Work done separating 2 masses -gravitation

Discussion in 'Physics & Math' started by Robittybob1, Nov 21, 2011.

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  1. Trippy ALEA IACTA EST Staff Member

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    And because he's gone and altered his bleeding post again:
    YES!

    Why?

    Because at no point did I claim that your solution was wrong.

    The only thing I commented on was the fact that the errors you asserted in my proof were in fact your own errors, resulting from you not following the prose, and not following the math.
     
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  3. Trippy ALEA IACTA EST Staff Member

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    I don't recall having suggested otherwise, although I would be tempted to argue that it began when he misrepresented my post.

    Right, but this is the third or fourth time in the last 48 hours that Tach has misrepresented one of my posts - something you should take into consideration (yes, mea culpa for carrying baggage from other threads).

    Perhaps it was a simple mistake, but look at it from my perspective for a moment. In the last 48 hours, he has outright lied about what I have said in other threads, and here he is again, misrepresenting something that I believed to be both straight forward, easy to understand, and unambiguous.

    Yes, and surely you're aware of how patient I generally am.

    See above.
     
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  5. Trippy ALEA IACTA EST Staff Member

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    Actually, you know what?

    Fuck it.

    I apologize for being rude.

    I still demand that you admit you were wrong, and that the errors you perceived in my derivation were your own failings.

    You know, there's a good reason I laid it out the way that I did, and that was so that if I cocked up somewhere, someone could point to the specific step where I erred, and we could then discuss rationally, calmly and sensibly what went wrong.

    And while you're at it, you can apologize for your own rude remarks.
     
    Last edited: Nov 22, 2011
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  7. AlphaNumeric Fully ionized Moderator

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    Let's everyone take a long slow breath and calm down..... before wailing on Reiku's sock puppet....

    No one needs to rush to Funk's rescue, he's a big boy and can look after himself.

    You've posted stuff about that and I don't believe you can do integrals. Reiku posted all kinds of crap and he couldn't multiply out (a+b)(c+d). Simply being able to spew out a bunch of equations you find in a paper on ArXiv or in a textbook on Google books doesn't make you a competent physicist. Just like putting on a lab coat doesn't make you a scientist.
     
  8. Aqueous Id flat Earth skeptic Valued Senior Member

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    No, I thought you were talking about a practical problem, like raising a bucket of water 3 ft. on earth, which is where mgh would be applied. I couldn't make the connection to gravitation at 100 MLY since it's unrealistic. By this I mean the gravity of the intervening cosmos buries the effects 100 MLY away.

    Actually I haven't yet understood what you're after. You want to go from a quantum distance to the span of 100 MLY? And reconcile the gravity at that range? That's what I'm not following. Long before you get to 100 MLY, the work has vanished to a hugely insignificant number.

    So the formula I gave you was to solve for two bodies being moved from a distance of a to b. There was some question about the definition of work and how to calculate it which continues here in an almost surrealistic debate.

    The definition of work, as Trippy pointed out, is the amount of energy expended to raise a system from an initial to a final level. The equivalent way to define work is that it's the product of force times distance. That's simplified, since, in the more general case, if force varies over distance, you need to integrate over that, which is equivalent to plotting force versus distance, then calculating the area under the curve. So that's where the discussion on calculus launched.

    Since the gravitational force is an inverse square law, as was pointed out way back, you are looking at the integral of 1 / (r **2). r is the variable, everything else is a constant, so G m1 m2 moves out in front the integral. r is the distance, and we are integrating over distance, so the 1 / (r **2) gets integrated.

    The integral of 1 / (r **2) is (-1 / r). This is the indefinite integral, so next you are going to use your start and end points (over which you are calculating the area under the curve) to calculate the definite integral.

    To calculate the definite integral, you plug in the endpoint (b) to the indefinite integral (-1 / r) to get (-1 / b). Then you do the same for the start point (a) to get (-1 / a). Finally you subtract the end from the start to get the span under the curve, and the answer to the definite integral: (-1 / b) - (-1 / a).

    Last you have your constants out front, so you get:

    - G m1 m2 (1/b - 1/a).

    or, you can write it as G m1 m2 (1/a - 1/b).

    Now suppose b= 100 MLY. The 1/b term is going to be extremely small, as you might expect. The 1/a term is problematic because you can't have a=0, which is another way of saying two particles (or bodies) can't occupy the same space.

    Generally, in problems of this kind, where there is one huge number and another hugely small number, you are dealing with an unrealistic or faulty assumption. So there should be some alarms going off.

    Also, if you picture the plot of (1/r**2), it descends from an asymptote at zero, a singularity, plunges quickly and tapers off slowly. This is the classical exponential decay. So obviously, you're effects 100 MLY out are insignificant.

    Anyway, I didn't follow most of the rest of what you were talking about concerning various particles or some ideas about the Schwarzchild radius or something, and also I don't get the significance of 100 MLY, or how earth enters into the discussion at all.

    So if you rephrase this question I'll try to say more. You would do better to pose a clear presentation of what you are assuming and what you are looking for.
     
    Last edited: Nov 22, 2011
  9. James R Just this guy, you know? Staff Member

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    Moderator note: 28 off-topic and/or insulting and/or irrelevant posts have been deleted.
     
  10. James R Just this guy, you know? Staff Member

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    This is 1st-year undergraduate stuff or earlier. I am surprised that Tach couldn't think of any easier way to solve the problem than to perform an integral from first principles. But then again, my assessment is that Tach can cope with some math but isn't too hot with physics. In fact, my impression is that he mostly thinks that physics is math.

    Also interesting that Tach couldn't recognise that Trippy's answer was correct apart from the sign change that Pete pointed out. Maybe that comes from being too tied up in one's own ego.

    I guess we can clock this up as another display of incompetence by Tach and move on.
     
  11. prometheus viva voce! Moderator

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    IMO since the OP has been answered this thread can be locked. If someone wants to extend it to a general relativistic analysis (!) let a mod know and we'll unlock it for you.

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