Two-body Newtonian Gravity (Sample worksheet for students) – Suggested Solutions

These are suggested solutions for this sample worksheet.

(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.

\begin{align} m\mathbf{r}(0)+M\mathbf{R}(0) & =\mathbf{0} \\ \mathbf{R}(0) & \label{eq-cm-pos}=-\frac{m}{M}\mathbf{r}(0)\\ \end{align}

(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.

\begin{align} \mathbf{R}(0) & =\left(-\frac{m}{M}r_{0},0\right)\\ \end{align}

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

The velocity should be along the y-axis, i.e.

\begin{align} \mathbf{\dot{r}}(0) & \label{eq-dir-vel}=(0,v)\\ \end{align}

for some value of \(v\) (either positive or negative). This way, the resultant force on (and thus acceleration of) the mass is perpendicular to the velocity of the mass – a necessary (but insufficient) condition for circular motion.

(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).

In other words, we want to solve for \(v\) in (\ref{eq-dir-vel}). The key is to realise that the larger (red) mass will have to circulate in the opposite direction around the origin, so that the separation of the masses is always a fixed distance of \(d=r_{0}\left(1+m/M\right)\). This allows uniform circular motion (at constant speed), since the force on the smaller (blue) mass is of a constant magnitude, and directed towards the origin. Apply Newton’s second law of motion, using Newton’s law of universal gravitation, and the expression for the acceleration of an object in circular motion.

\begin{align} \frac{GMm}{d^{2}} & =\frac{mv^{2}}{r_{0}} \\ v & =\sqrt{\frac{GM}{r_{0}}}\left(\frac{1}{1+{m}/{M}}\right)\\ \end{align}

(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.

Set the total momentum to zero, i.e.

\begin{align} m\mathbf{\dot{r}}(0)+M\mathbf{\dot{R}}(0) & =\mathbf{0} \\ \mathbf{\dot{R}}(0) & =-\frac{m}{M}\mathbf{\dot{r}}(0)\\ \end{align}

We see that taken together with (\ref{eq-cm-pos}), if the system is started up as such at time \(t=0\), then this is (mathematically) consistent with having

\begin{equation} \label{eq-cm-final}\begin{cases}\mathbf{R}(t)=-\frac{m}{M}\mathbf{r}(t)\\ \mathbf{\dot{R}}(t)=-\frac{m}{M}\mathbf{\dot{r}}(t)\end{cases}\\ \end{equation}

for all times \(t\).

(d)(i) Write down an expression for the momentum \(\mathbf{p}(t)\) of the smaller (blue) mass in terms of \(m\) and \(\mathbf{\dot{r}}(t)\).

\begin{align} \mathbf{p}(t) & =m\mathbf{\dot{r}}(t)\\ \end{align}

(d)(ii) Briefly explain how the momentum of the smaller (blue) mass is related to the force acting on it.

By Newton’s second law, the resultant force on a mass is proportional to the rate of change of its momentum. We conventionally set the constant of proportionality to one. If the resultant force acting on the mass is \(\mathbf{F}\), then we have

\begin{align} \mathbf{F}(t) & =\frac{d\mathbf{p}(t)}{dt}\\ \end{align}

(d)(iii) Discuss how, by thinking of both masses as a system, the total momentum of the system evolves with time.

There is no external force acting on the system. (The internal force acting on the smaller mass is equal and opposite to the internal force acting on the larger mass. This is guaranteed by Newton’s third law of motion, and explicitly expressed in Newton’s law of universal gravitation for the case of gravitational forces.) Thus the total momentum of the system is a constant and does not change.

(d)(iv) Explain which important property of the two-body interaction leads to the conclusion in (d)(iii).

The interaction produces equal magnitude and oppositely directed forces acting on each body.

(d)(v) Write down an expression for the magnitude of the force \(\mathbf{F}(t)\) acting on the smaller (blue) mass. Describe the direction of this force.

Since the total momentum of the system is constant, if initialised as zero, it will stay zero in the frame of reference used in the simulation. Thus, the centre-of-mass, if initialised at the origin, stays at the origin. This means that we have

\begin{align} \mathbf{R}(t) & =-\frac{m}{M}\mathbf{r}(t)\\ \end{align}

at all times \(t\), as in (\ref{eq-cm-final}). Thus we see that the two masses are always on opposite directions of the origin, so the attractive force of gravity is directed towards the origin. The magnitude is

\begin{align} F & =\frac{GMm}{r^{2}}\cdot\frac{1}{(1+m/M)^{2}}\;,\\ \end{align}

where \(r=|\mathbf{r}(t)|\) is the distance of the smaller (blue) mass from the origin. Note that the magnitude of the force changes if the distance \(r\) changes.