
Math & Physics Mathematical and physical problems of space flight and astronomy. 

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06022019, 01:35 AM  #1 
Orbinaut

Can I get the Vector components xyz given only this information?
hello,
I have these 3 dates and I know the position X Y Z of moon Amalthea relative to Jupiter and then I know the Velocity Vectors V for each one. I need to get the VX VY VZ components for each of the 3 datetimes. Do I have enough information to get those from what you see? I believe I need the instantaneous values. I am aware of VX = dx/dt, VY = dy/dt, VZ = dz/dt and I know VX + VY + VZ = V I am interested in the math behind finding those components given what I have. If you know the procedure can you tell me what it is? or shed some light on how I might solve the problem if its possible? thank you. 1979 MAR 05 00:00:00 178932.619 km 28063.045 km 17448.755 km V = 29.804114 km/sec 1979 MAR 05 00:02:00 178337.419 km 30878.143 km 18771.071 km V = 29.231286 km/sec 1979 MAR 05 00:04:00 177688.033 km 33683.859 km 20087.682 km V = 28.649604 km/sec thank you. I do have the answers if someone knew how to do it. I can check to see if your right. Last edited by ncc1701d; 06022019 at 01:45 AM. 
06022019, 01:48 AM  #2 
Passed the Turing Test

If I understand correctly, you have XYZ position information and a speed, and you want XYZ velocity information.
This can get really complicated if you care about a perfect answer. However I would say simply knowing that the orbit is nearly circular and nearly equatorial is enough information to have a pretty good idea of the velocity vector. You need to know what coordinate frame those XYZ values are in though. Edit: I'm thinking you may actually care about a more accurate answer given your three times are less than an orbit apart. I suspect you can in fact evaluate those three points together to get some orbital parameters out. Last edited by Quick_Nick; 06022019 at 01:55 AM. 
06032019, 01:16 PM  #3 
Orbinaut

I think this article describes the required math. It's not easy. https://science.larouchepac.com/gaus...erProblem.html

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06062019, 02:32 PM  #4 
thing

Hi,
thanks for the interesting problem to think about :) I can get the semimajor axis (a) and the standard gravitational parameter (mu) from the visviva equation; v^2 = mu*(2/r  1/a) But then I get stuck! Then I thought maybe it was an energyconservation problem (Potential + Kinetic = Constant) but I'm not getting anywhere with that. Since the times of the data points are given, I'm pretty sure that is necessary information. Needs some kind of insight regarding Kepler's 2nd law (radius sweeps out equal areas in equal times) maybe? I think that the shape of an ellipse is completely determined by three points (pos vectors), so theoretically you should be able to find the equation of the ellipse and the tangent at a given point  but the math for doing that in 3D is looking really gnarly, so I'm not even thinking about that :) Anyway, I'm having fun playing with this one. Cheers, Brian P.S. Your first two Velocity values are negative  typo? 
06062019, 06:08 PM  #5 
Orbinaut

the checkmarked answer I found here might provide the answer.
https://stackoverflow.com/questions/...elocityvector Although not sure its that simple yet. I am still researching. 
07122019, 02:35 PM  #6 
Orbinaut

