<p>In principle, Laplace's method (and Gauss's method) should be very accurate for determining orbits.</p>
<p>Of course, &quot;garbage in, garbage out&quot; applies, and if the observations were inaccurate, the results may also be inaccurate. However, you should not get great discrepancies. I would consider careful checking of your calculation, and perhaps comparing your results with those of an independent orbital determiner, to check for a mistake in the maths.</p>

CC BY-SA 3.0
2017-02-13T05:18:33.710Z
To double check, it's [this Laplace method](http://adsabs.harvard.edu/full/1991SvA....35..428K) and not [this Laplace method](https://en.wikipedia.org/wiki/Laplace_transform)?
uhoh
CC BY-SA 4.0
2017-01-13T22:30:43.590Z
2022-01-10T04:12:29.477Z
2022-01-10T04:12:29.477Z
James K
<p>Interesting project! Perhaps first determine convergence criteria for alternative methods being considered and if possible acquire more widely distributed observations of target's position. Then to the extent appropriate to examine how least squares estimates for various parameters of the selected method correlate (or not) with results obtained by applying that method to available observations of the orbiting object under study, one may decide to use consequent approximations of orbital elements as basis for examining differences in sensitivity to distance for combinations of feasible methods, observations, and elements. This is a suggestion, not a validated process.</p>

CC BY-SA 4.0
2022-01-10T01:34:20.677Z
2022-01-10T03:46:05.773Z
2022-01-10T03:46:05.773Z
Albert DiCanzio
<p>Recently I had to implement the Laplace method and apply it to 3 observations of Mars (10 days in between two observations). The results were pretty good, with discrepancies with real data well below 5% in most of the orbital elements.</p>
<p>I thought the Laplace method was very accurate, so I tried to apply it to an asteroid (CASLEO, 5387). The 3 observations were also taken at intervals of 10-days, but now the results widely diverge from the real orbital elements.</p>
<p>I expected it to be less accurate than in the Mars case since the observations covered a smaller proportion of the orbit, but some of the elements were quite far from expected.</p>
<p>So my question is: What is the accuracy of the Laplace Method? Is it really that sensitive to the distance?</p>

CC BY-SA 4.0
2022-01-10T04:13:52.413Z
Is the date of the “control” orbital elements the same as the date you made your measurements? Minor planet orbits change enough in comparatively “short” periods—at least compared to major planets. Maybe that’s part of the answer.
Pierre Paquette
CC BY-SA 4.0
2022-01-10T15:05:09.147Z
It'd be better to post some concrete examples, "less accurate", "smaller portion", "very accurate" are pretty abstract terms.  Can you post the actual data?
Greg Miller
<orbit><solar-system><asteroids><orbital-elements>
Accuracy of Laplace Method for determining orbital elements
Javier
# Accuracy of Laplace Method for determining orbital elements

(Asked by: username_0 on 2016-01-04)

Recently I had to implement the Laplace method and apply it to 3 observations of Mars (10 days in between two observations). The results were pretty good, with discrepancies with real data well below 5% in most of the orbital elements.

I thought the Laplace method was very accurate, so I tried to apply it to an asteroid (CASLEO, 5387). The 3 observations were also taken at intervals of 10-days, but now the results widely diverge from the real orbital elements.

I expected it to be less accurate than in the Mars case since the observations covered a smaller proportion of the orbit, but some of the elements were quite far from expected.

So my question is: What is the accuracy of the Laplace Method? Is it really that sensitive to the distance?

**Tags:** orbit, solar-system, asteroids, orbital-elements
> **Comments:**
> **username_1**: Is the date of the “control” orbital elements the same as the date you made your measurements? Minor planet orbits change enough in comparatively “short” periods—at least compared to major planets. Maybe that’s part of the answer.  
> **username_2**: It'd be better to post some concrete examples, "less accurate", "smaller portion", "very accurate" are pretty abstract terms.  Can you post the actual data?  

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## 2 answers

### Answer by username_3 on 2017-01-13. Score: 1

In principle, Laplace's method (and Gauss's method) should be very accurate for determining orbits.

Of course, "garbage in, garbage out" applies, and if the observations were inaccurate, the results may also be inaccurate. However, you should not get great discrepancies. I would consider careful checking of your calculation, and perhaps comparing your results with those of an independent orbital determiner, to check for a mistake in the maths.

> **Comments:**
> **username_4**: To double check, it's [this Laplace method](http://adsabs.harvard.edu/full/1991SvA....35..428K) and not [this Laplace method](https://en.wikipedia.org/wiki/Laplace_transform)?  

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### Answer by username_5 on 2022-01-10. Score: 0

Interesting project! Perhaps first determine convergence criteria for alternative methods being considered and if possible acquire more widely distributed observations of target's position. Then to the extent appropriate to examine how least squares estimates for various parameters of the selected method correlate (or not) with results obtained by applying that method to available observations of the orbiting object under study, one may decide to use consequent approximations of orbital elements as basis for examining differences in sensitivity to distance for combinations of feasible methods, observations, and elements. This is a suggestion, not a validated process.


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