*** Moon orbit is wrong according to Cornell University *** PIN

Wrong. I am an amateur astronomer. Saturn's exactly where it should be, that would not be the case if your claimed object existed. It doesn't exist, the evidence is conclusive.

TrES-2b is nowhere near our solar system, and indeed the evidence shows that there is no similar planet lurking at 200 AUs as you suggested.

"Practically saying?" So now you're putting words in my mouth? I have little patience for that.

How about sticking with what I actually said, which is that it doesn't even matter; it could be a jupiter mass black hole for all I care and reflect absolutely no light at all; it would still be detected through its gravitational perturbations, which would be even easier to see in the positions of the other planets. It doesn't exist.

If you want to know if its possible to directly detect a jupiter-like planet with the albedo of TrES-2b at 200 AUs distance from the sun, do the math. You'll find that such an object, given Jupiter's surface area and an albedo of 0.04% (the best fit solution of the data in the paper you alluded to), would reflect about 1.0928 x 10^11 Watts of light. That works out to approximately magnitude 23 or so at a viewing distance of 200 AUs.

The current record that I know of for dimmest object detected by an amateur astronomer is magnitude 28.5, but that's with an extremely long series of exposures. Even so, amateurs hunting for new asteroids find asteroids dimmer than magnitude 22 in 5 minute exposures; a 10 minute exposure or series of a few could easily pull it off with a half meter instrument (like GLP's main scope), particularly since this object would hardly move in that period of time and would thus remain on a single pixel collecting light, unlike most asteroids. In short, yes, amateurs could find that in deep exposures.

Read the original authors research who actually found the eccentricity anomaly; they studied the moon's position exactly, not just the eccentricity. I linked you to it, I know you said it doesn't work, but I know it works. Show me where the data is consistent with your claim of a highly inclined object.

Last Edited by Astromut on 08/18/2011 04:15 AM

Jupiter's albedo is 52% or 0.52 not 0.04% as you claim (typo? or did you mean .04?), and TrES-2b's albedo is less than 1% or less than .01, so definitely, if TrES-2b should be 200 au away, its "brightness" will be much much dimmer than magnitude 23 or even 28.5, or whatever magnitude Jupiter will have if its 200 au away, since its albedo is not as high as Jupiter's.



Last Edited by The Opened Scroll on 08/18/2011 07:06 AM

Now you're being deliberately obtuse; I specifically told you I calculated it using an albedo of 0.04%, in other words 0.0004, not 52% or 0.52. Jupiter itself would be much, much brighter than mag 23 at that distance, I calculated for TrES-2b. You are literally ignoring what I say and putting words in my mouth in place of what I say.

now its the moon of yesterday night ;)

Normalcy Bias is like a cancer to those observing something new...

Do you believe yourself to be observing something unusual in the moon's behavior?

I would like to see your calculations then :)

I believe Nibiru is a dark planet, with an orbital inclination almost perpendicular to the eccliptic plane, and with a very high eccentricity. These 3 combinations make it very difficult to detect Nibiru, whether directly or indirectly.

To help others imagine what a perpendicular orbital inclination with a very high eccentricity is like, try drawing the sun and the planets (with their orbits) on a paper. Now point your pen towards the center of the Sun and perpendicular to the paper, and start moving your pen closer and closer to the center of the Sun. This is an extreme case scenario wherein the eccentricity is set to infinity (straight line orbit) and the orbital inclination is perpendicular to the eccliptic plane.

Now notice that the tip of your pen never moves accross your paper nor does it cross the orbits of any of the planets. It simply moves closer and closer to the Sun.

Take note also that the tip of your pen (which corresponds to Nibiru) is always closest to the Sun, then to Mercury, then to Venus, then to Earth, etc. in this proper order.

Because of this, the gravitational acceleration (perturbations) caused by Nibiru will be strongest with the Sun, then Mercury, then Venus, etc. again in this proper order.

What will be the resulting perturbations on the planets if Nibiru's orbit is like this?

If this is the case, then Nibiru will cause the Sun to move perpendicularly with respect to the eccliptic plane, while the Sun drags the planets (and therefore the eccliptic plane) along with it.