Quote:
Let's suppose, as above, that you are given two (3vector) points on an Keplerian orbit and and that you know the (scalar) orbital speeds at those points, and respectively. Now calculate: In addition, calculate the semimajor axis at either the first or last point using: or Then (and here's the magic!), solve for the two roots of the quadratic equation: Call these roots and . These two numbers encode information about the two Keplerian arcs that join the points and with semimajor axis . Then, the radial and transverse components of the velocity vector at the are given by: and the radial and transverse components of the velocity vector at the are given by: As mentioned, this gives two possible solutions  corresponding to the two possible Keplerian arcs that pass through the initial and final points with setmajoraxis . For each solution, we calculate the eccentricity in the usual way. In practice, for problems involving nearly circular orbits, one value of will give a high eccentricity; and the other will give one close to zero. Obviously, we choose the solutions corresponding to the low orbital eccentricity  thereby removing the degeneracy. (N.B., I have a proof of the above but it's a little longwinded  too long for this post. But I can share if anyone is sufficiently interested)  Post added at 02:35 PM  Previous post was at 01:23 PM  Let's take a simple worked example: Suppose that we choose units such that with: Find the velocity vectors at and . Using the above algorithm we calculate: We next calculate the semimajor axis using or and in both cases we calculate the same semimajor axis . (You may find this value a little suspicious  but it's just a result of me setting up a test problem with simple parameters.) Next, we solve the quadratic: The two roots of this are and . With , the radial and transverse components at are: and the radial and transverse components at are: And with , the radial and transverse components at are: and the radial and transverse components at are: Based on this, if we go through the straightforward exercise of calculating the eccentricity with we find that the orbital eccentricity is 0.9796683; and with we find that the orbital eccentricity is 0.010000. Since I know that the orbit is nearly circular (i.e., so that the eccentricity is close to zero), the appropriate solution to use is the one corresponding to . So, finally, the radial and transverse components of the velocity vector at are ; and the radial and transverse components of the velocity vector at are . Last edited by MontBlanc2012; 07122019 at 02:10 PM. 
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07122019, 08:49 PM  #7 
thing

Hi,
wow! Thanks so much, I would never have got those later steps. Magic indeed 
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07132019, 03:38 AM  #8 
Orbinaut

Actually, it occurs to me that the solution to ncc1701d's original problem is actually quite trivial  given that we are given the time of flight on each of the arc segments (120 seconds each). This means that the information about speeds is superfluous and one just needs to use a standard Lambert Solver to solve the problem.
However, we can also use the maths set out above to do the same job of solving the Lambert Problem,  so, in outline, here's the solution: In SI units, the first and second Jupitercentric positions of Amalthea are: We also know that for Jupiter, the accepted value (in SI units) of the gravitational parameter is . And we know the time of flight is 120 seconds. As before calculate: Now set and as functions of an as yet unknown semimajor axis : where and are the two roots of the equation . As it turns out, for elliptical orbits, the transfer time from the initial to final points can be calculated as: So, now we have an expression that is a function of only. And since we know that the orbit of Amalthea is almost circular, we use a standard Newton rootfinding algorithm with an initial guess of 180,000 km. From this, we deduce that: and hence and Then, the radial and transverse components of the velocity vector at the are given by: and the radial and transverse components of the velocity vector at the are given by: And from this it is straightforward to convert back to the xyz coordinates of the original coordinate system. N.B. In case you were wondering what the orbital eccentricity is, we get this from the expression: i.e., very nearly a circular orbit. Yay!  Post added at 03:38 AM  Previous post was at 02:15 AM  And finally, just a quick comment on ncc1701d's summation of velocity vector components (V = VX + VY + VZ). Initially, I thought this was a typo and ncc1701d really meant V^2 = VX^2 + VY^2 + VZ^2  which is, by far and away the more conventional way of 'adding' velocity components. As it turns, out, I was wrong: ncc1701d actually did mean the straight arithmetic sum of the velocity components. How do I know this? Well, I can calculate the xyz representation of the radial and transverse unit vectors at that point. Although I'm not going to bother showing the intermediate calculations, in xyz coordinates the radial unit vector is: and the transverse unit vector is: So, the velocity vector at the first point in ncc1701d's list can be written as: So, at that point, VX = 4734.483; VY = 23495.74; and VZ = 11041.87. And the straight arithmetic sum of those is 29803.12 which is suspiciously close to the value given by ncc1701d of 29804.11. Last edited by MontBlanc2012; 07172019 at 03:43 AM. 
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07142019, 03:26 PM  #9 
thing

Thanks for the further insight.
I was on the wrong track to start with, since I had assumed the OP actually meant V^2 = VX^2 + VY^2 + VZ^2. Note to self: always read the question carefully! Cheers, Brian 
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07162019, 07:35 AM  #10 
Orbinaut

thank you everyone for all the feedback!


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