# Double-Crossing Orbits

## Newtonian orbits

Suppose we pick two points, A and B, in the vicinity of a massive object, O, that exerts a Newtonian inverse-square gravitational force. If we choose some fixed time T, how many different ways can a “test particle” P move from A to B in that time? That is, how many different ways can we choose the velocity of P when we place it initially at point A so that it arrives at point B after a time T has elapsed?

Or to put this another way, how many orbits can we find that cross each other at our two chosen points after our chosen time?

We will idealise O as occupying a single point, and we will assume it is so much more massive than P that we can ignore any gravitational pull P exerts on O. We will abbreviate the constant that determines the acceleration due to gravity from O as:

k = G M

so that the acceleration due to gravity at a location r relative to O is:

g(r) = –k r / |r|3
|g(r)| = k / |r|2

We will write rA and rB for the distances from O to the points A and B.

### Returning to the same point

First, let’s get one special, atypical case out of the way. Suppose A and B are chosen to be exactly the same point. For a test particle moving in a bound, elliptical orbit, the period of the orbit, τ, the semi-major axis of the ellipse, a, and the total specific energy (kinetic plus potential energy, per unit mass), E, of the particle, are related by:[1]

τ = 2π √(a3 / k)
E = –k / (2a)

and when the test particle is at point A, if its speed is vA then its total specific energy is:

E = –k / rA + ½vA2

It follows that we can make the orbital period of the particle, τ, precisely equal to our chosen elapsed time, T, if and only if we are able to give the particle an initial speed vA of:

vA(T) = √[2 (k / rA – ½k / T))]

which corresponds to:

E(T) = – ½k / T)
a(T) = k (T / (2 π))
τ(T) = T

The direction in which the particle is moving as it leaves point A will make no difference, so there is [almost] a whole spherical shell of velocities that will all cause the particle to return to point A after a time T. The only constraint is on the speed, but note that the formula we have given for vA(T) will only be the square root of a non-negative quantity if:

k / rA ≥ ½k / T)
T ≥ π √[½ rA3 / k]
2 a(T) ≥ rA

The last inequality here says that the full length of an ellipse with orbital period T must be at least enough to stretch from the central mass all the way to point A.

When we said there is almost a whole spherical shell of velocities that work this way, the exceptions would be velocities that point directly towards or away from the central mass, which give orbits that degenerate from ellipses into straight lines. In fact, even if T is less than the minimum value needed to match it to the full period of an elliptical orbit passing through point A, there is still a way for the test particle to leave A and return after time T. We simply need to throw the particle directly away from the central mass, with the correct speed so that it rises and falls back, along a straight line, in the desired time.

Suppose the body ascends from rA to some maximum distance r0 where it is instantaneously at rest. For allowed values of r, by conservation of energy we have:

½v(r)2 = k (1/r – 1/r0)
dr/dt = v(r) = √[2k (1/r – 1/r0)]

Here we have chosen a positive sign for dr/dt, to describe the ascent of the particle. We can integrate this with a simple substitution:

r = r0 (cos α)2
dr/dt = √[2k ((sec α)2/r0 – 1/r0)] = √[2k / r0] tan α
dr/dα = –2r0 cos α sin α
dt/dα = dr/dα / (dr/dt) = –√[2 r03 / k ] (cos α)2

t(α) = –√[r03 / (2 k) ] (α + cos α sin α) + C

Putting ρ = rA/r0, we have:

α(r0) = 0
α(rA) = arcos √ρ

If we choose C=0 so that t=0 when r=r0, we have the time it takes to ascend from rA equal to:

t(α(rA)) = √[r03 / (2 k) ] (arcos √ρ + √[ρ (1 – ρ)])

We want twice this time to be equal to T:

T = √[2 r03 / k ] (arcos √ρ + √[ρ (1 – ρ)])

In general this cannot be solved explicitly for r0, but the solution can be found numerically, and then used in our original formula for v(r) to give the speed at rA.

