View Issue Details ID: Category: Severity: Reproducibility: Date Submitted: Last Update: 2 [Memorandum Series] Memorandum #1: Keplerian Orbit Elements to Cartesian State Vectors feature have not tried 2013-04-30 10:35 2013-05-01 23:42 Reporter: anonymous Platform: Assigned To: DigNative OS: Priority: normal OS Version: Status: closed Product Version: Product Build: Resolution: won't fix Projection: none ETA: none Fixed in Version: Target Version: Summary: Memorandum 1, Algorithm step 1, version: 2013/01/30 Description: Hello, It's Frederic again, Ithe first step, setting or calculating M(t), the period 86400 s corresponds to the Earth, but a is not given ... In the inputs, as you mentions various possible values for MG = mu, may you give gemeric formulas and set of values (sun, earth, period for a satellite ...) Best Regards Frederic Baudrand fbaudrand@free.fr Tags: Steps To Reproduce: Additional Information: Attached Files:
Notes
(0000002)
DigNative
2013-05-01 20:41
(Last edited: 2013-05-01 23:42)

Hi Frederic,

thank you for your enhancement request for Memorandum No. 1. As far as I understand there are two points into your request, which I may answer separately:

#### Conversion of Days to Seconds

The factor $$86\,400$$ in equation 1 for the conversion of Julian days to seconds does not belong to Earth's rotation. This factor is the astronomical unit of time as defined by the IAU (1976) System of Astronomical Constants (1, p. 696). Because neither the initial epoch $$t_0$$ nor the actual computing epoch $$t$$ belong to any celestial body, this factor is independent.

#### Values for the Standard Gravitational Parameter $$\mu = GM$$

As mentioned in the input section of the algorithm, the Sun is assumed to be the central body for the conversion of Keplerian elements to state vectors. However, if someone want to use another central body, e.g. for calculating the state vectors of a satellite around Earth, he has to supply a value for the standard gravitational parameter of this central body within the desired uncertainty bounds.

Because the values for this standard gravitational parameter are different for each celestial body and --- in general --- this value can not be determined with high certainty, I will just give the current best estimate as published by the IAU Division I Working Group on Numerical Standards for Fundamental Astronomy (2). It is within the responsibility of the user to provide a reliable value for $$\mu$$.

The second part of your point is more general on the nature of $$\mu$$. The standard gravitational parameter $$\mu$$ is used, because the value of $$\mu$$ is known to greater accuracy than the Newtonian constant of gravitation $$G$$ or the mass of the celestial body $$M$$ due to the measurement principle. Because the value of $$\mu$$ is based on observations, I can't give a more general formula for this.

#### References

1. Seidelmann, P. Kenneth (ed.): Explanatory Supplement to the Astronomical Almanac. First paperback impression. University Science Books, Sausalito, California, USA, 2006. ISBN 978-1-891389-45-0.
2. IAU Division I Working Group on Numerical Standards for Fundamental Astronomy: Astronomical Constants: Current Best Estimates (CBEs). Online available at http://maia.usno.navy.mil/NSFA/NSFA_cbe.html#GMS2012. Retrieved 2013/05/01.

#### Closing Remarks

I hope I could answer your questions/enhancement requests. As I think this was mainly a misunderstanding, I have drawn the conclusion to slightly alter my memorandum to make the discussed things more ostensible. However, since my memoranda are meant to be wrap-ups, not extensive discussions of the related topic, I won't add more explanations to it.

Kind regards,

René

 View Issue Details ID: Category: Severity: Reproducibility: Date Submitted: Last Update: 1 [Memorandum Series] Memorandum #1: Keplerian Orbit Elements to Cartesian State Vectors minor have not tried 2013-04-30 10:11 2013-04-30 21:28 Reporter: anonymous Platform: Assigned To: DigNative OS: Priority: normal OS Version: Status: closed Product Version: Product Build: Resolution: fixed Projection: none ETA: none Fixed in Version: Target Version: Summary: Step 6 - Error in the transformation formula 8 and 9 Description: Hi, As I was looking for Keplerian orbit elements, I found your memo. Great work, with nice présentation, références. I thinks there is an error in the formulas 8 and 9, where o stands for Omega, the longitude of the ascending node Actual Formula is :  x*(cos(o)*cos(w) - cos(i)*sin(o)*sin(w)) - y*(cos(o)*sin(w) + cos(i)*cos(w)*sin(o)) x*(cos(w)*sin(o) - cos(i)*cos(o)*sin(w)) - y*(sin(o)*sin(w) - cos(i)*cos(o)*cos(w)) x*sin(i)*sin(w) + y*cos(w)*sin(i) The correct formula is :  x*(cos(o)*cos(w) - cos(i)*sin(o)*sin(w)) - y*(cos(o)*sin(w) + cos(i)*cos(w)*sin(o)) x*(cos(w)*sin(o) + cos(i)*cos(o)*sin(w)) - y*(sin(o)*sin(w) - cos(i)*cos(o)*cos(w)) x*sin(i)*sin(w) + y*cos(w)*sin(i) Best Regards Frederic Baudrand fbaudrand@free.fr Tags: Steps To Reproduce: Additional Information: Attached Files:
 Notes (0000001) DigNative    2013-04-30 21:28 Hi Frederic, thank you very much for your bug report. Of course you are right --- there was a minus sign, were a plus sign had to be. The typo has now been corrected; you can find the corrected version in my download center. I appreciate your efforts. Kind regards, René