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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.
explanation: The document is a detailed explanation of the motion of a particle under a central force, using polar coordinates and the concept of equivalent potential energy. It derives the equations of motion and explains how the problem can be reduced to a one-dimensional equation by considering the conservation of angular momentum. The text does not contain a specific question or an answer to a question.
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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.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a detailed explanation of the motion of a particle under a central force, using polar coordinates and the concept of equivalent potential energy. It derives the equations of motion and explains how the problem can be reduced to a one-dimensional equation by considering the conservation of angular momentum. The text does not contain a specific question or an answer to a question.
  
  Label: no-question