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Orbit change due to hypothetical mass change

  1. Apr 3, 2014 #1
    Hi everyone,

    I want to discuss the following question:

    What would happen to the distance between two objects with masses [itex] M [/itex] and [itex] m [/itex] with [itex] M >> m [/itex] if the central mass [itex] M [/itex] was time-dependent (e.g. the moon's orbit if the mass of the earth would increase slowly)?

    My theory goes like this:

    The orbit equation in the above case can be written as


    r = \frac{l^2}{m^2 \; \gamma} \frac{1}{1+e}, \ \ \ \ \ (1)


    where r is the separation distance between the two bodies and \theta is the angle that [itex]\mathbf{r}[/itex] makes with the axis of periapsis (also called the true anomaly). The parameter [itex]l[/itex] is the angular momentum of the orbiting body about the central body, and is equal to [itex]mr^2\dot{\theta}[/itex].
    The parameter [itex]\gamma[/itex] is the constant for which [itex]\gamma/r^2[/itex] equals the acceleration of the smaller body (for gravitation, [itex]\gamma[/itex] is the standard gravitational parameter, GM). The parameter e is the eccentricity of the orbit.
    (See: http://en.wikipedia.org/wiki/Orbit_equation)

    For making it easier, as a first approach, I would like to look at the special case of a circular orbit ([itex] e = 0 [/itex]) in which equation (1) reduces to:

    r = \frac{l^2}{m^2 \; \gamma}. \ \ \ \ \ (2)

    From equation (2) and the definitions of [itex]l[/itex] and [itex]\gamma[/itex] it follows that another valid expression for r is:

    r = \sqrt[3]{\frac{G \; M}{(d\theta / dt)^2} }. \ \ \ \ \ (3)

    Since the orbital period is [itex] T = \frac{2 \pi r}{v} [/itex] and the tangential velocity [itex]v[/itex] should be unaffected by a change of the central mass it follows that [itex] \frac{d \theta}{d t} \propto r^{-1} [/itex].

    Together with equation (3) I come to the conclusion that the following relation is valid:

    r \propto G \; M

    and since G is a constant:

    r \propto M.

    So a change of for example [itex] \frac{\Delta M}{M} = 10^{-11} [/itex] in the mass of the central body should lead to the same relative change in the distance r.

    Maybe this is one possible explanation for the 3.8 cm per year increase of the distance between the earth and the moon, but I'm not sure if my derivation is correct.

    Last edited: Apr 3, 2014
  2. jcsd
  3. Apr 3, 2014 #2
    Actually, I think the tangential velocity of the moon would be affected by a change in mass of the earth: it would change slightly as a radial component of velocity would be introduced briefly and then resume a new tangential speed in a new orbit.

    Also I think it's more complicated than what you have outlined since the earth-moon system has a constant angular momentum associated with it. So changing the mass of the earth would change the angular momentum of the system as well.

    And this is the actual reason the moon-earth distance is increasing: the earth's rotation is slowing down because of tidal friction (not increasing mass) and this transfers to the moon which increases it's orbit accordingly.
  4. Apr 3, 2014 #3
    This is one reason the moon-earth distance is increasing. But new measurements show that it can't be the only reason.

    Tidal friction explains only 2.9 cm/yr, but the Lunar Laser Ranging Experiment (LLRE) reports 3.8 cm/yr. So 0.9 cm/yr remain unexplained.

    (See: http://www.lpi.usra.edu/meetings/lpsc2013/pdf/2436.pdf)
  5. Apr 3, 2014 #4
    Your reference cites uncertainties that overlap in the values which suggests that maybe the discrepancy is just experimental error.

    Also, the 3.8cm number represents current measurements while the 2.9cm is based on geological measurements from 300 million years ago. It's possible that the rate of change has not been constant over that time period.
    Last edited: Apr 4, 2014
  6. Apr 4, 2014 #5
    For the LLRE measurements I can't see such an overlap.
    And I think the LLRE measurements are more reliable than the data taken from eclipse observations are.

    Yes, of course this is possible.
    But there is still the fact that for present values we have a discrepancy between the modelled value and the most exact measurement of about 30 %. So I think the search for additional explanations makes sense.
  7. Apr 4, 2014 #6
    Your reference cites the LLRE measurement as 3.82+/-0.7cm/yr.

    It also cites the tidal measurements as 2.9+/-0.6cm/yr.

    So it looks like there is an overlap.
    Last edited: Apr 4, 2014
  8. Apr 4, 2014 #7
    Ok, I see your point.
    The reference has obviously an error, because in the first lines the uncertainty of the LLRE measurements is 0.07 cm/yr (point before the zero) and later on it is 0.7 cm/yr.

    But still the 2.91 cm/yr which the detailed numerical simulation predicts is outside of the 3.82+/-0.7cm/yr.
    In the reference they write:
  9. Apr 4, 2014 #8


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    It should probably be noted here, that discussions of personal theories are not allowed on PF as per the rules.
    Similarly, papers sourced need to be from peer-reviewed publications. Conference proceedings usually do not fall into that category, and I can't find the article mentioned in post #3 as being published in any of the accepted journals.
  10. Apr 4, 2014 #9

    D H

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    I've closed this thread.

    First, a minor correction:
    The measurements from the Lunar Laser Ranging Experiment yield a recession rate of 3.82±0.07 cm/year, not 3.82±0.7.

    The current rate is high compared to the average over the last 620 million years, 2.17±0.31 cm/year (G.Williams, "Geological constraints on the Precambrian history of Earth's rotation and the Moon's orbit". [Reviews of Geophysics 38 (1): 37–60 (2000).)

    I don't know where the paper cited in post #3 got it's numbers, particularly the value from "detailed numerical simulations."

    The value of 2.17±0.31 cm/year is an average rate. The lunar recession rate strongly depends on the layout of the Earth's continents. The recession rate was low during periods of supercontinents and when the continents allowed a freer tidal flow. The recession rate is high right now because the Americas and Africa/Eurasia create two huge barriers to a free tidal flow. There's more tidal friction now than there has been in the past, so the recession rate is higher now than it has been on average.

    This site is not the place for speculative nonsense such as an expanding Earth. This thread is closed.
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