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Challenge 15: Worshipping the Moon

  1. Mar 8, 2014 #1
    Many of you don't know this, but as a young man, Greg once decided to worship the moon. He was so obsessed by the moon that he once decided to start following it. So at any given moment, he would check where the moon is and then walk in that direction. Greg has special powers so that he can see the moon always, even when it's "beneath the earth". Also, Greg can walk with a constant speed (much less than the moon of course) without ever resting. He can also walk over water, etc.

    The question to you is: where did Greg end up?

    Try to find an attractive mathematical model that descibes this situation but isn't too complicated. Then figure out what happens to Greg's walking path asymptotically. Is this dependent of where Greg starts? Prove it!
  2. jcsd
  3. Mar 8, 2014 #2


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    I can be missing something, but if he is fast enough, he is always exactly under the Moon - so all he does is circling the Earth all the time.

    Or does "much less than the Moon" mean "much less than the projection of the Moon position on the Earth surface"?
  4. Mar 8, 2014 #3
    Yes, we assume he can't catch up with the moon, so he can't be under the projection of the moon all the time.
  5. Mar 8, 2014 #4


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    The projection of the moon is moving around the globe roughly once per 24 hours, so its projection is moving roughly as fast as the speed of sound (+- factor 2).
    I am sure Greg can indeed move that fast. He is forum admin!
    It would also explain something else... (insider)

    Also, xkcd did something similar

    Some initial thoughts:

    Approximate the earth as a sphere. First note that the actual distance of the moon does not matter - Greg, the center of earth, the moon and its projection to the surface are always in one plane, and this plane is sufficient to determine the walking direction.
    Let's assume Greg does not walk at full speed, so the moons's projection is faster than him.
    Convert the rotation of earth to a rotation of the moon, roughly once per 24 hours (feel free to calculate a more precise number, I'll just call it "1 day" here).

    Model 1
    The simplest nontrivial model is a circular moon orbit directly above - or at - the equator. In this case, we can also add the orbit of moon to our rotation, as both happen in the same axis.
    Then we have two special cases:
    - Greg is at the equator - then he moves back and forth forever.
    - Greg is circling the north or south pole once in 1 day, and the moon is always towards the pole for him.
    For the general case: in a static Gregcentric coordinate system, the moon direction is perfectly symmetric within a day. Therefore, in first order, Greg will always make a circle with a period of 1 day. With a normal human walking speed of 5km/h, this circle has a radius of ~20km. Second-order effects are suppressed with a factor of 20km/6370km = 1/320, so it will take at least of the order of 1 year to shift this circle by 20km. Due to the east/west symmetry, I would not expect east/west shifts as second order effect.

    I did not find a clear definition of "young man", but it probably does not involve a timespan of more than 30 years. Therefore, chances are good he did not went so far away from his starting position. If the second order happens to be zero, chances are good he would be within 100 km of his origin.

    Model 2
    If we ignore the orbit of moon, but consider an inclination (so the moon is over some latitude != 0), we get the same case as for Sirius. Greg walks towards the appropriate pole, circling it forever (at least in the very long [strike]run[/strike] walk).

    Model 3
    Combine earth rotation and an inclined orbit of the moon. As seen from a non-moving observer, the moon now oscillates in its latitude every ~28 rotations. Half of the time, Greg is driven north, the other half he is driven south, as established in model 2. Again, symmetry tells us that any net motion, if it exists, has to be very small. Similar to model 1, I expect that we can still find Greg somewhere close to his starting position.

    Model 4
    Consider the eccentric orbit of moon. This is left as an exercise for the reader.
  6. Mar 8, 2014 #5


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    EDIT: much of the original version of the post was nonsense. Here's another attempt!

    Taking mfb's model 1, a circular orbit over the equator I get different conclusions. Work in a coordinate system fixed to the (rotating) earth.

    At a fixed instant in time, visualize the field of Greg's velocity vector, at all points on the earth's surface. The velocity field is simple: from any point on earth, Greg walks along a great circle to the point where the moon is directly overhead.

    If Greg is on the equator, he walks alternately east and west. But because of the moon's rotation, he will spend more time each day walking with the moon traveling "away" from him "towards" him, so his time-averaged position will continually circle the equator (at a low average velocity), with his east-west oscillations superimposed on that.

    At a latitude close to each pole, he can walk continuously west, with the setting moon always on the western horizon.

    At intermediate latitudes, the north-south component of his velocity changes 90 degrees out of phase with the east-west component. So he will probably walk on a path that looks like "nutation" of a rotating object that is precessing. In other words, on average a slow rotation in the same direction as the moon, with smaller "loops" superimposed on that.

    I say "probably", because the he walks for a longer time towards the west that the east, and it's not obvious to me whether he spends equal amount time walking towards the north and south, and whether the distances traveled north and south in each "loop" are equal. If they are not, he will probably end up either at the equator, or walking in a small circle round one of the poles.

    What happens very close to the poles, inside that small circle, and whether the continuous circle westwards around the poles is a stable situation, I'm not sure.
    Last edited: Mar 9, 2014
  7. Mar 9, 2014 #6


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    Ah, good point, so he does some net westwards motion.

