what would a planetary orbit look like if gravity had constant magnitude?

Question

Consider a unit-mass particle that is always experiencing a single unit-magnitude force towards the origin. This is a central force, but it is not one of the familiar ones, e.g. gravity whose magnitude is proportional to inverse distance squared, or a spring force whose magnitude is proportional to distance.

So the particle is always accelerating towards the origin with constant acceleration magnitude $1$. Stated as a differential equation, working in the $x$-$y$ plane, the particle's position as a function of time $\mathbf{p}(t){=}(x(t),y(t))$ satisfies: $$\ddot{\mathbf{p}}(t) = -\mathbf{p}(t)/\Vert\mathbf{p}(t)\Vert.$$ If we are additionally given initial position $\mathbf{p}(0)$ and velocity $\mathbf{v}(0)=\dot{\mathbf{p}}(0)$, then the function $\mathbf{p}$ is completely determined, and it can be easily computed numerically to any desired accuracy by simply iterating the following with small enough timestep $dt$: \begin{align} \mathbf{a} &\leftarrow -\mathbf{p}/\Vert\mathbf{p}\Vert \\ \mathbf{v} &\leftarrow \mathbf{v} + \mathbf{a}\,\,dt \\ \mathbf{p} &\leftarrow \mathbf{p} + \mathbf{v}\,\,dt \end{align}

My question: is $\mathbf{p}(t)$ a well-known function, and does it have a closed form?

Of course one case of this is a simple circular orbit of unit radius and speed: $$\mathbf{p}(0){=}(1,0), \,\, \mathbf{v}(0){=}(0,1) \,\,\Rightarrow\,\, \mathbf{p}(t)=\left(\cos t,\sin t\right).$$ More generally, a uniform circular orbit of any radius $r$ and speed $\sqrt{r}$ can be obtained: $$\mathbf{p}(0){=}(r,0), \,\, \mathbf{v}(0){=}(0,\sqrt{r}) \,\,\Rightarrow\,\, \mathbf{p}(t)=\left(r \cos\frac{t}{\sqrt{r}},r \sin\frac{t}{\sqrt{r}}\right).$$ and we check that the desired equation holds: \begin{align} \dot{\mathbf{p}}(t) &= \left(-\sqrt{r} \sin \frac{t}{\sqrt{r}}, \sqrt{r} \cos \frac{t}{\sqrt{r}}\right) \\ \ddot{\mathbf{p}}(t) &= \left(-\cos \frac{t}{\sqrt{r}}, -\sin\frac{t}{\sqrt{r}}\right) \\ &= -\mathbf{p}(t)/\Vert\mathbf{p}(t)\Vert. \end{align}

Another simple case is when the initial velocity is zero or collinear with the position and the origin; in this case it's a 1-dimensional problem and the position can easily be seen to be a simple piecewise quadratic function of time.

But what if the initial conditions are not so nicely aligned?

To get an idea of the shapes that are possible, I've made some plots, using gnuplot, of simulations using the simple evolution algorithm I described earlier, with $dt = 1/10000$ degree $\approx .00000175$.

Figure 1 shows five different initial states, each evolved from $t{=}0$ to $t{=}2 \pi$: $\,\,\mathbf{p}(0){=}(1,0)$, $\mathbf{v}(0){=}(0,v_{0 y})$ for $v_{0 y}{=}0.5,1,1.5,2,2.5$.

Figure 2 shows the one with $\mathbf{v}(0){=}(0,2)$ evolved farther, to $t{=}20\pi$.

Figure 3 shows it evolved even farther, to $t{=}60\pi$.

CLARIFICATION: I am ultimately interested in finding the simplest way of expressing $\mathbf{p}$ as a function of $t$. I.e. I really want to know "what this function looks like" rather than "what the curve looks like". Other parametrizations of the curve, and intuition about shape of the curve, are of interest only if they help lead to this answer.

UPDATE 2015/07/02:
It sure looks like a spirograph hypotrochoid, doesn't it? http://mathworld.wolfram.com/Spirograph.html .

Exploring this possibility, I found by binary search an initial velocity (0,1.662656) (probably accurate to only 4 decimal places or so) yielding a closed orbit in the shape of a 7-petalled flower, and then compared that sim result with the 7-petalled hypotrochoid having the same min and max radii; see Figure 4.

Conclusion: It's really close, but it's not a hypotrochoid. It moves too fast at the fast parts and too slow at the slow parts, and stays a bit too close to the origin during the in-between parts.

Answer 1

This is a nice question, and I don't have a comprehensive answer for you, but perhaps this will help a bit. For conservative forces, it's generally convenient to define the scalar potential $V$. The force will satisfy $\vec{F} = \vec{\nabla}V$, and your potential $V = r$ (yay, system is conservative). Since the kinetic + potential energy is constant, and $V$ is unbounded above, this means that the particle will always be in a bounded orbit around the body.

The orbits you've found are bounded, but not closed (on themselves), resulting in the flower-pattern trajectory. I do know that you only get closed orbits for potentials that are $\tfrac 1r$ (like ours), and $\tfrac 1{r^2}$ (where the force goes as $\tfrac{1}{r^3} \large)$, but I never really understood why (phase-space apparently).

Answer 2

Consider an orbit $$\gamma: \quad t\mapsto z(t)=r(t)e^{i\phi(t)}$$ of such a particle. We are interested in the polar representation $$\phi\mapsto r(\phi)\tag{1}$$ of the resulting curve $\hat\gamma\subset {\mathbb C}$. Denoting the differentiation with respect to $t$ by a $\cdot$ and the differentiation with respect to $\phi$ by a $'$ we have $\dot r=r'\>\dot\phi$.

The kinetic energy of the particle (of mass $1$) is given by $$T={1\over2}(\dot r^2+ r^2\dot\phi^2)={1\over2}(r'^2+r^2)\dot\phi^2\ ,$$ and its potential energy simply by $V(z)=r$. It follows that $$E={1\over2}(r'^2+r^2)\dot\phi^2 + r\tag{2}$$ is constant along $\gamma$. A second invariant is the angular momentum $$L=r^2\>\dot\phi\ ,$$ so that we can eliminate $\dot\phi^2$ in $(2)$ and obtain the following differential equation for the functions $(1)$: $${1\over2}(r'^2+r^2){L^2\over r^4} + r=E\ .$$ This can be rewritten as a standard first order $ODE$ in the following form: $$r'=\pm{r\over L}\sqrt{2Er^2-L^2-2r^3}\qquad(r>0)\ .\tag{3}$$ The ODE $(3)$ can be separated. In fact, one obtains the inverse functions $r\mapsto \phi(r)$ by a mere quadrature: $${d\phi\over dr}=\pm{L\over r}\bigl(2Er^2-L^2-2r^3\bigr)^{-1/2}\ .$$ Unfortunately the resulting integral is nonelementary for general values of $E$ and $L$.

Answer 3

This is an old post but the following remarks might be of interest. Firstly, if one writes the orbit equations in the (modified) polar form $rf(\theta)=1$, then the ode $$f+f´´=cf^\alpha$$ give all orbits for a force proportional to $ r^\beta$ with $\beta=-2-\alpha$. In the case of a constant force, this leads to the equation $$f+f´´=cf^2$$ which has the explicit solution $(\cos(\frac {3x}2))^{2/3}$ (MacLaurin). This, by dilation and rotation, supplies two thirds of the solutions (that is, a two parameter family within the full three parameter family). One can use numerical methods to get pictures of the other orbits (Mathematica can handle the general solution to the ode but in a form which probably isn´t of much use).