What known solar system object is farthest from the sun, currently?

One of the Voyagers.

dv said:

What known solar system object is farthest from the sun, currently?

I had to look it up, so I’ll keep my mouth shut.

Bubblecar said:

dv said:

What known solar system object is farthest from the sun, currently?

I had to look it up, so I’ll keep my mouth shut.

I was surprised. NEHOI.

Nemesis

dv said:

Bubblecar said:

dv said:

What known solar system object is farthest from the sun, currently?

I had to look it up, so I’ll keep my mouth shut.

I was surprised. NEHOI.

I’ll look it up, too. I’ve never heard of Nehoi, must be a new name.

Hold on, actually I don’t know how to look it up. Must rethink.
I was thinking the furthest SDO, but I don’t know how to convert ephemerates into distances.

It would be the first ever discovered non-periodic or near-parabolic comet wouldn’t it?
That would make it the Great Comet of 1680.

The maximum distance from the sun of the Great Comet of 1680 is 889 AU, but it’s nowhere near that distance away yet, so there may be an SDO that’s further.

The distance from the Sun to 1994TH is exactly 40.94 AU, so that’s a minimum distance on the maximum distance from the Sun.
Eris is further than 38.3 AU, but how much further?

I’m going to opt for the Great Comet of 1680.

> The maximum distance from the sun of the Great Comet of 1680 is 889 AU, but it’s nowhere near that distance away yet, so there may be an SDO that’s further.

> I’m going to opt for the Great Comet of 1680.

ephemerates ephemerides

There ought to be a very simple mathematical formula for the distance to objects in parabolic orbit, particularly so for objects like the Great Comet of 1680 with negligibly small perihelion distance.

Potential energy = Kinetic energy
Potential energy = G m_1 m_2 / r^2
Kinetic energy = m_2 v^2
So v = sqrt (G m_1) / r where m_1 is the mass of the Sun
When r is infinity then v is zero which is correct for a parabolic orbit

r = integral (v dt) = sqrt (G m_1) * integral (dt/r)
total time t = integral (dt)

Guess r = c * ln(t)
dr/dt = c/r
r = integral (dr) = c integral (dt/r)

So the heliocentric distance to the Great Comet of 1680, give or take a few percent, is r = sqrt (G m_1) * ln(t)
t = 1.072e10 seconds.
G m_1 = 1.327124e20 metres cubed per second squared
so
r = 2.6606e11 metres is only 1.78 AU, that can’t be right, I’m made a calculation mistake somewhere.

The distance must be much greater than 888.85 (AU) * 340 (years) / (9370/2) (years) = 64.5 AU, because the Great Comet of 1680 is periodic with period 9370 years and aphelion 888.85 AU.

You were right, this is fun.

mollwollfumble said:

> The maximum distance from the sun of the Great Comet of 1680 is 889 AU, but it’s nowhere near that distance away yet, so there may be an SDO that’s further.

> I’m going to opt for the Great Comet of 1680.

ephemerates ephemerides

There ought to be a very simple mathematical formula for the distance to objects in parabolic orbit, particularly so for objects like the Great Comet of 1680 with negligibly small perihelion distance.

Potential energy = Kinetic energy
Potential energy = G m_1 m_2 / r^2
Kinetic energy = m_2 v^2
So v = sqrt (G m_1) / r where m_1 is the mass of the Sun
When r is infinity then v is zero which is correct for a parabolic orbit

r = integral (v dt) = sqrt (G m_1) * integral (dt/r)
total time t = integral (dt)

Guess r = c * ln(t)
dr/dt = c/r
r = integral (dr) = c integral (dt/r)

So the heliocentric distance to the Great Comet of 1680, give or take a few percent, is r = sqrt (G m_1) * ln(t)
t = 1.072e10 seconds.
G m_1 = 1.327124e20 metres cubed per second squared
so
r = 2.6606e11 metres is only 1.78 AU, that can’t be right, I’m made a calculation mistake somewhere.

The distance must be much greater than 888.85 (AU) * 340 (years) / (9370/2) (years) = 64.5 AU, because the Great Comet of 1680 is periodic with period 9370 years and aphelion 888.85 AU.

You were right, this is fun.

Found my mistakes.

