Work done separating 2 masses -gravitation

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

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  1. Aqueous Id flat Earth skeptic Valued Senior Member

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    Robb

    I got lost when an extra radius popped up.

    I thought you were going from zero to 100 MLY

    the integral is evaluated between two points 100 MLY apart

    I thought that's what you asked

    and beyond that, the problem is unfathomable, because we can't be at zero

    so what's the other position, the one that can't be zero?

    because it's the one that's the killer as far as work is concerned.

    that inverse square law is a beast at close range


    for two bodies of mass m1 and m1 kg, starting from a separation of a meters and moving to a separation of b meters will require

    -G m1 m2 (1/b-1/a) Joules of work to get from a to b. (a>0)
     
    Last edited: Nov 22, 2011
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  3. OnlyMe Valued Senior Member

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    It was a joke Tach. You so often demand others use math to prove their point that I just asked for the math...

    At least.., I think you finally got it!
     
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  5. Reiku Banned Banned

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    I can sit in the space of an hour and type about 8 times as much as he did. It was such an elementary integral, the only question I have now is why it took him so long.
     
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  7. Reiku Banned Banned

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    Oh we covered this Tach. You were the one who came in here poking fun at the OP for knowing how to conduct an integral. Now if I can post equations describing how to swap timelike and spacelike events, or I know the mathematics behind neutrino's, dirac particles, photon behaviour, general relativity then I doubt anyone would believe I don't know how to conduct an integral on something.

    You on the other hand, can mess up integrals and have done in the past with Pryzk. I don't need to troll. If I have a question, I am honest enough to ask. You on the other hand prance around as if you know it all. For all we know, you submitted a question to Ask an Astrophysicist in the many hours of debating this without putting up anything.
     
    Last edited: Nov 22, 2011
  8. Robittybob1 Banned Banned

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    If "Zero" is a point in the centre of the neutron, well it doesn't come from there but still emanates from below the surface of the neutron. It would be like taking something from the upper core - mantle boundary of the Earth and taking it into space. It will require some work to dig it out and bring it to the surface. I hadn't taken that into consideration before either. I think there is an estimate for the size of a proton. R1 maybe about 0.6 of the radius of a proton.
    R2 would depend on the energy of the graviton.
    This is all very hypothetical but challenging.

    I think it surprised you a little when you thought of us being partly here on Earth and also a thin filament in space extending to max of 100 million light years that may or may not connect. But wasn't that the same with Newtonian gravity and with the String Theory in those cases everything was being connected. But at least in my concept it was kept to a single galaxy and not the entire universe.

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  9. Pete It's not rocket surgery Moderator

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    The two expressions are actually equivalent, except for a change of sign. The key is that r1 and r2 in Trippy's post are not the same as R1 and R2 in Tach's post.

    In [post=2860816]Trippy's post[/post], which describes the general case:
    r1 = initial distance between centres
    r2 = final distance between centres

    In [post=2860920]Tach's post[/post], which describes the special case in the OP:
    R1 = Earth radius
    R2 = Radius of the other mass.
    R1+R2 = initial distance between centres
    L = final distance between centres

    Substituting:
    r1 = R1 + R2
    r2 = L
    We find that:
    \(Gm_1m_2\frac{r_1 - r_2}{r_1.r_2} = -Gm_1m_2(\frac{1}{R_1+R_2}-\frac{1}{L})\)
     
  10. Reiku Banned Banned

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    Oh yes!

    Well spotted Pete.
     
  11. Robittybob1 Banned Banned

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    Thanks for the connection between the two versions. They may never stop being sparring bulls.

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    I'll try putting Trippy's version into an excel workbook and see if there are any sensible results from it. I'll be able to vary M1 and M2 on the condition M1 + M2 = Mn (Neutron mass) Ri is 0.6 of the radius of a neutron, G = Gravitational constant. so the output will be the R2 factor. Here goes!
     
    Last edited: Nov 22, 2011
  12. Robittybob1 Banned Banned

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    The results do not confirm my hypothesis - Failed. Thanks
     
  13. Trippy ALEA IACTA EST Staff Member

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    The only digging is yours. I did not associate m[sub]2[/sub] with the mass of the earth, explicitly, or implicitly, except for a side note where I pointed out how the equation relates to reality.

    You're wrong, and you're the only person that can't see it.

    Deal with it.
     
  14. Trippy ALEA IACTA EST Staff Member

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    Use mine, in spite of what Tach says it's correct - ignore Tach, his objections are based on emo-rage. At all steps in my derivation m[sub]1[/sub] and m[sub]2[/sub] were refering to the masses of your two test masses, and in all cases r[sub]1[/sub] is the initial distance between them, and r[sub]2[/sub] is the final distance between them.

    I really don't know what's crawled up Tach's ass.

    Maybe it's because I'm just a chemist.
     
