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Earth & moon rotate round center of mass

  1. Jun 9, 2009 #1
    1. The problem statement, all variables and given/known data
    Calculate the time the moon rotates once around the earth, and show that you have to add 4 hours to that since the moon & the earth rotate around their center of mass. hint: use the reduced mass: [tex]\overline{m}[/tex]=[tex]\frac{m_{1}m_{2}}{m_{1}+m_{2}}[/tex]
    mass of moon: m=0.0734E24[kg]
    mass of earth: 81 times mass of moon: M=81m
    radius of earth: Re=6.37E6[m]
    distance centers earth-moon: 60Re
    gravity constant: G=6.673E-11

    2. Relevant equations
    (1) F=[tex]\frac{GMm}{r^{2}}[/tex]
    (2) F=[tex]\frac{mV^{2}}{r}[/tex] (3) V=[tex]\frac{2r\pi}{T}[/tex]
    xc.m.=[tex]\frac{m_{1}x_{1}+m_{2}x_{2}}{m_{1}+m_{2}}[/tex]
    maybe: the momentum in the center of mass coordinate system: (4)[tex]\vec{p_{1}}[/tex]=m1[tex]\vec{V_{1}}[/tex]=[tex]\overline{m}[/tex][tex]\vec{V_{12}}[/tex], while V12 is the relative velocity between them, and [tex]\overline{m}[/tex]=[tex]\frac{m_{1}m_{2}}{m_{1}+m_{2}}[/tex] is the reduced mass.

    3. The attempt at a solution
    the revolution time when the earth is "stuck" in place and the moon rotates only: using equations: (1)=(2), (3): [tex]\Rightarrow[/tex] [tex]\frac{mV^{2}}{60R^{e}}[/tex]=[tex]\frac{81Gm^{2}}{(60R_{e})^{2}}[/tex] [tex]\Rightarrow[/tex] T=2,357,028[sec]=27.28[days], and the relative velocity is: V12=1018.8[m/sec].
    the distance of the center of their mass from center of earth: xc.m.=[tex]\frac{60R_{e}m}{82m}[/tex]=4,660,976[m]
    now, in the coordinate system of the moving center of gravity, i find the moon's tangential velocity, knowing that the relative velocity V12 is the same in all coordinate systems: according to equation (4) above: V1=[tex]\frac{\overline{m}V_{12}}{m_{1}}[/tex]=1006.3[m/sec]
    now, the revolution time of the moon round the distance to center of mass (3): [tex]\frac{2\pi(60R_{e}-x_{c.m.})}{1006.3}[/tex]=2,357,187[sec]

    this is only 159[sec] more, not 4 hours!
     
  2. jcsd
  3. Jun 9, 2009 #2
    Have you considered using Kepler's law for a rotating binary system?

    [itex] \Omega^2 = \frac{4\pi^2}{P^2} = \frac{G(m_1+m_2)}{R^3 }[/itex]

    (P is the period of time it takes for the binary to orbit one another)

    We know that: [itex] m_1r_1=m_2r_2 [/itex], so R is defined as [itex] r_1+r_2 [/itex]. r1 and r2 are the radii from the centre of mass of the two objects to the objects themselves.

    I'm not sure it'll work if they asked you to use the reduced mass, but it might work - it's how i'd go about trying to solve it anyway.
     
    Last edited: Jun 9, 2009
  4. Jun 10, 2009 #3
    Thanks, Faloren, but I only know the 3 Keppler laws for eliptical motion, not this one.
    and also this gives a better result: 2 hours difference, but not 4.
     
  5. Jun 10, 2009 #4
    did you use reduced mass AND non-infinite earth mass? I think one makes the other unnecessary.
     
  6. Jun 10, 2009 #5
    What is non-infinite mass? give me a clue, but i should use the reduced mass of both, the earth & moon as given by the formula.
     
  7. Jun 10, 2009 #6
    in the first calculation you treated the earth as being stationary. you treated it as though it had infinite inertial mass. (but not infinite gravitational mass)

    I'm not an expert but I'm pretty sure that when you use the reduced mass for the moon you still treat the earth as though it had infinite inertial mass. thats why its so much easier to use.
     
    Last edited: Jun 10, 2009
  8. Jun 10, 2009 #7
    I am not that expert either, and i don't expect you to explain to me what are infinite inertial mass & not infinite gravitational mass, but the formula:
    [itex]\overline{m}=\frac{m_{1}m_{2}}{m_{1}+m_{2}}[/itex]
    makes use of them both.
    I think this is not the direction, but using the moving center of mass coordinate system, in which this formula applies to.
     
  9. Jun 10, 2009 #8
    by infinite inertial mass I just mean that you treat it as though it were stationary
     
  10. Jun 10, 2009 #9
    By infinite mass, that's not entirely true - it's more like you treat the smaller object as having negligible mass, right?

    The idea being that if you're working out, say the orbit of the Earth about the Sun, you don't need to know the mass of the Earth because the sun's mass will dwarf it (i.e. as you say, it won't affect the orbit of the sun - the sun is for all intents and purposes stationary - although on an interesting note this is how astronomers detect exo-planets these days, by measuring stars wobbling).

    Oh, and inertial mass = gravitational mass, this was one of Einstein's assumptions when developing general relativity i believe - no experiment has yet found a difference between the two.
     
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