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    Re: Halley's Comet reaching aphelion...
    From: Antoine Couëtte
    Date: 2023 Nov 6, 06:29 -0800

    Interesting topic.

    With MKSA international units, GS = 1.327 E20 m³/s², and 1 AU = 1.496 E11 m.

    Assuming for Halley's Comet : semi-major axis "a" = 17.93 AU = 2.682 E12 m and eccentricity "e" = 0.967 9 .

    Assuming for its mass to be negligible in comparison with that of the Sun, then "nominal" - i.e. newtonian - speed "Vs" at aphelion is given by the following formula : Vs² = GS/a *(1-e)/(1+e) . See e.g. William Marshall Smart's Textbook on Spherical Astronomy § 65 (6th Ed. 1977, rep. 1980).

    We then derive Vs = 898 m/s = 1,746 kts (to be compared to the Earth speed equal to about 29,300 m/s or 56,978 kts at apogee).

    Since its eccentricity is not equal to exactly 1 - neither extreeeemely close to 1 - Halley's Comet speed is never equal to zero. In other words, no direct fall onto the Sun surface here.

    If there were no planetary perturbations, the revolution in its orbit would be exactly symmetrical. In other words its aphelion time and date would occur exactly halfway between two adjacent perihelia dates as per pure Newtonian Celestial Mechanics.

    But the actual aphelion date does not generally occur right at the middle of the above due to the influence of the outer planets which can greatly and significantly modify its orbit, up to even a few years in its revolution period. But ... did I understand your last question properly, Frank ?  Probably no, because I cannot exactly figure out your quantitative request for it : " Is there a proper quantitative explanation for the difference? ".

    Thanks in advance for your clarification to worrried frog here :-) .


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