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In early November, I proposed a little astronomical calculation question:
"How fast is Halley's Comet moving away from the Sun today in knots [because Halley was a sailor, so a speed in knots seems appropriate to me!]? And when will that speed be zero?"

Jumping to an answer, I calculate that the speed of fall of Halley's comet (now towards) the Sun is about FIVE knots. That's the speed today, around 1:00pm Eastern US time, and it's gaining about one knot daily.

Here's the message thread: https://navlist.net/find/halley-in-date-2023. Antoine Couette provided an excellent answer to a slightly different question, giving the total speed of the comet which includes a very large speed perpendicular to the line to the Sun. I was really trying to ask about, as I explained in a follow-up, the heliocentric radial speed. To put it a little differently, how rapidly is it rising from the Sun? When will that speed hit zero? And now that the date has passed, how rapidly is it falling towards the Sun?

Robert H. van Gent provided some great details on the date of this maximum distance, or aphelion, from three sources:

• JPL Horizons say 9 Dec 2023 at 0100 UT
• IMCCE says 7 Dec 2023 at 1800 UT
• Aldo Vitalgliano's Solex says 7 Dec 2023 at 1000 UT

Are any of these better than the others (see PS)? Hard to say. I saw a few brief articles and comments on this date. I'm fairly certain that most of them sourced a date from Wikipedia, and if you dig a bit in their links, the Wikipedia article's editors cheated a little bit on the rules of Wikipedia and did some "independent research" using JPL Horizons. I would suggest that part of the reason it's so difficult to be more specific about the date and time is because the speed I'm asking about, the heliocentric radial speed, the "in-out" speed relative to the Sun, is, in fact, quite small, less than one knot for more than 48 hours centered on the actual time of maximum distance.

I didn't expect any further action on this thread, and I intended to let it die, but then this morning I realized, I actually want to know how fast Halley's Comet is falling! Not thinking too hard on it, I came up with an efficient, fast (and I hope/believe correct!) way of analyzing this...

The comet's orbit is a long, thin (and somewhat variable) ellipse. Of course the Sun is at one end (focus) of the thin ellipse, and the comet reaches perihelion, closest approach to the Sun, at that end of the ellipse. It will arrive there in 2061. Right now, it's at the other end of the ellipse, beyond the distance to the planet Neptune at about 35 AUs from the Sun (1AU equals about 150 million km, nearly the mean distance of the Earth from the Sun). If we zoom on that top end of the ellipse, it's a shallow curve rising up and then bending down towards the Sun. We can treat the comet like a tossed snowball in my backyard. The variation of the Sun's gravitational pull way out there is insignificant over a few days. Just call it some constant acceleration, a. The comet travels on a shallow parabolic arc, and its horizontal and "vertical" (heliocentric radial) motions are decoupled, just like a tossed snowball here on Earth. The motion then obeys the simple equation y = (1/2)a·t², and the speed I'm looking for obeys v = a·t. No problem.

There are various ways to calculate a, the acceleration due to the Sun's gravity, out there at 35 AU from the Sun. We can start by calculating the acceleration here at the Earth. Basic kinematics tells us that the acceleration in circular motion is v²/r where v is the speed in the circle and r is the radius. Knowing the Earth's distance from the Sun and knowing that the Earth's speed in its orbit is 30km/sec (also easily derived, when in doubt, from the known circumference and the duration of the year in seconds), I can work out the acceleration. Then to account for the greater distance from the Sun, I have to divide that by 35² (Newtonian gravitation... inverse-square law). The result is just about 5(10⁻⁶)m/s². That's about two million times smaller than the acceleration of gravity (from the Earth's gravitation) at sea level on the Earth.

With the acceleration in hand, it's easy. The change in speed is just that acceleration multiplied by the number of elapsed seconds. Seconds in a day = 86400, so per day we get just about 0.43m/s. One last trick: convert to knots. And it's useful to know that meters per second, when doubled, very nearly (within a few percent) equals knots. That's good enough for me. So that amounts to about 0.9 knots per day or nearly 9 knots in ten days. Assuming aphelion occurred nearly at the JPL Horizons time and date, the speed right now is 5 knots inbound, and on Christmas Eve, ten days from now, it will be about 14 knots.

Frank Reed
PS: One more "prediction" of aphelion from a source that we would expect to be modestly less accurate: Stellarium. If you go into Stellarium and do a ctrl-F to find Halley's comet, you'll discover that it's fainter than magnitude 30 located in our sky right now not too far from Procyon (draw a line from Betelgeuse through Procyon... continue on that line about 40% further... it's right there). You can check its distance by visiting the Sun (select the Sun in your sky, then hit ctrl-g to "go" there, and then ctrl-f to find Halley). When I select Halley's Comet in Stellarium, it informs me that the orbital elements are out-of-date and need updating in my setup. That probably explains why it predicted aphelion around October 4 (two months too early). But this doesn't affect the "differential" behavior so we can look at changes in distance during the days around aphelion. This shows an inward acceleration of about 5(10⁻⁶)m/s², too. So that's consistent with everything above.