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Consider a system of \(n\) different masses \(m_i\), interacting with each other and being acted on by external forces. We can write Newton's second law for the positions \(\vec{r}_i\) of each of these masses with respect to a fixed origin \(\cal{O}\), thereby obtaining a system of equations governing the motion of the masses. \begin{align} m_1 \frac{d^2\, \vec{r}_1}{dt^2} &=\vec{F}_1+\;\; 0\;\, +\vec{f}_{12}+\vec{f}_{13}+\;\,\dots\;\, +\vec{f}_{1n}\nonumber\\ m_2 \frac{d^2\, \vec{r}_2}{dt^2} &=\vec{F}_2+\vec{f}_{21}+\;\; 0\;\, +\vec{f}_{23}+\;\,\dots\;\, +\vec{f}_{2n} \label{NewtonSystem}\\ \vdots\nonumber\\ m_n \frac{d^2\, \vec{r}_n}{dt^2} &=\vec{F}_n+\vec{f}_{n1}+\vec{f}_{n2}+\dots+\vec{f}_{n(n-1)}+0\quad\nonumber \end{align} Here, we have chosen the notation \(\vec{F}_i\) for the net external forces acting on mass \(m_i\) and \(\vec{f}_{ij}\) for the internal force of mass \(m_j\) acting on \(m_i\).

In general, each internal force \(\vec{f}_{ij}\) will depend on the positions of the particles \(\vec{r}_i\) and \(\vec{r}_j\) in some complicated way, making \((\ref{NewtonSystem})\), a set of coupled differential equations. To solve \((\ref{NewtonSystem})\), we first need to decouple the differential equations, i.e. find an equivalent set of differential equations in which each equation contains only one variable.

The weak form of Newton's third law states that the force \(\vec{f}_{12}\) of \(m_2\) on \(m_1\) is equal and opposite to the force \(\vec{f}_{21}\) of \(m_1\) on \(m_2\). We see that each internal force appears twice in the system of equations \((\ref{NewtonSystem})\), once with a positive sign and once with a negative sign. Therefore, if we add all of the equations together, the internal forces will all cancel, leaving: \begin{equation} \sum_{i=1}^n m_i \frac{d^2 \vec{r}_i}{dt^2} =\sum_{i=1}^n\vec{F}_i\label{NewtonCOM} \end{equation}

Notice what a surprising equation \((\ref{NewtonCOM})\) is. The right-hand side directs us to add up all of the external forces, each of which acts on a different mass; something you were taught never to do in introductory physics.

The left-hand side of \((\ref{NewtonCOM})\) directs us to add up (the second derivatives of) \(n\) “weighted" position vectors pointing from the origin to different masses. We can simplify the left-hand side of \((\ref{NewtonCOM})\) if we multiply and divide by the total mass \(M=m_1+m_2+\dots+m_n\) and use the linearity of differentiation to “factor out” the derivative operator: \begin{align} \sum_{i=1}^n m_i \frac{d^2 \vec{r}_i}{dt^2} &=M\frac{d^2}{dt^2} \left(\sum_{i=1}^n \frac{m_i}{M}\, \vec{r}_i\right)\label{CenterOfMass1}\\ &=M\frac{d^2 \vec{R}_{cm}}{dt^2}\label{CenterOfMass2} \end{align} We recognize (or define) the quantity in the parentheses on the right-hand side of \((\ref{CenterOfMass1})\) as the position vector \(\vec{R}_{cm}\) from the origin to the “center of mass” of the system of particles, i.e. \begin{equation} \vec{R}_{cm}=\sum_{i=1}^n\frac{m_i}{M}\, \vec{r}_i\label{CenterOfMass3} \end{equation} With these simplifications, equation (\ref{NewtonCOM}) becomes: \begin{equation} M \frac{d^2 \vec{R}_{cm}}{dt^2} =\sum_{i=1}^n\vec{F}_i\label{NewtonCOM2} \end{equation} which has the form of Newton's 2nd Law for a fictitious particle with mass \(M\) sitting at the center of mass of the system of particles and acted on by all of the external forces from the original system.

We can define the momentum of the center of mass as the total mass times the time derivative of the position of the center of mass: \begin{equation} \vec{P}_{cm}=M\frac{d\vec{R}_{cm}}{dt} \end{equation} If there are no external forces acting, then the acceleration of the center of mass is zero and the momentum of the center of mass is constant in time (conserved). \begin{equation} M\frac{d^2 \vec{R}_{cm}}{dt^2}=\frac{d\vec{P}_{cm}}{dt}=0 \label{MomentumConservation} \end{equation}

Notice that the entire discussion above applies even if all of the internal forces are zero \(\vec{f}_{ij}=0\), i.e. none of the particles have any way of knowing that the others are even present. Such particles are called non-interacting. The position of the center of mass of the system will still move according to equation \((\ref{NewtonCOM2})\).

• Practice evaluating line integrals;
• Practice choosing appropriate coordinates and basis vectors;
• Introduction to the geometry behind conservative vector fields.

In this unit, you will explore the classical mechanics of central forces, especially gravitational orbits like the earth going around the sun.

Motivating Questions

• What shapes can the orbits have?
• What are Kepler's laws and why are they true?
• What is an effective potential diagram and how can it be used to predict the shape of an orbit?

Key Activities/Problems

• Problem: Undo Formulas for Center of Mass (Geometry)
• Activity: Acting Out Effective Potentials (In class, only)
• Activity: Effective Potentials
• Problem: Hockey
• Problem: Scattering

Unit Learning Outcomes

At the end of this unit, you should be able to:

• List the properties that define a central force system.
• Calculate a reduced mass for a two-body system and describe why it is important.
• Use the solution (algebraic or geometric) to a reduced mass system to describe the motion of the original system.
• Describe the role that conservation of energy and angular momentum play in a central force system. In particular, where do these properties appear in the solutions of the equations of motion?
• Use an effective potential diagram to predict the possible orbits in a central force system: which orbits are bound or unbound? which are closed or open? where will the turning points be?

Equation Sheet for This Unit

• Classical Orbits Equation Sheet

• Recall the relationship between the sign of the dot product and the orientation of the vectors.
• Use graphical methods to estimate the value of a vector line integral.
• Lay the groundwork for thinking about conservative and non-conservative vector fields.

Use geometry to find formulas for velocity and acceleration in polar coordinates.

Students will estimate the work done by a given electric field. They will connect the work done to the height of a plastic surface graph of the electric potential.

These lecture notes covering week 8 of https://paradigms.oregonstate.edu/courses/ph441 include a small group activity in which students derive the Carnot efficiency.