In other words, such perturbations of the planets and the Sun relative to each other will be very difficult to detect. We may in fact wrongly conclude that these perturbations are part of the Sun's and the planet's natural movement along the Milky Way galaxy.

So how will these perturbations appear to us? It will appear like the eccliptic (along with the Sun and the planets) has moved perpendicularly with respect to the background stars. But since the background stars are very far away, this perpendicular movement of the eccliptic with respect to the background stars will be very difficult to detect.

Now since Nibiru is not perfectly perpendicular to the eccliptic (which as I said earlier is an extreme case scenario), we are in fact seeing small perturbations on the planets orbital eccentricities, but these perturbations will be much smaller than one will expect if Nibiru's orbital inclination with respect to the eccliptic is zero.

My point is this, the higher the orbital inclination of Nibiru with respect to the eccliptic and the more eccentric its orbit is, the less detectable will be its perturbations on the planets.

Also, the perturbations caused by Nibiru, will be more detectable on the Moon, than the outer planets like Saturn, Uranus, etc.

Last Edited by The Opened Scroll on 08/18/2011 04:51 PM

Fair enough, but you're not going to like the result. I was too conservative late last night with just how much light the object would be receiving and consequently reflecting. Here's all the numbers worked out for you. Enjoy!

Given
b = L/(4*pi*d^2)
where b = brightness (flux)
L = luminosity (in watts)
d = distance (in meters)
Solar luminosity = 3.839 × 10^26 watts
200 AU distance = 2.99196 × 10^13 meters
That works out to a solar flux of 0.034 watts/m^2 at 200 AUs distance, so that is the solar flux this hypothetical object would receive. Given a surface area of 6.1419 × 10^16 m^2 and divide that by two (since the object would receive light from the sun over half its surface at any given time), then multiply by the flux, that works out to a total of 1.044123 × 10^15 watts that the object receives. Given TrES-2b's best fit albedo of 0.04% (0.0004) it will reflect 4.176492 x 10^11 watts of light. Now plug that number back into the original formula again with the same 200 AU distance to calculate the flux received from this object by an observer at earth. The result is 3.712703 x 10^-17 watt/m^2

given that mx = -2.5log(Fx/Fox)

where mx = magnitude difference between object x and reference object
Fx = flux of object x
Fox = flux of reference object

Using the sun as a reference object, and given the sun's flux of 1370 watts/m^2 at earth, plugging in the flux for the object shows the difference between the sun's magnitude and the object's magnitude being 48.9175759. The sun has an apparent magnitude at earth of -26.74, so the magnitude that is 48.9175759 magnitudes dimmer is magnitude 22.1775759, even brighter than I originally approximated last night. It also happens to be about the same magnitude as asteroids that are detected by amateurs using half meter instruments in 5 minute exposures.

I'm still waiting on you to show how your claim of high inclination is compatible with the actual lunar laser ranging data presented by the actual researchers who measured it... Nonetheless, it would not prevent perturbations from being noticed in the outer planets anyway, but I'm endeavouring to model the situation realistically given your claimed solution to the data. I've done the calculations showing you're wrong 9 ways to sunday, now where is your evidence that your claimed inclination of this hypothetical perturber is even valid?

You are right, a jovian mass planet 200 au away should be magnitude 22.1775759 by now.

Which made me think, what if Nibiru is more massive than 1 Jupiter mass? Some say Nibiru is a brown dwarf, while others say it's a giant planet. The limit between a giant planet and a brown dwarf is 13 Jupiter masses. Thus a Nibiru with 13 Jupiter masses is neither a brown dwarf nor a giant planet, or maybe it's both :)

A planet with 13 Jupiter masses and 721.11 au away will give the same gravitational perturbation as a planet with 1 Jupiter mass and only 200 au away. (Gravitational force is directly proportional to the mass of the body and inversely proportional to the square of the distance).

Substituting this new distance of 721.11 au, will give an apparent magnitude of 27.74729. This means that if Nibiru is no more than 13 jupiter masses and no more than 721.11 au away, it should be barely detectable by now, that is, if astronomers are looking at the right direction.

