Two-body Newtonian Gravity (sample student worksheet)

Preliminaries

Simulation view. There are two massive objects (“bodies”) in the simulation, the smaller (blue) circle and the larger (red) circle. They are programmed to follow the laws of Newtonian dynamics (force proportional to the rate of change of momentum), with a gravitational force of attraction between them according to Newtonian gravity (force proportional to the product of masses and inversely proportional to the square of the separation).

Graphs. There are two plots shown. On the left, the kinetic energy and potential energy are plotted as a function of time, with total energy shown as a dashed line. On the right, the horizontal axis is the distance of the smaller (blue) mass from the origin, and the vertical axis shows the total energy (dashed line) and effective radial potential. The regions of the plot where the total energy is less than the effective radial potential are unphysical.

Screenshot of simulation

Notation

General conventions. We use italics to indicate scalar quantities. We use boldface to indicate vector quantities (in your handwritten work, you might wish to underline or draw a small arrow above such quantities). We use a dot over a symbol to indicate a time derivative.

More specific conventions. For symbols relating to the smaller (blue) mass, we generally use lowercase. We generally use uppercase for the equivalent symbols for the larger (red) mass. We use the following notation throughout this worksheet:

  • \(t\) for the simulation time

  • \(m\), \(M\) for the masses of the objects

  • \(\mathbf{r}(t)\), \(\mathbf{R}(t)\) for position vectors of the masses

  • \(\mathbf{\dot{r}}(t)\), \(\mathbf{\dot{R}}(t)\) for velocity vectors of the masses

Mathematical note. If students are having trouble with notation like \(\mathbf{\dot{r}}(0)\), help them realise that the “dot” only makes sense when applied to a function of \(t\). So conceptually you have to “dot” (take time derivative) first, before substituting the value \(t=0\) into the resulting expression.

Initial Parameters

Motivation. Let us work out the initial values of various parameters, before the simulation runs (which proceeds by numerically solving Newton’s second law of motion, which are implemented as a series of differential equations).

Assumptions. Assume that \(m,M\) are chosen. We want the centre-of-mass of the two-body system to be at the origin at all times, so that we centralise the simulation view on the origin (0,0) in Cartesian coordinates.

Positions. Consider the initial configuration at \(t=0\).

(a)(i) Write down an expression for \(\mathbf{R}(0)\) in terms of \(\mathbf{r}(0)\), \(m\), \(M\) such that the centre of mass of the two-body system is at the origin.

(a)(ii) The smaller (blue) mass is initially at \(\mathbf{r}(0)=(r_{0},0)\) in Cartesian coordinates. Write down \(\mathbf{R}(0)\) for the initial position of the larger (red) mass in Cartesian coordinates.

Motion (circular). This is a particularly simple two-body system, and so there is only a single interaction between the bodies. Suppose we want the smaller (blue) mass to move in a circle around the origin.

(b)(i) If so, explain what the direction of \(\mathbf{\dot{r}}(0)\) should be.

(b)(ii) If so, determine the magnitude of \(\mathbf{\dot{r}}(0)\), in terms of the masses, \(r_{0}\) and \(G\) (the universal gravitational constant in Newton’s law of gravitation).

Non-circular motion. The simulation allows the user to impart an initial radial velocity to the smaller (blue) mass, so that it does not follow a circular path. Thus the direction and magnitude of \(\mathbf{\dot{r}}(0)\) might be different from what was calculated in (b)(i) and (b)(ii).

(c) Write down a general expression for \(\mathbf{\dot{R}}(0)\) in terms of \(\mathbf{\dot{r}}(0)\), \(m\), \(M\) such that the centre-of-mass of the two-body system stays at the origin and does not move.<