In the image above, the red and blue curves trace the orbits of test particles moving under the gravitational acceleration due to a massive object (black dot) that all depart from the same point (black cross) in different directions with the same speed, and return to that point simultaneously. The green dot shows a particle that has been thrown directly upwards (at a slightly greater speed than the other particles) and falls back to the same location in the same time. [Note that when the animation repeats, the red and blue particles seem to continue along their trajectories, but the green particle is shown starting again rather than continuing to fall inwards, which of course is what it would do at later times.]

### Distinct points

We will now look at the more general situation, where our starting point A and finishing point B are distinct. And we will begin by asking a simpler question: what velocities can the test particle have at point A that allow point B to lie on its orbit at all, regardless of the travel time along that orbit between the two points?

Because the central mass, O, A and B are three distinct points, in general they will define a plane, and the motion of the test particles will be confined to that plane. [We will assume the three points are not colinear; if they are, the problem can be analysed as a 1-dimensional system, in much the same way as we did for the particle thrown straight up returning to its original launching point.] So we have a 2-dimensional space of initial velocities for the particle. Requiring the trajectory to pass through point B without saying when it does so just imposes a single constraint, so we would expect there to be a 1-dimensional subset of velocities, i.e. one or more curves in velocity space, which meet that condition.

Associated with every orbit are three conserved quantities: the total energy, the angular momentum, and the direction from the central mass to the periapsis (the point of closest approach). In a three-dimensional context the last two of these are vectors, but once we fix the plane of the orbit the angular momentum can be specified with a single number, its component perpendicular to the plane, and the direction to periapsis is specified by one angle. One vector that points from the central mass to the periapsis is the Laplace-Runge-Lenz vector:

A = m2 (v × Lk r/r)

where m is the mass of the body, and L is the angular momentum vector of the body, per unit mass. We don’t really care about the factor of m2, since we only need the direction of this vector to give us the direction from the central mass to the periapsis.

If we start with four variables, the x and y components of the test particle’s velocity at point A and at point B, and require the three conserved quantities to be the same at both points, we can eliminate the two velocities at point B, leaving us with a single equation in the two velocities vx, vy at point A:

k (rA rBxA xByA yB) – rA (vy xAvx yA) (vy (xAxB) + vx (yByA)) = 0

This is a quadratic in vx, vy, and in fact it turns out to be a hyperbola. We can express this in a more coordinate-independent manner, in terms of the angle δ from point A to point B as seen from the central object O, and the angle η between OA and the velocity at point A:

k rA rB (1 – cos δ) – rA2 vA2 sin η (rA sin η + rB sin(δ – η)) = 0

If we choose a coordinate system where point A lies on the positive x-axis, so xA = rA and yA = 0, the resulting equation has only linear terms in vx, so it has a single-valued solution for vx in terms of vy.

vx = [rA (xBrA) vy2 + k (rBxB)] / [rA yB vy]

The top panel in the image on the right shows this velocity hyperbola, for the case:

k = 1
rA = 1
(xB, yB) = (½, 1)

The dashed circle marks the limit of velocities for which the orbit is bound, i.e. the total energy is negative.

The bottom panel shows the orbit through point A and point B that corresponds to the currently marked point on the velocity hyperbola.

The velocity hyperbola has two asymptotes: the vx-axis, and the blue line vy = yB vx/(xBrA). As the velocity approaches the first asympote (i.e. as |vx| grows very large and vy approaches zero), we have orbits where the test particle moves rapidly in an almost straight line between point A and the central mass, where it takes a sharp corner around it and travels in another almost straight line to or from point B.

As the velocity approaches the second asymptote, we have orbits where the test particle moves very rapidly in an almost straight line directly between point A and point B, though of course its path is slightly curved into a hyperbola for all finite velocities.

