Sorry, but I'm not up to the math, (excuse me, *maths*) on this. Can someone calculate this?

Question

One of the participants here is working on a 2012 debunking site and I wanted to contribute something. I've tried to figure how far away an earth-sized planet would be if it had a 3700 year orbit around the sun and was 4 years away from us.

Assume it's orbit is highly elliptical so that it would come as far in as the earth's orbit. This can be approximate, but I wanted something a little better than what I could come up with on my own. Some of you are much better with the math than I am. If I use this information in an answer, I will credit your work in the source box.

Answer 1

I run a copy of Starry Night Enthusiast (planetarium program) on my computer and it has the option to add user-defined objects in the Solar System, so I have already programmed in "Nibiru" as an asteroid with the same size and albedo as the Earth. I took the orbital period as 3600 years (as this seems to be the most common figure on the crank websites) and set the perihelion to 1 A.U., occurring on December 21st 2012. I set the orbital inclination to 66.5 degrees so that it is over the South Pole at aphelion because the cranks seem to think that most of us can't see it in that direction.

The orbital elements come out as

Semi-major axis: 234.892 A.U.

Eccentricity: 0.9957427

Mean Anomaly (Epoch 2000.0) 358.7027 degrees.

The current (July 9th 2008) distance from the Sun is 14.184 A.U. and from Earth 13.725 A.U. at visual magnitude 7.44.

I might tinker with the orbital elements to bring it to within something like 0.01 A.U. of the Earth on December 21st 2012, just to make it more exciting, but I have yet to work out a suitable ascending node and argument of perihelion in order to achieve that.

Answer 2

Sounds like fun.  At least I finally know upon what basis this 2012 end of the world has.  A planet with a 3700-year orbit... never heard that one before.

I thought that it was based on the Mayan calender ending on that date.  Heck, I have a calender that ends on December 31, 2008 and I'm not worried.

No, don't know my math/maths well enough either.

Answer 3

From another question about this I answered:

>>>

Thanks to Kepler's really simple laws:

The semimajor axis is 250 AU, Eccentricity is 0.99997837 (We just assume it is massive enough to NOT get destroyed by suns gravity gradient).

Apohelion is 470 AU (about 20 times the distance to Pluto).

Where is it today, about 4.5 years before it should annoy us? (solve by iteration)

Exactly 15 AU away. That is closer than Pluto. Also it should move very fast and be very large, so it is hard to miss in any of the comet/NEO searches.

So close, it should already disturb our probes with it's gravity, but it doesn't. There is no unknown huge mass, which pulls our spacecraft towards it.

Conclusion: It can't exist.

<<<

The math was really plain Kepler, the position of the planet got solved by iteratively approaching the 4.5 years from the moment when it will pass perihelion.

Using a better tool for such calculations, like ORSA would maybe get better results, but as the hoaxers did not provide any information about the orbit plane or argument of perihelion, the accuracy of my ecliptic solution should be well enough.

Answer 4

That is alot

Answer 5

sorry, Brant, me either and I bet I have less excuse than you, as Celestial Mechanics was one of my courses in college.

I would say offhand that you do need to decide a few more parameters first.

perihelion in AU

eccentricity (guessing .97?)

mass

lots more would be nice, too. Like angle to the ecliptic.

I played with a few computer models, thinking about Niburu and LAUGHED... A Jupiter-sized planet passing between the orbits of Mars and Jupiter could toss Mars out of its orbit completely and in my model, made Saturn and Uranus switch orbits!

the important observation, I thought was... there was NO WAY such an encounter could have happened 3700 years ago and left 8 planets with such low eccentricities.  (Venus is d**n near a perfect circle)

Sorry, i failed you.

Answer 6

Here are some links to astronomical calculators.

Answer 7

thats 23,462,784,000,000 miles which is a light year (5,865,696,000,000 ) times 4. thats a far planet!!! thats as far as we are from the nearest star. with a orbit the size of a very long pencle in comparison. hoped that helped.

oh and thanks for anwsering my question.