Maths problem - planet alignments

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 Phil Lyon 15 Sep 2019

Me and a group of students used lowest common multiple to work out how often all the planets would be lined up as about once every 84 million years.

The problem with this approach is we're only taking whole orbits into account; when all the planets are in the same position as t=0.

Any ideas for improving this approach to find alignments in between, say after 38002.7 earth years etc.

I guess we need a "to within" figure, so let's say to within a degree?
 

 petemeads 15 Sep 2019
In reply to Phil Lyon:

Are you including Pluto?

In reply to petemeads:

> Are you including Pluto?

And don't neglect Uranus of course.

1
 petemeads 15 Sep 2019
In reply to Phil Lyon:

And are you just considering the plan view (from above) for alignment, because the orbits all have different inclinations to be taken into account if you want true alignment in 3 dimensions - goodness knows how often that has happened but it must be very rare, even allowing 1% solid angle...

 ablackett 15 Sep 2019
In reply to Phil Lyon:

I haven't had time to read all this, but it might answer your question.

https://www.mathpages.com/home/kmath161/kmath161.htm

 profitofdoom 15 Sep 2019
In reply to petemeads:

> And are you just considering the plan view (from above) for alignment, because the orbits all have different inclinations to be taken into account if you want true alignment in 3 dimensions - goodness knows how often that has happened but it must be very rare, even allowing 1% solid angle...

"When astrologers speak of the planets being aligned (something which doesn't really concern astronomers) they don't mean that the planets will actually all lie on a straight line at some instant of time. One calculation of alignments within around thirty degrees (about as close as they can get) shows that the last such alignment was in 561 BC, and the next will be in 2854. The eight planets plus Pluto are somewhat aligned every 500 years...."

http://curious.astro.cornell.edu/about-us/57-our-solar-system/planets-and-d...

 petemeads 16 Sep 2019
In reply to Phil Lyon:

Well worth reading the above suggestions. My contribution is to ignore Pluto, put the Sun at one end of the line and line up all the planets in order. Mercury orbits the sun 4 times a year. A one degree window, taking 6 hours, during which it will line up with Neptune, will happen 4 times a year. During each of these periods, the probability of any of the other 6 planets being in this one degree window is (1/360) to the sixth power, which is just over 1 in 2,000,000,000,000,000. You do get 4 tries per year though so it should happen every 500,000 trillion years on average...

 wintertree 16 Sep 2019
In reply to petemeads:

> You do get 4 tries per year though so it should happen every 500,000 trillion years on average...

Although the chaotic effects in orbital mechanics mean we can’t predict usefully beyond 5 million years or so...

 graeme jackson 16 Sep 2019
In reply to Phil Lyon:

There was one in 2001 as depicted in the Tomb raider documentary.  probably be another one along in a minute.

Lusk 16 Sep 2019
In reply to wintertree:

When you're sat in the observatory chair in the observatory in Myst, I'm sure can see the positions of all the planets and stars billions of years into the future, and past.

1
 petemeads 16 Sep 2019
In reply to wintertree:

Oops - meant billion, not trillion. Academic anyway because the sun takes out the nearest 3 planets in only 6 billion years as it goes red giant. Alignment of the rest then becomes much more likely/frequent, of course...

 DancingOnRock 16 Sep 2019
In reply to Phil Lyon:

I get 52million years. What values are you using for orbits? Are you using prime factors? 

 petemeads 16 Sep 2019
In reply to Phil Lyon:

What I should have said in my first response - who lined them all up at t=0 anyway?

OP Phil Lyon 16 Sep 2019
In reply to petemeads:

I assumed that at some point in the last few billion years, there was a straightish alignment and called that t=0, hoping to find out how long until the next one; assuming a regular pattern to it, this would be the frequency with which it occurred. I think I'm probably wrong on most assumptions in this problem.

Dancingonrock - yes, it was an exercise in taking prime factors to an extreme example.

 krikoman 16 Sep 2019
In reply to petemeads:

> What I should have said in my first response - who lined them all up at t=0 anyway?


God

1
 krikoman 16 Sep 2019
In reply to Phil Lyon:

Are you only interested in them lining up on one side of the sun, or simply forming a straight line with each other?

OP Phil Lyon 16 Sep 2019
In reply to krikoman:

I would have thought the calculation is easier if they're in line with the sun.

 birdie num num 16 Sep 2019
In reply to Phil Lyon:

The ultimate eclipse would consist of all planets, with all of their moons lined up on the same side of the sun.

1
 DancingOnRock 17 Sep 2019
In reply to Phil Lyon:

It would happen more often if you ignore  the sun as there are loads more situations where it could happen. 

I was assuming the OP was taking about all the orbits aligned. 

 krikoman 17 Sep 2019
In reply to Phil Lyon:

> I would have thought the calculation is easier if they're in line with the sun.


That's why I asked, even easier if it's all in order from inner to outer all on the same side i.e with the Sun at the extreme left or right.

 petemeads 17 Sep 2019
In reply to krikoman:

See above. The calculation in terms of probabilities suggests this will never happen within the 1 degree restriction and the life of the universe. Other alignments are much more likely but still very unlikely, and fleeting, if 1 degree is still required.

 tlouth7 17 Sep 2019
In reply to Phil Lyon:

The easiest way to find when all the planets (+the sun) will be in alignment is to consider pairs of planets. Specifically the 1st planet (A) and each of the others.

