The 12th Planet/Nibiru/ Zacharia Stitchen/ End Time/ The Gods of Eden & the Mayan Calender/Alternative Belifes
POSTED BY: piratejenny
UPDATED: Monday, June 5, 2006 10:46
VIEWED: 30595
PAGE 8 of 13
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I will forgo the definition of the elements of the equation for brevity (please hold your applause, I apologize for the size of my other posts).
NOTHING in this equation constitutes a conditional exclusion based upon distance. This equation is valid to the ends of the universe (or as far as the speed of light reaches since the formation of the objects it describes, depending on which scientist you believe). If m1 (mass of the Sun) is sufficiently greater than m2 (mass of a planet or comet), then the gravitational effect of m2 on m1 can be ignored and Newton's equations simplify into Kepler's. Even if you ignore the simplification, the ONLY affect it has on Kepler's equations is that it places center of the Sun off from one of the foci in the ellipse. This results in a trivial change in the calculation of the aphelion of the planet as long as its mass is much less than the Sun's.
My goodness. So for brevity you left out the definition of the elements. Let me help you out here.
Try this
G(m1m2)=F
--------
r^2
(Newton's law of Universal gravitation) where F is gravitational force, m1 and m2 are two masses, G is the universal gravitation constant and r THE DISTANCE BETWEEN THE MASSES.)
You will notice that like most forces it is an inverse square law, the force decreases as a square of the distance. Your kepler calculation showed a verylong eliptical orbit so r will get big and f will get very small.
You also said that escape velocity remains constant. It does from the same point in the gravity well however the further away from the center of gravity you get the smaller it becomes.
That's because the general equation for escape velocity is
v = square root ((2GM/r) where r in this case would be distance from the sun.
I suspect that you would have to take Kepler's equations and the escape velocity ones combine them and integrate for the entire orbit of X. If at any point the orbital speed exceeds solar escape velocity at that point the object will be gone. It's none trivial, which is why I'm not really up to trying it.
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Quote:
Originally posted by Fletch2:
You will notice that like most forces it is an inverse square law, the force decreases as a square of the distance. Your kepler calculation showed a verylong eliptical orbit so r will get big and f will get very small.
... but non-zero for any distance. My point is that only another force in the vicinity will alter the calculable affects of the Sun's force. Of course, we have zillions of other gravitational forces in our stellar neighborhood. But none of them are as strong as the Sun's at a mere .0074 LY from the Sun.
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You also said that escape velocity remains constant.
I'll have to challenge you to show me where I said that. I merely mentioned that if the orbiting body doesn't exceed's the Sun's escape velocity, and no other force is acting upon it, it won't escape. Perhaps my use of "velocity" made it seem I thought there was only one. If so, I apologize for that ambiguity. But I'll take solace in the fact that that ambiguity also exists in countless physics texts as well. When they say "escape velocity', they are referring to an infinite family of velocities along the trajectory.
At any rate, I'm well aware that the escape velocity varies by distance from the gravitational source. But the fact remains that if the object is not travelling at escape velocity at any point in its trajectory and no other force acts upon it, then it will not escape. Hence it is either in orbit, or on a collision trajectory with the gravitational source.
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I suspect that you would have to take Kepler's equations and the escape velocity ones combine them and integrate for the entire orbit of X. If at any point the orbital speed exceeds solar escape velocity at that point the object will be gone. It's none trivial, which is why I'm not really up to trying it.
As long as we are talking about a two-body system with no external energy/force being added, I disagree. Velocity is nothing more than kinetic energy. Distance from the Sun is potential energy. With no other inputs (no rockets firing or other bodies contributing gravitational force) he total energy of the object (whether its a planet or a spacecraft) is a constant: the sum of its kinetic and potential energies. If the total energy of the object exceeds a certain value, it will escape the Sun. If it is less than that value, it won't.
Hence, for any given distance from the Sun (potential energy) there is a corresponding kenetic energy (escape velocity) that will allow the object to never return to the Sun. If it has that escape velocity, then we won't ever see it again. If it doesn't, it comes back. You only need to snapshot its velocity at that distance to know the answer.