Notice also, that I did not increase the surface area of Nibiru as I increased its hypothetical mass, which means, this solution requires that Nibiru should be much denser than Jupiter. And I just realized, that with this kind of density, Nibiru will be either a planet of Uranium or of Gold :)

Last Edited by The Opened Scroll on 08/20/2011 06:06 AM

Moving the goalposts. You're assuming a brown dwarf would have an albedo of 0.04%, right? How is that valid? You're no longer talking about a hot jupiter somehow being 200 AUs from our sun, you're talking about a brown dwarf. In fact, as Nibiru is often claimed to be a very "cold" object, that would translate to a higher albedo due to cloud formation.

[link to cdsweb.cern.ch]



In addition to the problem with trying to assume TrES-2b's albedo can be applied to a brown dwarf 721.11 AUs away, you're still going to see perturbations of the outer planets at that distance. That would make its presence quite obvious.

Last Edited by Astromut on 08/20/2011 10:29 AM

It's a planet made of Um - Unobtainium, which is what gives it the magical properties PXtards ascribe to it.

You seem knowledgeable see if you can help me out here:
If it can be at 721.11 AU at some point in its orbit as you're speculating, and it swings into the inner solar system getting within 0.50 AU to Earth as the fantasy goes -

What is its minimum orbital period? IOW, the shortest possible time for 1 orbit.

What is the minimum eccentricity?

If it is 1.5 years from its 'devastating 1st pass', what is the maximum distance it can be from the sun today?

Use any inclination you like, but I'd suggest 0 so your answers don't get the Nibirutards even more mad at you than they will be when you answer the above.

I await with bated breath.


R.

And who made the claim that Nibiru is a very "cold" object? Besides, I believe these people meant "cold" relative to stars and not planets :) And that makes Nibiru a very hot planet.

Btw, TrES-2b, the planet discovered to have an albedo < 1% is a jovian planet and not a small terrestrial planet like earth. (In fact it is slightly more massive than Jupiter).

In comparison, a jovian planet is like 317.18 earth masses, while a brown dwarf (which I am using the same albedo as TrES-2b) is around 13 Jovian masses only. So comparing the properties of a brown dwarf with TrES-2b makes more sense than comparing the properties of a brown dwarf with earth.

Last Edited by The Opened Scroll on 08/20/2011 06:42 PM

Oh please Mr R., don't ask me to do your homework for you :o

Um-hmmm. It never fails.
Every time I've asked one of the more intellegent PXers to do some simple high school physics calculations they suddenly become very reticent to display their knowledge (or lack) even though they've been promoting PX with what they think are very intellectual and scientific arguments.

It's almost like they haven't the first clue about what they are speaking and are just talking out their butts.

Those questions are 11th grade questions, and not very difficult at that.
The period should take you about 30 seconds to compute.
The eccentricity should take about 45 seconds.
The current distance is a little tricky and may take a number of minutes.

That's if you know what you're talking about and not just spewing BS.

I think it's clear which side of that equation you land on.


R.

I'm not comparing it to earth. It's claimed all the time to be colder than the planets, "which is why we can't see it" just look around this forum to see that claimed repeatedly. Again, it could be be a 13mj black hole for all I care, we would still see its perturbations. It doesn't exist, what part of that dont you understand? Now, why don't you answer reality's questions?

We are seeing its perturbations, which is why Lorenzo Lorio posted his research about it, saying that a planet x scenario will give a non-vanishing long term variation on the Moon's eccentricity - which is the kind of perturbation on the Moon which he observed. :)

Oh, and is Reality420 your uncle or is it the other way around?

Last Edited by The Opened Scroll on 08/21/2011 10:14 PM

We are not seeing perturbations in the outer planets we should see if such an object were responsible, which is why lorenzo called it highly unrealistic and ruled it out.

I hate to tell you pal, but evolution happens to species not individuals. We are all going to die save a relative few underground. Preps have been made for this. I will be remaining on surface to help others live if they can/die less painfully. I'm sure that I will die as well.

