Is eccentricity a vector?

[Rareden]

Greetings,
Messing around making a space sim game, but in the orbital calculations, atm im a bit stuck with if eccentricity is actually a vector depicted in a 1x3 matrix with I J K
But ive often seen it refereed to as a scalar/magnitude.
Im using the formula here 

Edited October 18, 2017 by Rareden

[Zeiss Ikon]

I'm used to seeing eccentricity as a scalar, calculated from scalars (scalar lengths of semi-major and semi-minor axes, which can be measured directly).  I'm no mathematician, but you're probably getting something akin to the vector that defines the semi-major axis in 3-space, with the scalar eccentricity mixed in.

[p1t1o]

I dont think its a vector since it can be expressed with a single figure, eg: "-26.1degs"

 

[Rareden]

17 minutes ago, Zeiss Ikon said:

I'm used to seeing eccentricity as a scalar, calculated from scalars (scalar lengths of semi-major and semi-minor axes, which can be measured directly).  I'm no mathematician, but you're probably getting something akin to the vector that defines the semi-major axis in 3-space, with the scalar eccentricity mixed in.

indeed, problem is i have to calculate those variables along with the 6 standard orbital elements from only position and velocity vectors

16 minutes ago, p1t1o said:

I dont think its a vector since it can be expressed with a single figure, eg: "-26.1degs"

 

Is what i suspected, have just been getting greatly confused by it being refereed to as a vector. or that could be the magnitude of said vector

Edited October 18, 2017 by Rareden

[Zeiss Ikon]

If you have a single well-defined position and velocity, and know the value and radius of the local gravitational field, you've got everything you need to calculate an orbit -- including finding the orbital elements.  Astronomers usually need three observations to define an orbit, because they can't directly measure velocity, and get position only as a direction from the point of observation.

[Rareden]

4 minutes ago, Zeiss Ikon said:

If you have a single well-defined position and velocity, and know the value and radius of the local gravitational field, you've got everything you need to calculate an orbit -- including finding the orbital elements.  Astronomers usually need three observations to define an orbit, because they can't directly measure velocity, and get position only as a direction from the point of observation.

I can get absolute position in XYZ relative to planet center as well as players current velocity, strength of gravity at their distance.
got the equations in C++, just trying to work out the issues such as my periapsis being a negative value that wont go higher than 15km, and my Apoapsis continuing to climb despite ceasing acceleration

[YNM]

It can be a scalar or a vector depends on how you're describing your orbit.

Usually, there are 6 orbital elements :

- Semi-major axis (a) : the "size" of your orbit.

- Eccentricity (e) : how squished the orbit is.

- Inclination (i) : how tilted your orbital plane is wrt reference plane.

- Longitude / Right Ascencion of Ascending Node (Ω) : Where your ascending node (point where you move from "below" reference plane to "above") is wrt a reference point

- Argument of Periapse (ω) : where your periapse is wrt. the ascending node, along your orbital plane; alternatively Longitude of Periapse (Ω+ω)

- True anomaly (ν) : Where you are wrt the periapse along the orbital plane. There are also eccentric anomaly ("idealized" to a circle) and mean anomaly (% time of orbit).

In this case, everything is a scalar, though you could argue it's a spherical coordinate as well.

 

In your case, I think the eccentricity vector is simply where direction of the periapsis is from the system center. It combines eccentricity (by how long the vector is) with inclination, LAN/RAAN and Arg. of Periapse (by the direction it's pointing in) into one entity. I guess it eases the drawing for the orbit, as you can extract where the orbital plane is from both the radial and velocity vector (or radial and eccentricity or velocity and eccentricity), and then you can make where the Pe is (multiply with sma), and where everything else is in the cartesian/rectangular system.

Edited October 18, 2017 by YNM

[Gilph]

I would vote no, as it is part of the patched conic branch of mathematics. All of the terms are scalars and not vectors. It is also the same regardless of orientation, which would not be the case if it were vector based (I think).

[p1t1o]

13 minutes ago, Gilph said:

I would vote no, as it is part of the patched conic branch of mathematics. All of the terms are scalars and not vectors. It is also the same regardless of orientation, which would not be the case if it were vector based (I think).

Agreed. You are not describing a spacecraft but a complete orbit. A single numerical figure describe the orbits eccentricity in its entirety, it doesnt matter that the longitude of the periapsis could be any value, the value describing eccentricity does not change. Two orbits identical except they are each retrograde to the other, both have the same eccentricity too. You can even have completely different orbits with exactly the same eccentricity, described by the same single figure.

Theres no reason it should be a vector, unless you package extra information into the statement, like "eccentricity in the direction of..."

[PB666]

25 minutes ago, Gilph said:

I would vote no, as it is part of the patched conic branch of mathematics. All of the terms are scalars and not vectors. It is also the same regardless of orientation, which would not be the case if it were vector based (I think).

eccentricity is a geometric parameter that defines the relative shape of an ellipse.

