Did I give a value for the hypothetical eccentricity of Nibiru's orbit, or its hypothetical perihelion distance? So how do you know if it will collide with the sun? And as I mentioned in my previous post, I was giving an extreme case scenario just to make it easier to see that Nibiru's perturbations on the outer planets don't have to be more significant or detectable than its perturbations on the Moon.
A less extreme scenario wherein the perihelion distance will be slightly farther than Mercury's orbit would mean that Nibiru will be closest to Mercury first, then Venus, then the Earth and the Moon, then the Sun, then Mars, then the outer planets, in that proper order. And still this means that Nibiru's perturbations will be strongest with the inner planets and the Sun than the outer planets. And that's just the point I'm trying to make.
"It will fall straight at the sun"? Are you saying that an object whose major axis is nearly perpendicular to the eccliptic plane will fall straight at the sun, while a similar object having the same eccentricity and perihelion distance but whose major axis is along the eccliptic plane is less likely to "fall straight at the sun"? lolz
Quoting: Anti-GLP Effect You really are a witless wonder blowing smoke and thinking you're technically competent. I'd figured you for an 11th grader but now I'm convinced you're an 8th grader scribbling diagrams on paper thinking you've got it all figured out.
You don't you dolt. That's why you can't do any simple calculations and why you can't see the laughable errors of your thought process, which is what Astronut is trying to explain to you. I don't know why he bothers with a 14 year old troll who thinks they know something. You're at least 8 years behind in education with the people you're crossing swords with you arrogant cretin.
Typical PXer.
Try this on for size nitwit:
You're fantasy numbers:
Current distance = 721.11 AU
Perihelion d = 0.387 AU (Mercury)
Inclination = 90
result in:
Minimum Orbital Period = 6851.8 years
Minimum e = 0.998927
Minimum distance of the nodes = 0.771 AU
Current distance to arrive in 1.5 years = 6.99 AU
The Nibirutards aren't going to be happy with your minimum orbital period being greater than 6850 years. Kinda' throws their doom of for another 5000 years.
Did you notice that you fantasy crosses the ecliptic at 0.771 AU? How's that jibe with your statement that "Nibiru will be closest to Mercury first"? Think your 14 year old brain may be missing something?
And how about it's current distance for it to arrive in 1.5 years of 6.99 AU? What magnitude would it be today, genius? You've tried that calculation earlier in this thread.
This is the same problem you PXtards get into starting with Ninny Lieder (and Sitchen) going back to 1997 on sci.astro.
What you droolers don't realize it that as soon as you give a couple of the correct orbital parameters you've constrained the orbit to something that can be calculated.
Ninny Lieder finally realized that on sci.astro and quickly invented her - "ZetaPhysics©" and "ZetaMath©" which had no calculations and consisted of essentially 'whatever delusion Ninny Lieder vomits is the way it is'.
Oh, and the preschool wail of, "all human math and science is wrong... 'cause I said so."
You can now spit out some more of you're ignorant thoughts on orbits. You may fool the other pre-schoolers, but you're providing a few chuckles to those that graduated high school and actually took math and physics... not to mention college science graduates.
Drool on kook. Drool on.
Done with you, and I suggest Astro deny you any more info as well.
R.
