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?
Are you including Pluto?
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...
I haven't had time to read all this, but it might answer your question.
> 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...
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...
> 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...
There was one in 2001 as depicted in the Tomb raider documentary. probably be another one along in a minute.
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.
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...
I get 52million years. What values are you using for orbits? Are you using prime factors?
What I should have said in my first response - who lined them all up at t=0 anyway?
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.
> What I should have said in my first response - who lined them all up at t=0 anyway?
God
Are you only interested in them lining up on one side of the sun, or simply forming a straight line with each other?
I would have thought the calculation is easier if they're in line with the sun.
The ultimate eclipse would consist of all planets, with all of their moons lined up on the same side of the sun.
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.
> 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.
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.
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.
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.
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.
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...
> 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
Indeed. Got a couple of photos somewhere (by Simon Pollard?), should post them in the historic category. It was 1978 if I remember right...
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
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 )
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.