Newb needs info from Cosmologist Superbrains

Dec 22, 2019
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Hi
And thank you for taking time to read this

Halley's Comet.

One end of the gravitational orbit of HC is Sol but what is the gravitational anomaly for it's return? A mini black hole?

I ask as I find it unrealistic to believe that a high velocity and massive object would suddenly decide to loop-the-loop and return to our solar system.
Frequency variations re appearance could be attributed to how much of the coma is re-absorbed by HC being the nearest mass with the heaviest pull.
I have found nothing to help me with this and this has boggled my limited mind for some time.

All replies will be gratefully received even from geocentrists or Copernicans (poor old Aristarchus of Samos, bless him, came up with the same theory 1,800 years earlier)

I will endeavour to reply to all posts ;o)

Ar Bliss U
 
Dec 11, 2019
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According to Isaac Newton's law of gravity: F(g) = (G*m1*m2)/r^2 where m1 = Mass of the Sun and m2 = Mass of HC, r is the distance between the Sun and HC and G is the gravitational constant. Highly elliptical cometary orbits are not essentially different from the more or less circular orbits of the planets (which are actually mildly elliptical too). So HC doesn't need anything on the "far end" of its orbit to turn it around and send it back towards the sun. Good ol' Sir Isaac suffices....
 
Jan 27, 2020
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According to Einstein, the mass of the Sun distorts space time and jolly old Mr Sun sits in a solar system centered gravity well which deforms space time beyond the orbit of the comets in the Oort Cloud.
The Einstein tensor is:
\mathbf {G} ={\frac {8\pi G}{c^{4}}}\mathbf {T} ,


The quantity G (which measures curvature) is equated with the quantity T (which measures matter content) and G is the gravitational constant of Newtonian gravity, and c is the speed of light from special relativity.

As a result, all the planets, planetessimals, asteroids and comets fall around this space time distortion.
 
Dec 22, 2019
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Yup, very tricky - I had to take my shoes and socks off just to keep up with the numbers.
The reason that I posted this question is I believe that we are afraid for the answer.
I obvious accept the gravitational formulae given but as the further away an object is and with the velocity that HC travel, Sol would not have the required gravitational force to produce the apparent slingshot effect to produce the predicatbility of HC's friendly visits.
Thoughts?
 
Jun 9, 2020
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I ask as I find it unrealistic to believe that a high velocity and massive object would suddenly decide to loop-the-loop and return to our solar system.

It is in solar orbit, as are the planets. It dives into the sun's gravitational well and climbs back out...but not fast enough at the bottom to be unbound. Its apogee is bound. Toss a ball up; it comes down. Solar escape velocity is 384 miles/second at the sun.

All Sky Fireball Network

If you think you can cheat...it had better be a wowser, re Noether's theorem.
 
Jun 9, 2020
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"It's going to be hard to determine the length of any solar orbit, without knowing Sol's velocity. "

Orbit dynamics are determined center-of-mass. The "length: of an orbit is the perimeter of its ellipse with the sun's center of mass at one focus. The massive sun does not move. General relativistic effects like orbital precession, frame dragging, nutation, Wick rotation , and such don't matter in your application. Example,

Sources of Mercury (extreme case) perihelion precession
Amount (arcsec/Julian century)Cause
532.3035Gravitational tugs of other solar bodies
0.0286Oblateness of the Sun (quadrupole moment)
42.9799Gravitoelectric effects (Schwarzschild-like), General Relativity
−0.0020Lense–Thirring precession
575.31Total predicted
574.10±0.65[7]Observed
 

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