- Potential energy = Kinetic energy
- Potential energy = G m_1 m_2 / r
Kinetic energy = m_2 v^2
So v = sqrt (G m_1 / r) where m_1 is the mass of the Sun
When r is infinity then v is zero which is correct for a parabolic orbit

r = integral (v dt) = sqrt (G m_1) * integral (dt/sqrt®)
total time t = integral (dt)

Guess r = c t^(2/3)
dr/dt = (2/3)c t^(-1/3) = (2c/3)/sqrt(r/c) =(2/3)c^(3/2)/sqrt®
r = (2/3)c^(3/2)* integral(dt/sqrt®)
(2/3)c^(3/2) = sqrt (G m_1)
(4/9)c^3 = G m_1
c = ((9/4) G m_1)^1/3

So the heliocentric distance to the Great Comet of 1680, give or take a few percent, is r = ((9/4) G m_1)^1/3 * t^(2/3)
t = 1.072e10 seconds.
G m_1 = 1.327124e20 metres cubed per second squared
so
r = 3.2495847e13 metres.
1 AU is 1.4960e11 metres.

So the heliocentric distance to the Great Comet of 1680, give or take a few percent, is 217.2 AU
That looks about right, and is further out than any known SDO.

Okay well near as I can tell, the solar system object that we know about that is farthest from the sun right now is probably…

C/42 K1

Caesar’s comet.

Now taken to be some 800 AU away

dv said:

Okay well near as I can tell, the solar system object that we know about that is farthest from the sun right now is probably…

C/42 K1

Caesar’s comet.

Now taken to be some 800 AU away

I was going to take a stab and suggest that.

dv said:

Okay well near as I can tell, the solar system object that we know about that is farthest from the sun right now is probably…

C/42 K1

Caesar’s comet.

Now taken to be some 800 AU away

Well if nothing else it has made us aware of Comet Caesar who’s history is brief but interesting.

JudgeMental said:

dv said:

Okay well near as I can tell, the solar system object that we know about that is farthest from the sun right now is probably…

C/42 K1

Caesar’s comet.

Now taken to be some 800 AU away

I was going to take a stab and suggest that.

Look people, I just cannot be here 24/7; but when I’m away I do expect others to make a stand and say something when Boris comes out with shit like this. Otherwise he’ll just continue willy nilly and we’ll be swamped.

sibeen said:

JudgeMental said:

dv said:

Okay well near as I can tell, the solar system object that we know about that is farthest from the sun right now is probably…

C/42 K1

Caesar’s comet.

Now taken to be some 800 AU away

I was going to take a stab and suggest that.

Look people, I just cannot be here 24/7; but when I’m away I do expect others to make a stand and say something when Boris comes out with shit like this. Otherwise he’ll just continue willy nilly and we’ll be swamped.

and i would have gotten away with it…

sibeen said:

JudgeMental said:

dv said:

Okay well near as I can tell, the solar system object that we know about that is farthest from the sun right now is probably…

C/42 K1

Caesar’s comet.

Now taken to be some 800 AU away

I was going to take a stab and suggest that.

Look people, I just cannot be here 24/7; but when I’m away I do expect others to make a stand and say something when Boris comes out with shit like this. Otherwise he’ll just continue willy nilly and we’ll be swamped.

Sorry, sibeen.

Good job, Boris! (Thumbs up)

dv said:

sibeen said:

JudgeMental said:

I was going to take a stab and suggest that.

Look people, I just cannot be here 24/7; but when I’m away I do expect others to make a stand and say something when Boris comes out with shit like this. Otherwise he’ll just continue willy nilly and we’ll be swamped.

Sorry, sibeen.

Good job, Boris! (Thumbs up)

et tu brute

SCIENCE said:

dv said:

sibeen said:

Look people, I just cannot be here 24/7; but when I’m away I do expect others to make a stand and say something when Boris comes out with shit like this. Otherwise he’ll just continue willy nilly and we’ll be swamped.

Sorry, sibeen.

Good job, Boris! (Thumbs up)

et tu brute

:)

dv said:

Okay well near as I can tell, the solar system object that we know about that is farthest from the sun right now is probably…

C/42 K1

Caesar’s comet.

Now taken to be some 800 AU away

Check wikipedia.

“July 44 BC … In the absence of accurate contemporary observations (or later observations confirming an orbit that predicts the earlier appearance), calculation of the comet’s orbit is problematic and a parabolic orbit is conventionally assumed. (In the 1800s a possible match was speculated which would give it a period of about 575 years. This has not been confirmed because the later observations are similarly insufficiently accurate.) The parabolic orbital solution estimates that the comet would now be more than 800 AU (120 billion km) from the Sun.”

Fine. I didn’t include that because I didn’t know about it.

It isn’t confirmed to be a near-parabolic comet. It doesn’t make it onto https://en.wikipedia.org/wiki/List_of_near-parabolic_comets and even most near parabolic comets can have apogees closer than 800 AU.

But it could be that far away.

UIVMM, the greatest altitude ever reached by a non-American is 475 km, attained by Belyayev and Leonov on the Voskhod 2, some 55 years ago.