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Gravitational pressure dependence on volume

  1. Jun 3, 2016 #1
    1. The problem statement, all variables and given/known data
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    2. Relevant equations


    3. The attempt at a solution

    Consider a very thin shell of width dr at a distance r from the center . The volume of this shell is ##4 \pi r^2 ## . Mass is ## m = 4 \pi r^2 \rho## .

    P is the pressure at distance r . Gravitational acceleration at distance r < R is ##g' = \frac{GMr}{R^3}##

    Doing a force balance on this thin shell .

    ##P(4 \pi r^2) - (P+dP) (4\pi r^2) = mg'##

    ## - dP (4\pi r^2) = (4 \pi r^2 \rho)\frac{GMr}{R^3}##

    Integrating under proper limits ,

    $$ P(r) = \frac{GM\rho}{2R} \left ( 1 - \frac{r^2}{R^2} \right )$$ . V is the volume of the star .

    Now , at the surface P(r = R) = 0 .

    Is the question wrong ? Should the question be asking about the pressure "at the center" instead of "on the surface" ?


    Thanks
     

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  3. Jun 3, 2016 #2

    phinds

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    Derive the equation for the gravitation attraction of a point on the surface in terms of radius and mass and then express radius as a function of volume. And you'll have to get rid of the mass term by expressing it in terms of volume and density (density is a constant you can ignore for the purposes of this problem)
     
    Last edited: Jun 3, 2016
  4. Jun 3, 2016 #3
    Isn't gravitational pressure 0 at the surface ? Is the problem statement correct ?
     
  5. Jun 3, 2016 #4

    phinds

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    If it were, what would keep the star from just evaporating from the outside inward ? Just figure out the gravitational force pointing in and the pressure is the same force pointing out.
     
  6. Jun 3, 2016 #5
    That gives option 3) as answer .
     
  7. Jun 3, 2016 #6

    phinds

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    not what I got, but I was pretty off-the cuff w/ the math. You're probably right.
     
  8. Jun 3, 2016 #7
    Do you get option 3) as answer ?
     
  9. Jun 3, 2016 #8

    phinds

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    Uh ... "not what I got" was a direct response to your statement that you got #3.

    If you were careful with the math, you probably got it right. Don't worry about what I got.
     
  10. Jun 3, 2016 #9
    But I surely need to worry about how you have interpreted gravitational pressure . Sorry , but I think you have got it wrong . Gravitational pressure is 0 at the surface . It is not the same thing as gravitational force .
     
  11. Jun 3, 2016 #10

    phinds

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    Well, there again you may be right. I'm no expert. I just looked up gravitational pressure and it showed a diagram with "gravitational pressure" pointing out exactly balancing "gravity" pointing in.

    Also, I go under the assumption that a problem statement is not nonsense (which is not ALWAYS right but certainly ALMOST always).
     
  12. Jun 3, 2016 #11
    Last edited: Jun 3, 2016
  13. Jun 3, 2016 #12

    TSny

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    ##P(4 \pi r^2)## does not give the total force on a spherical surface. The net force due to pressure on a spherical surface will be zero. The pressure force on a small element of the spherical surface is opposite in direction to the force at a diametrically opposite point.

    Instead, consider a small, thin cylinder of height ##dr## oriented as shown below and located a distance ##r## from the center of the star.

    Yes, I think the problem meant to ask about the pressure at the center.

    There is also an ambiguity in the question. When you are determining how the pressure at the center varies with volume, does the total mass of the star remain constant as the volume varies, or does the density remain constant as the volume varies?
     

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  14. Jun 3, 2016 #13
    Thanks for replying :smile:
    I get same result as in OP $$P(r) = \frac{GM\rho}{2R} \left ( 1 - \frac{r^2}{R^2} \right )$$
    Ok
    I think the later i.e density remains constant as the volume varies .
     
  15. Jun 3, 2016 #14

    TSny

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    OK, I think that's right.

    If so, I don't get one of the choices.
     
  16. Jun 3, 2016 #15
    No . Looking at the options , I think it is the mass that remains constant and density varies .

    If density remains constant then ## P (center) ∝ V^{\frac{2}{3}}## . But if mass remains constant then ## P (center) ∝ V^{\frac{-4}{3}}## which is an option given in the problem statement .
     
  17. Jun 3, 2016 #16

    TSny

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    OK. Good.
     
  18. Jun 4, 2016 #17
    I am attaching the solution provided with this problem . I can't make a head or tail of it o_O. Can you understand how it is being done in the solution ?
     

    Attached Files:

  19. Jun 4, 2016 #18
    Nope the gravitational acceleration at distance r<R is ##\frac{G\rho V_r}{r^2}## where ##V_r## is the volume of a sphere with radius r.
     
  20. Jun 4, 2016 #19
    ##\frac{GMr}{R^3}## and##\frac{G\rho V_r}{r^2}## are equivalent :smile:
     
  21. Jun 4, 2016 #20

    haruspex

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    No, I don't understand the reasoning there. That dW would be the work required to change the volume by dV at constant pressure P. Why that should relate the pressure at the centre of the star to the total gravitational PE I have no idea.

    Other than that, it looks like you have your answer.
     
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