6
$\begingroup$

Using a 3D Cartesian coordinate system if we know for certain that a craft will pass through two points in it's orbit (let's call them p1 and p2) and we also know the mass of the central body it is orbiting what parameters (such as Keplerian orbital elements) can we know about an orbit if at all any?

$\endgroup$
10
$\begingroup$

If you only know two points on the orbit, and the mass and position of the central body, all you can determine for certain is its orbital plane.

An example of two very different orbits that pass through a chosen $P_1$ and $P_2$
An example of two very different orbits that share two points.

More geometrically: Given a focus $F_0$ and any two points $P_1$ and $P_2$, you can define a hyperbola with foci at $P_1$ and $P_2$ that passes through $F_0$

All points on that hyperbola are valid locations for a second focus, $F_1$ of an orbit around $F_0$ that passes through $P_1$ and $P_2$. If $F_1$ is on the same lobe as $F_0$, it's a hyperbolic trajectory. If $F_1$ is on the other lobe, it's an ellipse.

The set of orbits and trajectories around $F_0$ that pass through a chosen $P_1$ and $P_2$
A set of orbits and trajectories that pass through a pair of chosen points

Ellipses are in green, Hyperbolas in Blue.

GeoGebra Graph Here: $P_1$ and $P_2$ are freely dragable, $F_1$ is draggable to anywhere along the focal hyperbola.

For an answer on why this works, see Ellipses given focus and two points on Mathematics Stack Exchange.

As a result, you can't determine inclination, or longitude of the ascending node, because you don't know which of those points the body visits first, or the travel time between them, so you can't tell which of the directions along the line of intersection between the orbital plane and the reference plane is towards the ascending node, but of the Keplerian orbital parameters, you have two choices for the values on those.

You don't have enough information to determine orbital eccentricity, argument of periapsis, semimajor axis, or eccentricity, or mean anomaly at epoch.

Without more information, a wide variety of orbits meet the "passes through these two points" criteria.

$\endgroup$
12
  • 1
    $\begingroup$ @Sam No, not really. There exist prograde and retrograde orbits that pass through P1 first, then P2. $\endgroup$
    – notovny
    Oct 9 at 13:27
  • 2
    $\begingroup$ Trivially, it also gives a lower limit to semi-major axis, and a minimum eccentricity. Semi-major axis and eccentricity are also no longer completely independent, though they still have a shared degree of freedom. $\endgroup$ Oct 9 at 13:36
  • 3
    $\begingroup$ @Sam Famously, determining a conic section requires 5 points, though the location of the central body counts as one of them. You thus need two more. $\endgroup$ Oct 9 at 13:41
  • 1
    $\begingroup$ @SE-stopfiringthegoodguys I think you might be able to staple down eccentricity, semi-major axis, and true anomalies with central body position and three points on the orbit, if you know the order they're each next passed in; at that point, using the polar equation $r=\frac{a(1-e^2)}{1+e \cos \theta}$ you have three equations and three unknowns. The radial distances are given, and the true anomalies of the three points can be written in terms of each other. $\endgroup$
    – notovny
    Oct 9 at 14:08
  • 3
    $\begingroup$ @notovny I think this deserves a separate question: space.stackexchange.com/questions/55295/… $\endgroup$ Oct 9 at 14:41
4
$\begingroup$

If two position vectors and the time between them are known, a conic section orbit may be determined using Lambert's method.

If the position and velocity relative to the observer are available (as is the case with radar observations), these observational data can be adjusted by the known position and velocity of the observer relative to the attracting body at the times of observation. This yields the position and velocity with respect to the attracting body. If two such observations are available, along with the time difference between them, the orbit can be determined using Lambert's method, invented in the 18th century. See Lambert's problem for details.

Source: https://en.wikipedia.org/wiki/Orbit_determination

In celestial mechanics, Lambert's problem is concerned with the determination of an orbit from two position vectors and the time of flight, posed in the 18th century by Johann Heinrich Lambert and formally solved with mathematical proof by Joseph-Louis Lagrange. It has important applications in the areas of rendezvous, targeting, guidance, and preliminary orbit determination.

The precise formulation of Lambert's problem is as follows: Two different times t1, t2 and two position vectors r1 ,r2 are given.

Source: https://en.wikipedia.org/wiki/Lambert%27s_problem

The two points on the orbit are needed as three dimensional cartesian coordinates. If the two points are given in polar coordinates, they may be transformed to cartesian coordinates.

$\endgroup$
6
  • 2
    $\begingroup$ You say position vectors, but doesn't the quoted except say "and velocity"? $\endgroup$ Oct 9 at 15:27
  • $\begingroup$ @SE-stopfiringthegoodguys Velocity means here the time between the two position vectors should be given too. $\endgroup$
    – Uwe
    Oct 9 at 15:41
  • 1
    $\begingroup$ That's an unusual meaning of "velocity", so I'm not convinced. $\endgroup$ Oct 9 at 15:43
  • 1
    $\begingroup$ @SE-stopfiringthegoodguys you can use either a single position and the instantaneous velocity at that point, or two positions and the time difference between them – which in the limit of Δt → 0 becomes the same thing. $\endgroup$ Oct 9 at 22:03
  • 3
    $\begingroup$ For the time between two points to uniquely determine velocity at either / both points, as @SE-stopfiringthegoodguys was asking about, it would need to be true that there aren't multiple equal-time paths between those points (with gravity from a central body as the only acceleration). That seems very believable, but it isn't so obvious that it should go without saying. Fortunately the answer is clear about Lambert's Problem being solved with two times and two positions. (If you just knew the magnitude of the time diff, there'd be 2 solutions, same orbit in different directions.) $\endgroup$ Oct 10 at 2:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.