If you have a point mass in which the system center and point mass are virtually indistinguishable then any 3 coordinate 3-vector velocity produces a single eccentricity. (In a 2-body system three dimensions can be reduced to 2 dimensions). Noting that for an point mass moving around another point mass in space, the relative states all points and velocity reference frames are valid, we can give the central point mass (CPM) a coordinate of 0,0,0 and a velocity of 0,0,0 and that at any moment we can give an object a two dimensional position of X = 0 and Y (r) = distance from CPM, a radial velocity of dr/dt, a CPM relative speed and dX/dt which is the (Speed² - (dr/dt)²)^0.5. For any motion of movement the triangle of the distance traveled as one side and the CPM is exactly the same. Roughly this is 0.5 * Y * dX/dt* t. From this you can calculate a (semi-major axis).   Also you can derive moment of Radians/sec.     dr²/dt = (radians/sec)²r- Mu/r^2. r1 = r0 +dr/dt + dr2/dt. Since total energy has to remain the same no matter where the object is and thermodynamic energy is determined the integral of _{r to inf}rMu, then dX²/dt can be calculated from SQRT(KE₀-dPE) = speed and new dX/dt is derived as above.

IOW non-inertial objects orbit with no regard to eccentrity, theoretically if the warping of spacetime about an object did not exhibit perfect radial symmetry, object orbits would not be elliptical and eccentricity could not define position. This is in-fact reality, as objects move around earth the density and elevation of the earth shifts slightly, consequently the orbits are not elliptical.

If you were to look at this from a more macro - quantum perspective you would say that the eccentricity of an orbit falls within a confidence interval in which all positions relative to the average positions at whatever sigma you like. This might have some pertinence for objects orbiting a black hole. Another example, the two neutron stars that merge both spinning rapidly themselves with each very powerful magnetic fields. Objects in that environments are not expected to exhibit true keplarian motion even when 3-body solutions are given.

 

To simplify, eccentricity is an approximation of a vector quantity just like centripetal force and gravity it is a faux parameters. Eccentricity is used because many orbital calculations begin with:

given:

a central point mass
assume it is infinitely greater than an orbit object.

When you are given these two the assumption is that space-time exhibits perfect radial symmetry, as a consequence mu is perfectly dependent or radius.

 

Edited October 18, 2017 by PB666

[cfds]

This vector looks like the Laplace–Runge–Lenz vector.

This vector points from the "centre" to the periaps and is, for 1/r potentials, a preserved quantity. This is the reason why there is no periaps movement for exact solutions of Kepler's problem and also why the energy levels in the Hydrogen atom do only depend on quantum number n but not l.

[Rareden]

switched to this equation, 
however the problem im encountering with this is that once the orbit starts to circularize and AP Peri switch, e becomes a negative value which kills the equation since you cant square root a negative.

Edited October 20, 2017 by Rareden

[vitekc45c]

Try using absolute value of e.

[YNM]

14 hours ago, Rareden said:

switched to this equation, 
however the problem im encountering with this is that once the orbit starts to circularize and AP Peri switch, e becomes a negative value which kills the equation since you cant square root a negative.

wait... whats ε ?

 

IMHO you should use equations that eases the work in the coordinate you're using. The first one was OK for cartesian I think.

Edited October 20, 2017 by YNM

[Bill Phil]

I just asked a trajectory team manager who works for NASA MSFC, the answer is yes, eccentricity is a vector.

[PB666]

4 hours ago, Bill Phil said:

I just asked a trajectory team manager who works for NASA MSFC, the answer is yes, eccentricity is a vector.

And they always carry some extra fuel to make corrective burns. lol. Anyway its the n-body problem, you cannot have a perfect calculation so why not just get a good enough calculation and correct at some later point.

 

[Pand5461]

On 20.10.2017 at 4:10 AM, Rareden said:

switched to this equation, 
however the problem im encountering with this is that once the orbit starts to circularize and AP Peri switch, e becomes a negative value which kills the equation since you cant square root a negative.

That's strange. You cannot ever get a negative under the root with physically correct values. I agree with @YNM, you need to stick to the OP formula and use the magnitude of that vector as scalar eccentricity (and direction if you need to find the SMA orientation).

See this stackexchange post to see how to get a full set of orbital elements from the cartesian representation.

[YNM]

4 hours ago, Pand5461 said:

That's strange. You cannot ever get a negative under the root with physically correct values.

I suppose it has to do with the "flipping". Which is why I ask what ε is. The rest shouldn't be a problem.

[Pand5461]

6 hours ago, YNM said:

I suppose it has to do with the "flipping". Which is why I ask what ε is. The rest shouldn't be a problem.

Right, ε is the only term that can change the sign. But the specific energy is not supposed to do that near circularization.

@Rareden are you sure you use ε= v²/2 - μ/r in your calculations?

[Rareden]

On 10/22/2017 at 6:27 AM, Pand5461 said:

Right, ε is the only term that can change the sign. But the specific energy is not supposed to do that near circularization.

@Rareden are you sure you use ε= v²/2 - μ/r in your calculations?

one i was using for mechanical energy is - mu / 2 * SMA

edit: went back to OP equation, made some changes and it works properly now TY all

Edited October 25, 2017 by Rareden