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Two masses attached by a rod orbiting around the Earth

  1. Nov 14, 2015 #1
    Hello guys! I hope you can give me a hand with this one

    1. The problem statement, all variables and given/known data
    A satellite consisting of two masses attached by a rigid-massless rod of length L, are orbiting around the earth at a distance R from the centre of the earth. During the entire movement the rod stays oriented in the radial direction. Consider the earth to be stationary and the gravitational attraction between the masses to be negligible.
    IMG_20151114_143600.jpg
    A) Find the angular velocity of the satellite and the force the rod exerts on each mass.
    2. Relevant equations
    Force of gravity:
    For mass 1→ Fg1= -G·Me·m1/R^2

    For mass 2 → Fg2= -G·Me·m2/(R+L)^2​

    3. The attempt at a solution
    First thing I did was to consider the angular frequency of both masses to be the same, so:
    For mass 1 → T1 = 2π/ω1

    For mass 2→T2 = 2π/ω2​

    Thus angular velocity ω must be the same for both masses, (is this correct?)

    From the conservation of angular momentum (as there is no net torque), considering the centre of momentum to be the centre of the earth, I have the following:
    L=L1+L2 ⇒ L=(m1·R^2 + m2·(R+L)^2)·ω​
    And as L is constant then:
    ω=L/(m1·R^2 + m2·(R+L)^2)
    Finally from the equations of motion I got:
    For mass 1→ ∑F = Fg1 + Tension1 = m1·a=m1·R·ω^2

    For mass 2→ ∑F = Fg2 + Tension2= m2·a=m2·(R+L)·ω^2
    So, from this two equations I seem to have everything I need, but I think I'm missing something here and I just can't see what it is
    Thank you guys for your help!

     
    Last edited: Nov 14, 2015
  2. jcsd
  3. Nov 14, 2015 #2

    mfb

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    2016 Award

    Staff: Mentor

    Sure.

    I don't think introducing angular momentum helps, as you do not know its value.

    You can express the accelerations (they will be different) in terms of the angular velocity and the given distances.
     
  4. Nov 14, 2015 #3
    Thanks for your help!
    That's really helpful. So from both equations I'll get:
    For mass 1→a1 = R·ω^2
    For mass 1 →a2 = (R+L)·ω^2
    and then
    Tension1 = (G·Me/R^2 + R·ω^2)·m

    Tension2 = (G·Me/(R+L)^2+(R+L)·ω^2)·m

    (m1=m2=m)
    But are tension 1 and 2 the same (in absolute value) but opposite? If so then I think that's all, I can sum the two equations above and solve for ω and then replace to find the tension.

    Praise be unto He
     
    Last edited: Nov 14, 2015
  5. Nov 14, 2015 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Sure.
    Right.

    Helium is important!
     
  6. Nov 14, 2015 #5
    Great, thank you!

    I agree, Helium is important!
     
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