And I already explained in one of my previous posts how a hypothetical orbit whose major axis is very nearly perpendicular to the eccliptic plane will make Nibiru closest to the Sun, then Mercury, then Venus, then Earth and the Moon, and then the outer planets, in that proper order.

This means that the gravitational perturbations of Nibiru will be greatest with the Sun, then Mercury, then Venus, then Earth and the Moon, and then the outer planets, also in that proper order.

Right now, the observed increase on the Moon's eccentricity is only (9 ± 3)×10^-12 per year, and this is very small. And this is also why a very sensitive instrument is needed to measure this very small increase.

In comparison, the Earth and the Moon is only 1 au away from the Sun, while Saturn's aphelion is 10.115 au away from the Sun. This means that whatever perturbation Nibiru is causing Saturn's eccentricity, or the other outer planets, it will be even smaller than (9 ± 3)×10^-12 per year.

Thus, my conclusion that a planet x with mass not less than 13 jupiter masses, and not less than 721.11 au away, with an orbit whose major axis is nearly perpendicular to the eccliptic plane, can be a viable solution to the anomalous increase on the Moon's eccentricity.

Last Edited by The Opened Scroll on 08/22/2011 04:57 PM

The orbit you describe would result in a collision with the sun. In order to be that "perpendicular" to the ecliptic, it would have to fall straight at the sun and ultimately collide with it. A realistic, highly eccentric, "perpendicular" orbit still results in large perturbations of the planets.

Even if an object were to perturb the planets in the order you describe due to an orbit with such high eccentricity that it actually does end in a solar collision, the result would still be apparent perturbations of the outer planets; the perturbation to earth would result in an apparent perturbation of the other planets from our point of view.

You have just demonstrated that you really don't know that much about how orbits and astronomy work. An object perturbing things on a solar system scale is not going to perturb the moon's position relative to the earth nearly as strongly unless it's very close to the moon and earth; both objects will feel very nearly equivalent gravitational pulls since they're at very similar locations in the solar system. That is why it would take such a close and massive object to produce even a perturbation so tiny that you need lunar laser ranging to see it.

Completely wrong, as I already demonstrated earlier. This shows me that not only are you ignorant of orbital dynamics and astronomy, you have no desire to learn.

Last Edited by Astromut on 08/22/2011 08:45 PM

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

Last Edited by The Opened Scroll on 08/22/2011 10:43 PM

What is the value then? If you're going to describe things non-specifically, I will take them in the worst possible way. Otherwise, state a specific claim so that I can show you what nonsense it is by the numbers.

Actually, they would, for the reasons I already outlined, which shows conclusively that you don't know what the heck you're talking about.

You still don't get it. Under no circumstance would such a perturber not produce a detectable effect in the apparent positions of the outer planets. I don't care if the perihelion distance is inside or outside mercury's orbit. You're demonstrating repeatedly that you don't understand astronomy and have no desire to, rather, you prefer to live in a fantasy land where a massive perturber can show no effect in the apparent positions of the outer planets by approaching "perpendicular" to the ecliptic (which again, is inconsistent with the moon data you cited).

You are intentionally misquoting me. I did not say Nibiru will show no effect on the outer planets, I was saying the gravitational perturbations on the outer planets will be less than the inner planets, since Nibiru is closer to the inner planets than the outer planets :)

Last Edited by The Opened Scroll on 08/23/2011 12:08 AM

You still refuse to learn. If the perturber is closer to the inner planets than the outer planets, it still doesn't matter; the problem is that you still have a differential gravitational pull on the inner vs outer planets, thus the apparent positions of the outer planets will still be off just the same. It doesn't matter if it's because earth is moving out of position or the outer planets are moving out of position, their apparent positions will still be off. You still fail to acknowledge this because if you did you'd have to admit your theory about just such an object actually existing is wrong.

Last Edited by Astromut on 08/23/2011 09:20 AM

Show me your figures and calculations then :)

Last Edited by The Opened Scroll on 08/23/2011 05:53 PM

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.