I believe that 3 points is the minimum to determine an orbit. (3 observations * 2 coordinates each = 6 = number of orbital parameters.). So Find_Orb plots an orbit exactly through all 3 points, and RMS would be 0 because there is no difference between the original points and the resulting curve.

Andy

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Andy Puckett, PhD puck@...
Director - Planetarium & Visualization Theater
Asst Professor of Astronomy 907-786-1838
Univ. of Alaska Anchorage CPSB 202P
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http://www.uaa.alaska.edu/planetarium/shows
https://twitter.com/UAAplanetarium
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On Jan 29, 2011, at 12:45 AM, "Alessandro" <alessandro_odasso@...> wrote:

> Here is a newby question for people who are interested in orbit determination.
>
> Let's imagine that an object MYOBJ has been measured only three times, according to this hypothetical MPCReport:
>
> MYOBJ C2011 01 11.43745 11 05 20.01 +38 02 33.7 17.00R 857
> MYOBJ C2011 01 11.47912 11 04 30.30 +38 05 12.4 16.58R 857
> MYOBJ C2011 01 11.52055 11 03 40.79 +38 07 47.7 17.93R 857
>
> Based on these three observations, what would you say about the orbit of such an object ?
>
> I tried to use the program FIND_ORB on this example but I am not yet able to use it understanding what it finds out.
> For example: when I press the autosolve button, the program determines an orbit with RMS error = 0 ?
>
> Here is what I find:
> MYOBJ
> Perihelion 2010 Nov 18.173749 TT = 4:10:12 (JD 2455518.673749)
> Epoch 2011 Jan 12.0 TT = JDT 2455573.5 Earth MOID: 0.2550
> M 72.84720 (2000.0) P Q
> n 1.32869204 Peri. 9.25958 0.59463100 0.39030866
> a 0.81944503 Node 301.60625 -0.78034732 0.06968394
> e 0.8931816 Incl. 55.62134 -0.19357690 0.91804318
> P 0.74/270.94d H 18.7 G 0.15 q 0.08753180 Q 1.55135825
> From 3 observations 2011 Jan. 11 (2.0 hr); RMS error 0.000 arcseconds
>
> Thanks,
> Alessandro
>
>


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