The shape of a fountain

Consider the following:

What points can be reached by a particle with given initial position and speed (but with any direction of velocity) moving under the influence of some gravitational field?

Or, less precisely but more evocatively,

What does a fountain spewing water in all directions look like?

The answer depends on the gravitational field, of course; we’ll mostly consider a uniform field in this post, with a brief jaunt away from that at the end.

Before we dive into the math, here’s a little visualization of the situation we’re considering.

You can imagine that if we were to draw the particles shooting out at every possible angle, their paths would fill in a solid area of some yet-to-be-determined shape.

Uniform gravity

We can work in a two-dimensional space, since there are (at most) two distinguished directions for any given particle: the direction of the field and the direction of the initial velocity. The particle can never leave the plane containing those two directions and the starting point (or the line, when the directions are parallel). If you want more dimensions, rotate appropriately.

To avoid carrying around a bunch of constants, let’s assume that the particle is fired from the origin of a coordinate system at unit speed, with unit acceleration in the negative $y$ direction. Then, with the angle of firing suitably specified by $\theta$, the initial velocity of the particle is $cos\theta$ in the $x$ direction and $sin\theta$ in $y$; after applying acceleration, we get that the position of the particle at time $t$ is

$$\begin{array}{cccc}& x(t)& =tcos\theta & \\ & y(t)& =tsin\theta -\frac{t^2}{2}.& \end{array}$$

Below are three very different-feeling ways of getting to the same answer. Since this is a blog post and not a paper, I’m going to sometimes play a bit fast and loose with technical details (alternate solutions to equations, dividing by zero in edge cases, etc.). It all works out in the end.

(There is some overlap here with https://math.stackexchange.com/questions/1495086/ and https://physics.stackexchange.com/questions/63129/, and presumably other content elsewhere, though I worked through all this independently.)

Algebraic solution

Let’s start by considering a point $(x,y)$ in the plane and directly determining whether it’s possible to reach that point.

We need there to be some $t$ and $\theta$ such that

$$\begin{array}{cccc}& x& =tcos\theta & \\ & y& =tsin\theta -\frac{t^2}{2}& \\ & & =\sqrt{t^2-x^2}-\frac{t^2}{2}& \\ & t^2-x^2& ={\left(y+\frac{t^2}{2}\right)}^2& \\ & & =y^2+y{t}^2+\frac{t^4}{4}& \\ & 0& =\frac{t^4}{4}+t^2(y-1)+(y^2+x^2).& \end{array}$$

This is a quadratic equation in $t^2$; for there to be any real solutions, we need the discriminant to be nonnegative.

$$\begin{array}{cccc}& 0& \le (y-1{)}^2-(y^2+x^2)& \\ & & =1-2y-x^2& \\ & y& \le \frac{1-x^2}{2},& \end{array}$$

meaning that all points on or below the parabola defined by $y=\frac{1-x^2}{2}$ are reachable.

This solution is apparently the standard one, according to one person on Stack Exchange, but I like it a bit less than the others here. It does have a compact writeup, though, and the helpful property of demonstrating that each point on the boundary can be reached in exactly one way, while points on the interior have two (because the determinant is strictly positive and so there are two solutions to the quadratic equation).

Polar solution

Consider a ray starting from the origin. How should we fire the particle so that it hits the ray as far as possible from the origin? If we figure that out for every ray, that gives us our overall answer in polar coordinates.

Suppose the ray is rotated by an angle $\varphi$ counterclockwise from the positive $x$-axis. Then, if we shift our coordinates to put the ray along the positive $x$-axis, we now have gravity rotated by $\varphi$ clockwise from $(0,-1)$, which brings it to $(-sin(\varphi ),-cos(\varphi ))$.

