating frame, the motion of the particle can be described as constrained by the potential energy function $$V'$$, where

$V' = V + \frac{L^2}{2mr^2}. \tag{21.2.4}\label{eq:21.2.4}$

This is the equivalent potential energy. If we divide both sides by the mass m of the orbiting particle, this becomes

$\Phi' = \Phi + \frac{h^2}{2r^2}. \tag{21.2.5}\label{eq:21.2.5}$

Here $$h$$ is the angular momentum per unit mass of the orbiting particle, $$\Phi$$ is the potential in the inertial frame, and $$\Phi '$$is the equivalent potential in the corotating frame.
The passage describes the mathematical formulation of the motion of a particle under a central force, including the derivation of the equation of motion and the concept of equivalent potential energy. It does not contain any specific mathematical questions or answers to such questions.
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# 21.2: Motion Under a Central Force

I consider the two-dimensional motion of a particle of mass $$m$$ under the influence of a conservative central force $$F(r)$$, which can be either attractive or repulsive, but depends only on the radial coordinate $$r$$. Recalling the formula $$\ddot{r} - r \dot{ \theta }^2$$ \)for acceleration in polar coordinates (the second term being the centripetal acceleration), we see that the equation of motion is

$m \ddot{r} - m r \dot{ \theta }^2 = F(r). \tag{21.2.1}\label{eq:21.2.1}$

This describes, in polar coordinates, two-dimensional motion in a plane. But since there are no transverse forces, the angular momentum $$m^2 \dot{ \theta }^2$$ is constant and equal to $$L$$, say. Thus we can write Equation $$\ref{eq:21.2.1}$$ as

$m \ddot{r} = F(r) + \frac{L^2}{mr^3}. \tag{21.2.2}\label{eq:21.2.2}$

This has reduced it to a one-dimensional equation; that is, we are describing, relative to a co-rotating frame, how the distance of the particle from the centre of attraction (or repulsion) varies with time. In this co-rotating frame it is as if the particle were subject not only to the force $$F(r)$$, but also to an additional force $$\frac{L^2}{mr^3}$$. In other words the total force on the particle (referred to the co-rotating frame) is

$F'(r) = F(r) + \frac{L^2}{mr^3}. \tag{21.2.3}\label{eq:21.2.3}$

Now $$F(r)$$, being a conservative force, can be written as minus the derivative of a potential energy function, $$F = - \frac{dV}{dr}$$. Likewise, $$\frac{L^2}{m^3}$$ is minus the derivative of $$\frac{L^2}{2mr^2}$$. Thus, in the co-rotating frame, the motion of the particle can be described as constrained by the potential energy function $$V'$$, where

$V' = V + \frac{L^2}{2mr^2}. \tag{21.2.4}\label{eq:21.2.4}$

This is the equivalent potential energy. If we divide both sides by the mass m of the orbiting particle, this becomes

$\Phi' = \Phi + \frac{h^2}{2r^2}. \tag{21.2.5}\label{eq:21.2.5}$

Here $$h$$ is the angular momentum per unit mass of the orbiting particle, $$\Phi$$ is the potential in the inertial frame, and $$\Phi '$$is the equivalent potential in the corotating frame.
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The passage describes the mathematical formulation of the motion of a particle under a central force, including the derivation of the equation of motion and the concept of equivalent potential energy. It does not contain any specific mathematical questions or answers to such questions.

TYPE: NO_QUESTION