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Hydrostatic equilibrium of a cylindrical area

  1. May 28, 2013 #1

    Consider a cylindrical region (length dr, end area dA) at a distance r from the center of the sun

    Density =ρ(r)
    Volume = Length x area = dr.dA (How is the formula for volume of a cylinder works here?)
    Mass= Density x Volume = ρ(r).dr.dA

    Now, computing F(grav) = -G.m(r).dm/r^2.......we get F(grav)=-g(r)ρ(r)drdA. I understand up to that.

    But the net pressure equation shows:

    F(press)=(P(r+dr))- P(r))dA)

    Can you please explain me how the above equation arrives?

  2. jcsd
  3. Jun 5, 2013 #2
    F(press)=(P(r+dr))- P(r))dA)

    Remember that force is pressure multiplied by area. This appears to be saying that the net force on the cylinder is the force on the top of the cylinder pushing downwards minus the force on the bottom of the cylinder pushing up.
  4. Jun 5, 2013 #3
    Hello colin456,

    Thanks a lot for the help. I could now understand the equation.
  5. Jun 6, 2013 #4
    Just one thing to be careful about. Do not forget about the sign of the force. Your gravity equation has a minus in it, which means the force is negative. Since gravity points towards the middle of the star, that means that positive force is by definition away from the centre of the star. However, the force from the pressure equation as you wrote it is going to be negative since P(r) is greater than P(r+dr) which means that it also will be directed towards the centre of the star, which would be wrong. I am sure you would notice this when you put the two equations together, but it is worth being clear about.
  6. Jun 10, 2013 #5
    Hello colin456,

    Sorry for the late reply. As you have written the 'minus' sign '-' means the force is negative. Correct me if I am wrong: the '-' sign is also as gravity is attractive not repulsive. Right? Yes the force from the pressure eqn.is -ve.
  7. Jun 10, 2013 #6
    There is no rule that says that negative force is attractive. The minus sign does not mean that gravity is attractive. It means that gravity is pointing in the negative r direction. This is what you want since you are using spherical coordinates with r=0 is defined as the centre of the star, and positive r values are pointing away from the centre of the star. The force in the pressure equation needs to be positive since it must be in the opposite direction to the gravity force.
  8. Jun 10, 2013 #7
    Can you tell me one thing: F=-G(m1.m2)/R^2. What does -G denotes? Pointing downwards?
  9. Jun 10, 2013 #8
    G is just a number. It stands for 6.67384 × 10-11 m3 kg-1 s-2 and is therefore always positive. The masses, m1 and m2, are also always positive, and R^2 must be positive because it is the square of a real number. Therefore, in order for the force to be negative, which it needs to be to point towards the centre of the coordinate system, the negative sign is necessary.
  10. Jun 10, 2013 #9
    Ok, clear.

    Thanks a lot.
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