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Roche limit

  1. Oct 1, 2016 #1
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    1. The problem statement, all variables and given/known data

    Show that d=(9M/(4*pi*p))^1/3 is an approximation for the roche limit. Note that x/d <<1
    with M = mass of the primary
    p = density of the secondary
    x= distance of a test particle from the center of the secondary (in part a) of the task one should give the motion equation for X. X is a particle that is situated at distance x from the center of the secondary (on a line that goes from the center of the secondary to the center of the primary)
    m = mass of the secondary

    2. Relevant equations


    3. The attempt at a solution
    In the critical point the tidal force excerted from the primary and the gravitational force that hold the secondary together are equal and therefore the secondary breaks. So I tried to put these forces equal and to solve the equation for d, but it did not work out.
     
    Last edited: Oct 1, 2016
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  3. Oct 1, 2016 #2

    gneill

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    Staff: Mentor

    Hi physicstudent_B, Welcome to Physics Forums.

    Can you show us the details of what you tried and point out where you think it went wrong or failed?
     
  4. Oct 2, 2016 #3
    Thank you for your message and for welcoming me to the Physics Forum.
    So the first part of this question was to write the equation of motion for X:
    I think that there are two forces, who effect particle X: the gravitational force from the secondary FD and the gravitational force from the primary FO
    FO+FD = mx * w2d (D is orbiting around O with constant velocity and D is not moving)
    FO = (G*mx*M)/(d-x)^2
    FD= (here I am not so sure, because obviously X is situated somewhere inside of D if I understood the question right, but from another exercise I remember a formula that you can use to calculate the gravitational force the Earth exerts on a body of mass m that is inside the Earth somewhere at distance r from the center of mass: F=mgr/R (R= radius of the Earth) --> so I tried to use this formular to express the force the primary exerts on the particle X :
    FD=(mx*G*m*x) / R^3
    Now I tried to put FO=FD and resolve for d, but I got stuck
    I think I should use the binominal approximation for calculatin FO (so that 1(d-x)^2 = 1/d^2 * (1+x/d), also I think that I should express the mass in terms of volume (4/3*pi*r^3) * density, but I don't get close to the term I should get, so I wonder if there is something wrong with how I calculate the two forces.
     
  5. Oct 2, 2016 #4

    mfb

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    You can assume that the small test mass is on the surface of the object. While it does not matter for the calculation, the object cannot fly away if it is inside.

    Be careful with the signs. The gravitational forces point in opposite directions. If you want to add them as in your first equation one has to be negative (which one?).

    In the first equation: The radius of the orbit of x is not d.

    You are missing two equations:
    w as function of the given parameters.
    A formula relating m/R3 and the density.

    Once you have those and fixed the sign issue and the radius, you should be able to solve for d.
     
  6. Oct 2, 2016 #5
    Thank you. So with the signs right it should be:
    FO-FD= mx*w2*d
    and the formular relating m/R3 and density would be m/R^3 = p * 4/3 * pi
    I have not yet an idea for w as function of the given parameters.
    And if X is on the surface of O, the x would be the same as the radius of O?
     
  7. Oct 2, 2016 #6

    mfb

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    Your mass m is in a circular orbit around the bigger mass.
    Right.
     
  8. Oct 3, 2016 #7
    OK, w2=GMm/d3
    Then I could solve for d using the equation of motion of the particle X.
    But I did not yet completely understand: I thought that FO and FD would be the same at distance d, so that FO-FD=0, but that is obviously not the case?
     
  9. Oct 3, 2016 #8

    mfb

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    You'll see if you can neglect the orbital motion later. You cannot just assume it to be negligible.
     
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