How large can planets be?

Discussion in 'Astronomy, Exobiology, & Cosmology' started by birch, Feb 1, 2011.

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  1. James R

    James R Just this guy, you know? Staff Member

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    Well, constant gravitational potential.

    Unstable in that the shell may drift into the star? That's what was said above.
     
    James R, Feb 6, 2011
    #41

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  3. kevinalm

    kevinalm Registered Senior Member

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    And I can choose the gauge to make the potential inside the shell some constant other than zero. The significant property of the potential inside the shell is that it is constant, therefor the gradient is zero. I know it's a minor point but strictly speaking you should refer to it as a constant.

    For Newtonian physics, the customary gauge is to set zero at infinity. Then the sum of the (negative) potential energy and the kinetic energy tells you whether the object is bound or not. Less than zero bound, greater than not.
     
    Last edited: Feb 6, 2011
    kevinalm, Feb 6, 2011
    #42

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  5. orcot

    orcot Valued Senior Member

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    Ones 1 side of the dyson sphere get's to close to the sun, wouldn't it's surface recieve more solar radiation propelling it backwards like a solar sail?
     
    orcot, Feb 6, 2011
    #43

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  7. James R

    James R Just this guy, you know? Staff Member

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    No. The net radiation pressure would be zero regardless of the Sun's location inside a spherical Dyson sphere. It's a consequence of the inverse-square law for intensity of the light.
     
    James R, Feb 6, 2011
    #44
  8. prometheus

    prometheus viva voce! Registered Senior Member

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    0 = a constant. You're both nitpicking.


    I also explained why the shell is unstable - it's due to the shell theorem, which is a consequence of the fact that the flux through a closed surface is proportional to the mass enclosed within it, which is Gauss' law.

    In all of physics it's customary to set up your problem in a way that is convenient. Would you set the potential = 0 at infinity for a Newtonian ballistics problem? It would obviously be a very stupid thing to do.

    I'd be very grateful if you would both stop trolling.
     
    prometheus, Feb 6, 2011
    #45
  9. James R

    James R Just this guy, you know? Staff Member

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    Yes, and I know you understand the point. There's no need to get all defensive about that small slip you made. Just admit your error and move on. No need to insist on being right.

    My original question to you was to ask you to explain what you meant by "unstable". You didn't explain that; somebody else did. Then you came in again saying that the explanation given wasn't what you'd meant at all, but you went on to repeat the same thing anyway.

    Also, trying to bamboozle me with talk of Gauss's law doesn't get you out of the initial question I asked you. I know you know what you're talking about. You don't have to prove it to me. What wouldn't hurt is your giving me a little credit too.

    It is, however, customary to set the zero of gravitational potential at infinity when you're talking about the gravitation of stars and planets (when you're not on the surface). Agree?

    I really don't see why you have a problem with a simple request for clarification. Maybe you've spent too much time dealing with cranks.
     
    James R, Feb 7, 2011
    #46
  10. prometheus

    prometheus viva voce! Registered Senior Member

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    I think it would be fair to say that I wasn't completely right but I wasn't wrong either. Is that fair?

    Sorry I don't have time to check up on what's happening here all that often. The explanation given by billvon was right, but it didn't explain the reasons for the instability, which I did explain.

    I'm not trying to bamboozle anyone. I rather like Gauss' law because it's easy to visualise and is extremely powerful, so I thought it was worth mentioning.

    Yes, but we're talking about the interaction between a shell and an internal star. All we care about is what happens inside the shell so it makes sense to me to set the zero gravitational potential position at the shell (which in turn means it's zero everywhere inside the shell as you know). Maybe I should have been more explicit about that, but that's my thinking.


    This is almost certainly true.

    Please Register or Log in to view the hidden image!

    Is everything clear enough now?
     
    prometheus, Feb 7, 2011
    #47
  11. James R

    James R Just this guy, you know? Staff Member

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    All good, prometheus.
     
    James R, Feb 8, 2011
    #48
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