Calculus (with heavy doses of approximation) becomes barely sufficient when you have more than 2 bodies in the system. I have no intention of going there either.
[Edited to clarify my statement on total energy of the object in question.]
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Physicists solved the 3+ body problem a year or two ago. I don't know the details, but I think it went something like - take the two largest bodies, find the centroid of their gravity contribution, using that as a single body, do the same calculation between it and the third largest body, etc ...
Nearly everything I know I learned by the grace of others.
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Cool! I'd like to know more about the solution.
I feel a google coming on...
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YEAH !!!
Nearly everything I know I learned by the grace of others.
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Quote:
Originally posted by danfan:
Thanks for the book reference. I may try to track it down if I can catch up on my other reading.
I dug the book out of my garage last night. The author's name is Brian Greene.
You're welcome on my boat. God ain't.
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Quote:
Originally posted by rue:
Never mind!
I should have read further before I posted since my answer was here: "I never really wanted to debate Stitchen's work I just wanted to disguss it"
Hi PJ,
I guess I didn't frame my question well. What I wanted to know was this: do you want to discuss Stichen's writings or the ideas? My guess is that you want to discuss the writings with others who are familiar with them. OTOH I suspect you're not interested in discussing the ideas (the existence of planet X for example), which can be debated without the writings (eg through astronomy, physics etc instead).
Am I correct in this?
Sincerely,
Rue
Nearly everything I know I learned by the grace of others.
your right I wanted and still wwould like to discuss Stitchen's writting with those who are familar with his work, as for his ideas for example the existence of planet X, there has been an interesting discussion about that going on, I only interjected at it's impossiblity. But I think its an interesting discussion and would have liked to added more to it but since I'm not very scientifically minded I thought I would leave it to those who have more of an inclination to that type of thought.
P.J
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Quote:
Originally posted by rue:
Physicists solved the 3+ body problem a year or two ago. I don't know the details, but I think it went something like - take the two largest bodies, find the centroid of their gravity contribution, using that as a single body, do the same calculation between it and the third largest body, etc ...
Maybe I'm being ignorant, but I'm not sure if that's correct. Consider the sun-earth-moon system: if you take the centre of mass between the two largest objects, the sun and the earth, it would be very close to the sun and you could take away the sun and the earth and consider an imaginary object with the combined mass of the sun and the earth being located at this centre of mass.
The centre of mass of the imaginary object and the moon is then the point around which the moon would revolve. Disregarding the stability of the moon's orbit around this point in space, I don't see a way to explain the moon's cycloid-shaped orbit as it moves around the centre of mass of the new system (the cycloid-shaped orbit of course coming from it revolving around the earth and the earth at the same time revolving around the sun).
Other people can occasionally be useful, especially as minions. I want lots of minions.
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Just a caveat: "if you take the centre of mass between the two largest objects, the sun and the earth" The earth is not the second largest (massive) object in the solar system.
That aside, you need to take into account distance between objects (mutual gravitational interaction). I don't have the exact formula but that is accounted for.
Nearly everything I know I learned by the grace of others.
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Quote:
Originally posted by rue:
Just a caveat: "if you take the centre of mass between the two largest objects, the sun and the earth" The earth is not the second largest (massive) object in the solar system.
No, I was talking about the 3-body problem consisting of the Sun, the Earth and the Moon. The gravitaional influence of the other bodies in the solar system can be assumed to be almost negligible in this context. And if not, then let's just assume, hypothetically, that there are no other bodies in the solar system. I still don't see how the proposed solution method would work.
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That aside, you need to take into account distance between objects (mutual gravitational interaction). I don't have the exact formula but that is accounted for.
Maybe I misunderstood the proposed solution, but I don't think there was any mention of the mutual distances actually being in the solution. As far as I remember, it was something like "take the centre of gravity of the two largest ones and go on from there". In fact:
Quote:
Physicists solved the 3+ body problem a year or two ago. I don't know the details, but I think it went something like - take the two largest bodies, find the centroid of their gravity contribution, using that as a single body, do the same calculation between it and the third largest body, etc ...
So yeah, I still don't think the method as you've described it works.
Other people can occasionally be useful, especially as minions. I want lots of minions.
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