Having determined which velocities at point A are required for an orbit to pass through point B at all, we will now examine the time it takes to travel along the resulting orbit from A to B. We can express both the location on the orbit, and the time since the particle was at the periapsis, in terms of a parameter, ξ, as follows.[2] For a bound, circular or elliptical orbit:

a = k / (–2 E)
e = √[1 + 2 E L2 / k2] < 1
r = a (1 – e cos ξ)
t = √[a3 / k] (ξ – e sin ξ)
x = a (cos ξ – e)
y = a √[1 – e2] sin ξ

For an unbound, hyperbolic orbit:

a = k / (2 E)
e = √[1 + 2 E L2 / k2] > 1
r = a (e cosh ξ – 1)
t = √[a3 / k] (e sinh ξ – ξ)
x = a (e – cosh ξ)
y = a √[e2 – 1] sinh ξ

For a zero-total-energy, parabolic orbit:

p = L2 / k
r = ½p (1 + ξ2)
t = ½ √[p3 / k] ξ (1 + ⅓ ξ2)
x = ½p (1 – ξ2)
y = p ξ

The formulas above apply to a coordinate system where the central mass is at the origin, the periapsis occurs on the positive part of the x-axis, and the angular momentum is positive, so the particle moves counterclockwise when we look down on the xy plane in its usual orientation. Of course, we want to be able to deal with a multitude of possible orbits, so we need to be able to transform them into appropriate coordinates to use these formulas. Identifying whether the angular momentum is positive or negative given the particle’s velocity at point A is easy; this just depends on the sign of vy. The Laplace-Runge-Lenz vector that we mentioned previously gives us the direction from the central mass to the periapsis.

The blue curves in the image on the left plot the time to travel along the orbit from A to B, for the same choice of points as in the previous image, as we vary vy. The light blue rectangles mark the range of velocities where the orbit is bound and cyclic, and here the curve is double-valued, because we plot both negative times [the time when the particle was last at point B] and positive times [the time when the particle will next be at point B], given that it is at point A at t=0. The green curves give the orbital period for the bound orbits.

The values of vy for which the corresponding kinetic energy exactly cancels the potential energy –k/rA, giving a total energy of zero, are:

vy = ±√[k rB (1 – cos δ) / (rA (rA + rB ± 2 √(rA rB) cos ½δ))]

(Here δ is the angle from point A to point B.) These are the boundaries of the light blue rectangles.

Of course we could add any integer multiple of the orbital period to the relevant sections of these curves and again obtain times when the particle was at point B. The image on the right includes some of the extra curves we obtain this way.

We can now see how the number of different ways to move along an orbit from A to B in a time of T will change as we start with T just above zero and increase it. Because we are doing Newtonian physics there is no upper limit on the velocity we can give our test particle, but we still rule out literally infinite velocities, so we don’t allow T=0.

• For small values of T, our only options will consist of two hyperbolic orbits, one of which is close to a straight line path from A directly to B, while the other bends very sharply around the central mass.
• As we increase T, each of these hyperbolas will eventually become parabolas, where the curve crosses into one of the light blue rectangles from the right in our plot. In our example, the values of T where this happens are quite close for the two cases, but they are not identical.
• The two parabolas will become two ellipses, and for some range of T there will be precisely two elliptical orbits that yield the required travel time.
• Eventually, T will start crossing the minima of the stacks of curves within the light blue rectangles; each time it equals one of these values, there will be a distinct new elliptical orbit we can use, where we complete one or more full orbits before arriving at B, and once it exceeds the minimum value, there will be two new elliptical orbits. Note that although the minima of the successive curves in the two different rectangles are quite close, they are not identical.

So there will always be at least two orbital paths from A to B in any given time T > 0, and as we increase T, the number of possible paths will keep increasing. The animation on the left shows these orbits for the same choice of A and B as we have used throughout this section.

To be clear, for the bound orbits with periods less than T, the test particle will also be passing through point B one or more times before finally arriving there at time T. Our goal is not to rule out any earlier visits, but to identify all the choices that allow the particle to be at the chosen point at the specified time.

## References

[1] Mechanics by L.D. Landau and E.M. Lifshitz, Butterworth-Heinemann, 1976. Equations (15.6), (15.8).

[2] Landau and Lifshitz, op. cit. Equations (15.4), (15.6), (15.10), (15.11), (15.12).

Science Notes / Double-Crossing Orbits / created Monday, 20 February 2023
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