  15. Trippy ALEA IACTA EST Staff Member

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    Thankyou Pete.
     
  16. Trippy ALEA IACTA EST Staff Member

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    And awaiting the public apology from Tach on the grounds that the only thing I have had to say about his commentary was that he was wrong about mine - which has been demonstrated by Pete.
     
  17. Robittybob1 Banned Banned

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    I used your equation but the amount of energy in the gravitational potential would not have been enough to stop the "graviton" going forever. If anything it is the Strong Force factor that would need to be deducted first. Newtonian physics did not account for any restricted gravitational range.

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  18. Tach Banned Banned

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    R1 is not the radius of the Earth, it is simply the radius of the first body. The OP stipulated that the experiment is conducted far away from the Earth. As an aside, this stipulation excludes any derivation using stuff like \(E=mgh\), a simple one line integration of the variable force \(f=\frac{Gm_1m_2}{r^2}\) is sufficient, as already explained in post 2. As such, there was no justification for Trippy to instruct the OP to discard all the pre-existing solutions as incorrect.

    ..meaning that the mechanical work in Trippy's proof is negative, when in reality, is positive, \(W=Gm_1m_2(\frac{1}{R_1+R_2}-\frac{1}{L})\).
    This is obvious, since one needs to expend energy in order to separate the two bodies.
     
    Last edited: Nov 22, 2011
  19. Trippy ALEA IACTA EST Staff Member

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    This is because I got r[sub]1[/sub] and r[sub]2[/sub] the wrong way around in the final equation.

    oops

    Meaning nothing more than this step:
    Should have been:
    So, at some distance \(r_1\) we have \(E_p_1\) and at some distance \(r_2\) we have \(E_p_2\), and the minimum work required is \(E_p_1-E_p_2\). which gives us:
    4.1 \(w= \frac{Gm_1m_2}{r_1} - \frac{Gm_1m_2}{r_2}\)

    :shrugs:

    And the final equation becomes:
    4.2 \(w= Gm_1m_2\frac{r_2 - r_1}{r_1.r_2}\)

    Which, in reretospect I should probably have spotted sooner.

    Oh well, mea culpa.

    You still have some serious apologizing to do though Tach.
     
    Last edited: Nov 22, 2011
  20. Robittybob1 Banned Banned

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    Modified quote above. Are the substitutions that Pete said he made are they valid? Did anyone check these substitutions?

    I couldn't use the second equation for there was no such thing as the radius of the second body. What is the radius of a graviton?
     
  21. Tach Banned Banned

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    So, you bungled your solution, were rude to me:

    in the context of my solution being correct all along and you still demand that I apologize to you?
     
    Last edited: Nov 22, 2011
  22. Pete It's not rocket surgery Moderator

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    Tach's got a point, Trippy.
    In this thread, the flaming between you and Tach began with you.

    Yes, Tach misinterpreted your post and said it was wrong for the wrong reasons.
    But that was a simple mistake which didn't warrant a rude response.

    Yes, Tach is never backward in escalating a flame war, and we know there's recent history involved... but I think we all have a responsibility to at least try to maintain politeness as best we can (though god knows it can be bloody hard!)

    Anyway... my advice is for both of you to just let it go, or apologize to each other.

    (I know, I know... I'm no saint either.)
     
  23. Trippy ALEA IACTA EST Staff Member

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    I bungled nothing. I made a minor that changed the sign, but not the magnitude of the work done.

    You FAILED to correctly indicate my error.
    You FAILED to understand my derivation.
    You FAILED to see how your answer related to mine.
    In short you FAILED to correctly understand the prose content of my post, and you FAILED to understand the quantitative content of my posted.

    You asserted on two seperate occasions that I was wrong, but neither of those assertions were correct, and my rudeness was in response to those false assertions.

    Not one of those assertions has been born out, so yes, I think I am entitled to an apology, and when you apologize, publicly, I will consider apologizing for my rudeness.

    Meanwhile, the mistake I made was this:

    When I was contemplating that particular step, I initially had it as E[sub]p1[/sub]-E[sub]p2[/sub], however, ironicaly, I changed it.

    I initially had it that way around, because I thought it was correct, because in the recesses of my mind I remembered that E[sub]p[/sub] effectively behaves as a negative number, approaching zero as you approach infinity. However, for some bizzare reason, I'm still not sure why, I convinced myself that it should be the other way around, because the box on my desk had more GPE than when it was in the basement, and I had to do work to carry it up the steps. In that moment, it escaped me that that was only true when you have an essentially uniform gravitational field, and so I went with what I posted, and was in too much of a rush to double check it.

    mea culpa

    None of which detracts from the fact that all of the errors you pointed out were failures in your ability to follow what I posted. It took Pete's proof that they were equivalent for you to see the error that I actually made.
     
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