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Old 10-08-2019, 02:45 PM
DPRK is offline
Join Date: May 2016
Posts: 3,967
Originally Posted by Frankenstein Monster View Post
Someone else should have a go at this, I speak French extremely poorly. But according to slightly corrected Google translate, the argument is:

So yes he finds parabolic orbits from two endpoints and a varying point inbetween and notes that they all then conflict with the rest of the observations. Or something.
I believe your description is consistent with what he writes. First he calculated a parabolic orbit matching two observations and having a perihelion of 10 AU, and noted that it did not match the intermediate observations. Then he tried 8 AU, and it also did not work. "Whence it is easy to conclude that if one tried to find a parabolic orbit which at the same time satisfied the observations of 9th April 1781 and 21 October 1782 and also agreed with a third, intermediate, observation, then that parabolic orbit would be found to be in conflict with a great number of observations by at least 30 minutes: so that it is incontestably proved that a parabolic orbit can absolutely not satisfy the observations; and the errors even becoming so considerable that it is right to presume that ellipses whose eccentricities are slightly considerable find themselves completely excluded."

So, it does not sound like he is deriving rigorous error bars on the value of the eccentricity, more like claiming it cannot be too great and still agree with all the observational data. [Actual eccentricity of Uranus's orbit: about 0.046]. Indeed, on page 74 he discusses fitting parabolic orbits to the data of 17 March 1781 and 23 January 1782, and notes there is no way to match them up with a couple of intermediate observations, but notes that ellipses will produce smaller errors, the more observational data the more shapes can be ruled out, and that it will take several years of observation to determine the precise value of the eccentricity.