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=GM

so that the acceleration due to gravity at a location ** r** relative to

(g) = –rk/ |r|r^{3}

|(g)| =rk/ ||r^{2}

We will write *r*_{A} and *r*_{B} for the distances from *O* to the points *A* and *B*.

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 √(a^{3}/k)

E= –k/ (2a)

and when the test particle is at point *A*, if its speed is *v*_{A} then its total specific energy is:

E= –k/r_{A}+ ½v_{A}^{2}

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 *v*_{A} of:

v_{A}(T) = √[2 (k/r_{A}– ½^{⅓}(πk/T)^{⅔})]

which corresponds to:

E(T) = – ½^{⅓}(πk/T)^{⅔}

a(T) =k^{⅓}(T/ (2 π))^{⅔}

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 *v*_{A}(*T*) will only be the square root of a **non-negative quantity** if:

k/r_{A}≥ ½^{⅓}(πk/T)^{⅔}

T≥ π √[½r_{A}^{3}/k]

2a(T) ≥r_{A}

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 *r*_{A} to some maximum distance *r*_{0} where it is instantaneously at rest. For allowed values of *r*, by conservation of energy we have:

½v(r)^{2}=k(1/r– 1/r_{0})

dr/dt=v(r) = √[2k(1/r– 1/r_{0})]

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

r=r_{0}(cos α)^{2}

dr/dt= √[2k((sec α)^{2}/r_{0}– 1/r_{0})] = √[2k/r_{0}] tan α

dr/dα = –2r_{0}cos α sin α

dt/dα = dr/dα / (dr/dt) = –√[2r_{0}^{3}/k] (cos α)^{2}

t(α) = –√[r_{0}^{3}/ (2k) ] (α + cos α sin α) +C

Putting ρ = *r*_{A}/*r*_{0}, we have:

α(r_{0}) = 0

α(r_{A}) = arcos √ρ

If we choose *C*=0 so that *t*=0 when *r*=*r*_{0}, we have the time it takes to ascend from *r*_{A} equal to:

–t(α(r_{A})) = √[r_{0}^{3}/ (2k) ] (arcos √ρ + √[ρ (1 – ρ)])

We want twice this time to be equal to *T*:

T= √[2r_{0}^{3}/k] (arcos √ρ + √[ρ (1 – ρ)])

In general this cannot be solved explicitly for *r*_{0}, but the solution can be found numerically, and then used
in our original formula for *v*(*r*) to give the speed at *r*_{A}.

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.]

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:

=Am^{2}(×v–Lk/rr)

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

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 *v*_{x}, *v*_{y} at point *A*:

k(r_{A}r_{B}–x_{A}x_{B}–y_{A}y_{B}) –r_{A}(v_{y}x_{A}–v_{x}y_{A}) (v_{y}(x_{A}–x_{B}) +v_{x}(y_{B}–y_{A})) = 0

This is a quadratic in *v*_{x}, *v*_{y}, 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*:

kr_{A}r_{B}(1 – cos δ) –r_{A}^{2}v_{A}^{2}sin η (r_{A}sin η +r_{B}sin(δ – η)) = 0

If we choose a coordinate system where point *A* lies on the
positive *x*-axis, so *x*_{A} = *r*_{A} and *y*_{A} = 0, the resulting equation has only linear terms in *v*_{x}, so
it has a single-valued solution for *v*_{x} in terms of *v*_{y}.

v_{x}= [r_{A}(x_{B}–r_{A})v_{y}^{2}+k(r_{B}–x_{B})] / [r_{A}y_{B}v_{y}]

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

k= 1

r_{A}= 1

(x_{B},y_{B}) = (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 *v*_{x}-axis, and the blue line
*v*_{y} = *y*_{B} *v*_{x}/(*x*_{B} – *r*_{A}). As the velocity approaches the first asympote (i.e. as |*v*_{x}| grows very large and *v*_{y} 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/ (–2E)

e= √[1 + 2EL^{2}/k^{2}] < 1

r=a(1 –ecos ξ)

t= √[a^{3}/k] (ξ –esin ξ)

x=a(cos ξ –e)

y=a√[1 –e^{2}] sin ξ

For an unbound, hyperbolic orbit:

a=k/ (2E)

e= √[1 + 2EL^{2}/k^{2}] > 1

r=a(ecosh ξ – 1)

t= √[a^{3}/k] (esinh ξ – ξ)

x=a(e– cosh ξ)

y=a√[e^{2}– 1] sinh ξ

For a zero-total-energy, parabolic orbit:

p=L^{2}/k

r= ½p(1 + ξ^{2})

t= ½ √[p^{3}/k] ξ (1 + ⅓ ξ^{2})

x= ½p(1 – ξ^{2})

y=pξ

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 *x**y* 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
*v*_{y}. 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 *v*_{y}.
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 *v*_{y} for which the corresponding kinetic energy exactly cancels
the potential energy –*k*/*r*_{A}, giving a total energy of zero,
are:

v_{y}= ±√[kr_{B}(1 – cos δ) / (r_{A}(r_{A}+r_{B}± 2 √(r_{A}r_{B}) 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.

[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).