If they start in alignment at t=0 then planets A and B will next be in alignment when the difference in their swept angles = a multiple of 180° (or 360° if you prefer the Sun to be at one end of the line)

This occurs when 180°/(angular speed A - angular speed B) = t(A,B) where t(A,B) is the time period of this pair aligning.

Find t(A,B), t(A,C) etc

Then do the least common multiples on those time periods to find the overall result.

Edit: I get 6*10^11 years, mostly driven by the insanely long orbital periods of the outer planets.

Edit 2: I was rounding to one decimal place of a day, to avoid the irrationality issue in post 6.

Post edited at 10:37
 JLS 17 Sep 2019
In reply to Phil Lyon:

Maths? Pah!

If we know...

Planet                   Orbit time in months

======             ===============

Mercury                3

Venus                   7

Earth                     12

Mars                      23

Jupiter                  142

Saturn                   354

Uranus                 1009

Neptune              1979

Pluto                     2977

Then there is a simple mechanical way to solve this problem...

Multiple the numbers above by 10 two give a suitable number of gear teeth for cogs to drive a small model of the solar system. I.e. Earth's cog will have 120 teeth. The system drive cog will also have 120 teeth. A dab of red paint on the initially aligned teeth will be all that's needed to track the planetary motion. Spin the drive cog and note the motion of the dabs of paint. One revolution of the drive cog will represent one Earth year. Simply count the number of revolutions of the drive cog it takes until the paint dabs realign. It might take a wee while so probably best engage a slave or some sort of wind turbine to rotate the drive cog for you. If you are feeling flush a DC motor and nuclear powered generator will speed things up.

 Michael Hood 17 Sep 2019
In reply to JLS:

Pluto is no longer considered to be a planet   Now known as a "dwarf planet" - is that PC?

Apparently it didn't meet all the new criteria; specifically, hadn't cleared its neighbourhood of other rubbish. Incidentally there's a larger dwarf planet than Pluto in the Kuiper Belt, which I think was why they brought out a new planet definition otherwise we might have them popping up all over the place out there.

How about the problem being constrained to visible conjunctions...

So for the naked eye that's only 5 because you want the sun out of the way to be able to see them (although it'll still be just below the horizon because of Mercury); i.e. all except Uranus & Neptune. And I think bigger than 1 degree would be ok for naked eye, needs a bit of discussion as to how big a field of view to allow, but I reckon if you could see all 5 planets close together at the same time it would be pretty impressive, would 20 degrees be ok?

If you use binoculars or a telescope so you can see Uranus as well (no mirrors allowed ), then you need a conjunction of 6 and the field of view is defined by the binoculars/telescope. That'll probably mean cutting the angle down to about 5 degrees.

Similarly for Neptune with a conjunction of 7.

Edit: Actually, Mercury is damn difficult to see with the naked eye because of the sun. I'd be impressed if I could see the other 4 close together (i.e. Venus, Mars, Jupiter, Saturn), so how often would that conjunction of 4 occur within 20 degrees, or say 10 degrees, etc.

Post edited at 16:38
 petemeads 18 Sep 2019
In reply to Michael Hood:

Follow the link offered by profitofdoom above and choose the  National Solar Observatory analysis - this should answer most of your questions about visible alignments and angular separations...

 Michael Hood 18 Sep 2019
In reply to petemeads:

> Follow the link offered by profitofdoom above and choose the  National Solar Observatory analysis - this should answer most of your questions about visible alignments and angular separations...

Thanks Pete, looks like September 2040 is a good time (must set an alarm to remind myself); only have to survive another 21 years

P.S. I remember you doing a very early ascent of Profit Of Doom, one of the occasions one of my ropes went up something much harder than I could

 petemeads 18 Sep 2019
In reply to Michael Hood:

Indeed. Got a couple of photos somewhere (by Simon Pollard?), should post them in the historic category. It was 1978 if I remember right...

 Michael Hood 18 Sep 2019
In reply to petemeads:

5/9/1978, Si Pollard and Ian Riddington there as well. My log notes "Pete led Profit of Doom, very impressive, it was also wet at the time." Can't have been raining but I remember it being one of those misty, overcast, slow drying days. I also remember you doing the threading two wires thing to get the crucial runner in.

The days of the mk 1 Escort RS

In reply to others: sorry about the hijack

 Tom Valentine 18 Sep 2019
In reply to Phil Lyon:

It's amazing what you can cram into a degree. Last month I was gobsmacked to find M81 and M82 galaxies both visible through binoculars and small telescope,  at the same time because they are only separated visually by half a degree. Even weirder is the fact that they are both the same distance away (more or less    )

 petemeads 19 Sep 2019
In reply to Michael Hood:

If you go to Heavens-above website (free, need a username) you can see the September 2040 alignment in the Solar System view - it's pretty impressive, but the Sun is not in the line.


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