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Integration constants, gravitational potential of sphere

  1. Dec 13, 2014 #1

    Vir

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    1. The problem statement, all variables and given/known data
    So I'm calculating the gravitational potential of a sphere at at point P. R = radius of sphere, r = distance from center of sphere to point P. I'm looking at two scenarios; r > R (1) and r < R (2). So I have the following integral:

    \begin{equation} V(r) = \int \frac{-GM}{Rr} ds
    \end{equation}

    3. The attempt at a solution
    Now, for (1) I integrate from r-R to r+R. However for (2) I'm thinking I have to integrate from -(R-r) to R+r, but this gives me the same constants as in (1). To my understanding the potential is supposed to be constant for all r < R, but then my constants are giving me the wrong answer.

    If I remove the - and integrate from R-r to R+r I get a correct answer, but this makes no sense to me. That would mean I was integrating from the center of the sphere out to the edge, effectively ignoring half of the sphere? What am I missing?
     
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  3. Dec 13, 2014 #2

    PeroK

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    I'm confused about what you are trying to do. ##\frac{-GM}{r}## is the gravitational potential for a solid sphere at a point outside the sphere, a distance r from the centre.

    What are you trying to calculate?
     
  4. Dec 13, 2014 #3

    Vir

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    I'm trying to arrive at the gravitational potential at a distance r < radius of sphere(R). The integral is

    \begin{equation}
    V(r<R) = \int \frac{-GM}{2rR}ds,
    \end{equation}

    but I'm not sure how to select my integration limits.
     
  5. Dec 13, 2014 #4

    PeroK

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    Is this a hollow sphere? And are you trying to integrate over a surface?
     
  6. Dec 13, 2014 #5

    Vir

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    Yes, a hollow sphere. I'm integrating small circles of the surface of the sphere. I guess this image describes it well. 500px-Shell-diag-1.png
     
  7. Dec 13, 2014 #6

    PeroK

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    That's a point r outside the sphere. In that diagram, you first need to work out the gravitational potential of the ring marked in blue. So you're going to need to take θ into account and calculate dM as well.
     
  8. Dec 13, 2014 #7

    Vir

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    Ok. So
    \begin{equation}
    dV(r<R) = \frac{-GdM}{S}
    \end{equation}

    \begin{equation}
    dM = \frac{M}{4\pi R^2}d \theta R Sin \theta R 2 \pi
    \end{equation}

    \begin{equation}
    dM = \frac{M}{2}Sin \theta d \theta
    \end{equation}

    \begin{equation}
    S = \sqrt{R^2+r^2-2Rr Cos \theta}
    \end{equation}

    \begin{equation}
    \frac{dS}{d \theta} = \frac{Rr Sin \theta}{S}
    \end{equation}

    Inserting in dV(r<R):

    \begin{equation}
    dV(r<R) = \frac{-GM Sin \theta S ds}{2SRr Sin \theta}
    \end{equation}

    \begin{equation}
    dV(r<R) = \frac{-GM}{2Rr} ds
    \end{equation}

    \begin{equation}
    V(r<R) = \int \frac{-GM}{2Rr} ds
    \end{equation}

    Is that right?
     
  9. Dec 13, 2014 #8

    PeroK

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    Not quite. If you think in terms of the force exerted by the ring, what is the direction of the force?
     
  10. Dec 13, 2014 #9

    Vir

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    When the ring is to the right of P the force points left, and vice versa.
     
  11. Dec 13, 2014 #10

    PeroK

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    You need to think more carefully. Remember that force is a vector.
     
  12. Dec 13, 2014 #11

    Vir

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    Yea, ok. I was thinking since each point on the circle has a point on the opposite side where the perpendicular component of the forces would cancel out the net force would be parallell to the axis through the center of the circle and P. Not so?
     
  13. Dec 13, 2014 #12

    PeroK

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    Yes, the perpendicular forces cancel. But what about the magnitude of the remaining force?
     
  14. Dec 13, 2014 #13

    Vir

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    That would depend on the mass at P right? In any case it's
    \begin{equation}
    F = -\frac{GMm}{(r-R Cos \theta)}
    \end{equation}
     
  15. Dec 13, 2014 #14

    PeroK

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    Take a look again at your diagram and make s a vector. That's the direction of the force. So, you have a "vertical" component and a "horizontal" component. The magnitude of the horizontal component is not the magnitude of the full force.
     
  16. Dec 13, 2014 #15

    Vir

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    Then I don't know. You asked about the magnitude of the remaining force, isn't that the horizontal component of the total force?
     
  17. Dec 13, 2014 #16

    PeroK

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    Yes, but the vertical force is ##Fsin(φ)## and the horizotal force is ##Fcos(φ)##
     
  18. Dec 13, 2014 #17

    Vir

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    Right. So what do I do with this?
     
  19. Dec 13, 2014 #18

    PeroK

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    You need to factor ##cos(φ)## into your equation. That's all your missing, I think.
     
  20. Dec 13, 2014 #19

    Vir

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    I don't know. My integral gives me the right answer for R>r:
    \begin{equation}
    V(r>R) = \frac{-GM}{2rR} \int_{r-R}^{r+R} ds
    \end{equation}

    Simplifies to:

    \begin{equation}
    V(r>R) = \frac{-GM}{r}
    \end{equation}

    I think the point is that when r<R then the rings both to the left and right of P have a positive distance to P, since the potential is supposed to be negative. I guess one has to think more in terms of polar coordinates? Thus, when r<R you have to integrate from R-r(closest side of sphere) to R+r(far side). This gives

    \begin{equation}
    V(r<R) = \frac{-GM}{2rR} \int_{R-r}^{R+r} ds
    \end{equation}

    Simplifies:

    \begin{equation}
    V(r<R) = \frac{-GM}{R}
    \end{equation}
     
  21. Dec 13, 2014 #20

    PeroK

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    I think you've got two mistakes cancelling out for the problem outside the sphere. But, they won't cancel inside the sphere.

    You've got ##dV = -GdM/S## but that assumes that all the mass in the ring is concentrated at one point a distance S. You need to take cos(φ) into account. The potential must be:

    ##dV = (-GdM/S)cos(φ)##

    If you imagine being very close to the ring, then there would be almost no potential. In fact, when φ = π/2, the potential will be zero.
     
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