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An oscillating mass within an orbiting spacecraft question

  1. Sep 21, 2016 #1
    1. The problem statement, all variables and given/known data

    A wire stretches from one side of an Earth-orbiting space station to the other through the centre of mass of the station, such that it always points along a radial line to the centre of the Earth. A bead is threaded on the wire and initially rests 1.0 m from the centre of mass position, on the side away from the Earth.
    Ignoring friction between the wire and bead, determine the time it takes for the bead to move 2.6 m further away from the Earth, given the space station orbits at an altitude of 216 km.


    2. Relevant equations
    [itex]\left |\vec{g}\right |=\frac{GM}{r^2}[/itex]

    3. The attempt at a solution
    I've not managed to make any mathematical progress since I don't know how to apply the physics.

    Specifically: ignoring the orbit for now, the bead wouldn't undergo SHM about the centre of mass (COM). Therefore I think that it must be a superposition of the bead attraction to the CoM and its 'centrifugal' tendency to move away from the Earth, is the equation of motion I need to set up to then solve for time. But I don't have a clue as to how to quantify centrifugal motion, given that the centripetal based Newton's II Law is not applicable here (I think at least!) ?
     
  2. jcsd
  3. Sep 21, 2016 #2

    The Buttered Cat

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    Start by asking what laws of physics is the bead obeying. Usually F=ma is a good place to start, and you know there is no angular acceleration since it is stuck on the wire. Can you write Newtons second law for the bead?
     
  4. Sep 21, 2016 #3

    haruspex

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    Suppose the bead is, at some point in time, x further away from Earth than is the mass centre of the satellite.
    ##\ddot x## is then the radial acceleration of the bead relative to the satellite.
    If the satellite's orbit has angular rate ##\omega##, write an equation relating its centripetal acceleration to the gravitational force on it.
    Write the corresponding equation for the bead.
     
  5. Sep 22, 2016 #4
    Thanks both. Will have a proper think and post my resulting equations in the morning, and will then likely benefit from some further guidance :)
     
  6. Sep 23, 2016 #5
    Edited RE haruspex's post below:

    Symbols: {R, M, m_s, m_b} = {satellite orbital radius, Earth's mass, satellite mass, bead mass}

    Here's where I've got to: assuming a circular orbit of the satellite, N2L says: [itex]\frac{GMm_s}{R^2}=m_sr\omega^2[/itex]
    and N2L applied to bead says: [itex]\frac{Gm_sm_b}{(R+x)^2}=m_b\ddot{x}[/itex]

    which clearly cancel to:
    [itex]\frac{GM}{R^3}=\omega^2[/itex] & [itex]\frac{Gm_s}{(R+x)^2}=\ddot{x}[/itex]
     
    Last edited: Sep 23, 2016
  7. Sep 23, 2016 #6

    haruspex

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    No. I defined x as the extra distance beyond R. ##\ddot x## is therefore the relative radial acceleration, not the total acceleration. And the distance from the Earth's centre should be R+x.
     
  8. Sep 23, 2016 #7
    NB I edited my previous post to included R+x etc.

    Please could you clarify what you mean by the above sentence.

    I still don't understand how one can incorporate the attraction the bead feel towards the satellite COM. Something occurred to me: the gravitational coupling of the bead and the space station should be ignored. Otherwise the satellite would also oscillate and the whole thing would get pretty hideous! It wouldn't be SMH since the restoring force due to gravitational attraction is ~1/r^2.
    Is this reasoning valid?

    I'm still pretty lost!
     
    Last edited: Sep 23, 2016
  9. Sep 23, 2016 #8

    haruspex

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    Take the bead as being at distance R+x from Earth's centre, x<<R.
    Let the orbital rate be ω.
    You have an equation relating R, ω, G and M. This gives you an expression for ω. If you were to write the same equation for r+x you would get a different ω, but ω is constarined to be the same for both. So instead, write the centripetal acceleration of the bead that corresponds to the satellite's ω (but at the bead's radius) and the acceleration provided by Earth's gravity. The acceleration of relative distance (I should not have called it relative acceleration earlier), ##\ddot x##, is the difference between the two.
     
  10. Sep 24, 2016 #9
    Ok thanks, I still don't follow the justification for establishing these equations, when the observer is in the rotating frame (sat in the satellite measuring the bead's motion). I'll have to leave it I think. But thanks anyway.
     
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