In this coordinate system,

$$\begin{array}{cccc}& x(t)& =tcos\theta -\frac{t^{2}sin\varphi }{2}& \\ & y(t)& =tsin\theta -\frac{t^{2}cos\varphi }{2}.& \end{array}$$

If the particle hits the ray at time $t_{hit}$, we need $t_{hit}>0$ and $y(t_{hit})=0$, so $0=sin\theta -t_{hit}cos\varphi /2$ or $t_{hit}=2sin\theta /cos\varphi$. Then

$$\begin{array}{cccc}& x(t_{hit})& =\frac{2sin\theta cos\theta }{cos\varphi }-\frac{2{sin}^2\theta sin\varphi }{{cos}^2\varphi }& \\ & & =\frac{1}{cos\varphi }\left(sin2\theta -(1-cos2\theta )tan\varphi \right).& \end{array}$$

To maximize this, set the derivative to 0:

$$\begin{array}{cccc}& 0& =cos2\theta -sin2\theta tan\varphi & \\ & tan2\theta & =cot\varphi .& \end{array}$$

We can go a step further and write out $\theta$ in terms of $\varphi$, but all we need here is that this implies that $sin2\theta =cos\varphi$ and $cos2\theta =sin\varphi$.

$$\begin{array}{cccc}& x(t_{hit})& =1-\frac{tan\varphi }{cos\varphi }+{tan}^2\varphi & \\ & & =\frac{{cos}^2\varphi -sin\varphi +{sin}^2\varphi }{{cos}^2\varphi }& \\ & & =\frac{1-sin\varphi }{1-{sin}^2\varphi }& \\ & & =\frac{1}{1+sin\varphi }& \end{array}$$

which is the polar version of the same parabola.

Falling circles solution

Instead of taking a particle with a particular $\theta$ and seeing how it moves across time, do the reverse: fire all possible particles at once (sweep $\theta$ across all possible values) and see where they end up at a particular value of $t$. So, rearranging the equations of motion, we have

$$\left[\begin{array}{c}x(t)\\ y(t)\end{array}\right]=t\left[\begin{array}{c}cos\theta \\ sin\theta \end{array}\right]+\left[\begin{array}{c}0\\ -t^2/2\end{array}\right],$$

i.e., the set of points is a circle of radius $t$ centered at $\left(0,-t^2/2\right)$. Going back to the animation from before, but with the particles now all shooting out at the same time, shows this nicely:

To figure out the boundary traced out by these circles over time, consider the circles at times $t$ and $t+dt$ for infinitesimal $dt$. The second circle is larger by $dt$ and lower down by $tdt$, since $\frac{d}{dt}\left(-\frac{t^2}{2}\right)=-t$. The point where these circles meet the envelope is the same as where a line tangent to both circles meets them.¹

The situation is shown in the diagram below.

The centers of the circles are the dots in the lower left, the dashed line is the common tangent, and the long diagonal line segments are radii to the tangent points. The difference in size and location between the two circles manifests itself in the orange triangle: the two radii are both perpendicular to the tangent, so they’re parallel to each other and we can drop a perpendicular from the first center to the second radius to chop off the excess length of $dt$.

To find the contact point with the tangent line, consider the blue triangle below.

The blue and orange triangles are similar right triangles and the legs of the blue triangle are axis-aligned, so the desired point is offset from the center by $\left(\sqrt{t^2-1},1\right)$, putting it at $(x,y)=\left(\sqrt{t^2-1},1-t^2/2\right)$. To convert that from parametric to the Cartesian form, we have

$$\begin{array}{cccc}& y& =1-\frac{t^2}{2}& \\ & & =1-\frac{x^2+1}{2}& \\ & & =\frac{1-x^2}{2}& \end{array}$$

once again.

Point source gravity

Instead of a uniform gravitational field, suppose we have an attractive inverse-square field centered at a point somewhere in the plane. I’ve done some numerical simulations that seem to indicate that the set of reachable points is now an ellipse whose foci are the origin and the gravitational center. Something I didn’t mention about the uniform field case is that the parabola bounding the set of reachable points has the origin as its focus, so the ellipse makes sense as a generalization of that case—a parabola is an ellipse with one focus moved out to infinity. Similarly, if the field is repulsive instead, it looks like the boundary is a hyperbola with foci at the two special points.

Those would be very pleasing outcomes if true; I feel like there must be some direct, intuitive explanation for them, but I don’t know what it is (or even what a not-necessarily-intuitive explanation might be).

Footnotes