    With the moon velocity V and his velocity v, we have ##\frac{\pi r}{V+v}## time eastwards and ##\frac{\pi r}{V-v}## time westwards, for a net motion of ##v \pi r \left( \frac{1}{V-v} - \frac{1}{V+v} \right) \approx 1d \frac{v^2}{V} = (1d)^2 \frac{v^2}{2 \pi r}## where 1d is 1 day.

    With v=5km/h, this gives 360m net motion per day or 128km per year.
  8. Mar 10, 2014 #7


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    There seems to be a useful symmetry. Instead of regarding the moon-seeker as chasing the point under the moon, one can consider him as a fleeing from its antipodal point. His path as he flees from an antipodal point will be the time-reversal of the path he would use when chasing the moon.

    Suppose that the moon-seeker is in the Northern hemisphere near the equator when the moon['s projection] makes its point of closest approach x meters away to the south. The following is an attempt to prove that his path makes a stable cycle.

    Divide the cyclic path into four unequal parts. The first segment of the path is a chase of a retreating moon. It starts at the point of closet approach. It ends roughly six hours later when the moon and its antipodal point are equidistant from the moon seeker. Call this segment "A". The second segment of the path is a flight away from an approaching antipode. It starts at the end of segment A and ends when roughly six hours later when the antipode makes its point of closest approach.

    Claim: The paths described by these two segments are mirror images of one another. It follows that his latitude of the point of closest approach at the beginning of segment A is the same as his latitude at the point of the antipode's closest approach at the end of segment B.

    The same reasoning applies for segments C and D where the man is first fleeing from the retreating antipode and then chasing the approaching moon. The starting and ending latitudes must be the same. At the end of the complete cycle when he is once more at the point of closest approach to the moon it follows that he is at the same latitude where he started.

    It then follows that a wide range of paths are stable. There is no tendency for them to converge on a single asymptotic path.
    Last edited: Mar 10, 2014
  9. Mar 10, 2014 #8


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    Just to point that out: the cycle is stable in N/S direction, but not in E/W direction.
    I agree with your analysis - assuming Greg is not close enough to the north pole to circle it.

    We can extend this model to cover model 2, where the moon is over the northern hemisphere. At the transition between A and B, the path would be symmetric if the antipodal point would "jump" towards the northern hemisphere (where the moon orbit is). The true antipodal point is more to the south, so the path taken by Greg goes more to the north. Therefore, A+B gives a net northwards motion. The same argument applies to C+D. And we can reverse it to cover a moon at the southern hemisphere. Greg ends up close to the pole in the hemisphere of the orbit.

    I see that Randall did not have the westwards motion in his analysis, hmm...
  10. May 1, 2014 #9


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    I will perhaps respond in more detail on how a sliding crank mechanism attached to the moon and earth would show pre-cognitative determination, in that minimizing the distance of subject with the projection of the moon on the earth's surface may not be the best plan of action but I have presently not yet attempted the calculations.

    The crankshaft would be the earth's centre. The connecting rod would pass throught a fixed phantom location at a distance of the moon to earth or similar location to be determined. The position of the subject would be the intersection of the connecting rod and the earth's surface. The crank length would be the radius of the earth and would thus be the projection of the moon as the earth rotates.

    The test subject would then not have a constant walk velocity but as variable as the velocity of the intersection of the connecting rod and earth surface. In this way, test subject would show intelligence in the regard that once the projection of the moon has surpaced a certain location it would be advantageous to reverse direction and attempt to meet the moon projection coming from the other direction rather than to continue in the same direction for an ever farther away increasing projection.

    The wobble of the moon's orbital plane can also be addressed by the crank having a lateral movement perpendicular to the rotational movement.

    A slowly rotaing phantom location ( rather than fixed ) can take into account the revolution of the moon.

    I suppose this could be considered a Ptolemaic system.
  11. Dec 18, 2014 #10
    Greg's direction vector relative to the Earth's center I will call n. The Moon's direction vector relative to the Earth's center I will call m. It points to the sub-Moon point on the Earth's surface. Since Greg tries to follow the Moon, he tries to move in direction (m - n). Since he is confined to the Earth's surface, we must project his direction vector:

    (m-n) - n*(n.(m-n)) = m - n*(n.m)

    Since he moves at constant angular speed r relative to the Earth's surface, we get this equation of motion for him:
    $$ \frac{d{\vec n}}{dt} = r \frac{{\vec m} - {\vec n} ({\vec n} \cdot {\vec m})}{\sqrt{ 1 - ({\vec n} \cdot {\vec m})^2}} $$
    This is difficult to solve, even for a simplified case of the Moon's motion, it orbiting above the Earth's equator in a circular orbit. That is actually a good approximation of the orbits of several other planets' moons around their planets. But in that case, we can decompose n into components related to m and dm/dt:
    $$ {\vec n} = L {\vec m} + \frac{M}{\omega} \frac{d{\vec m}}{dt} + \frac{N}{\omega} {\vec m} \times \frac{d{\vec m}}{dt} $$
    for the Moon having angular velocity ω and with L2 + M2 + N2 = 1. One gets from it
    $$ \frac{dL}{dt} = r \sqrt{1 - L^2} + \omega M ,\ \frac{dM}{dt} = - r \frac{LM}{\sqrt{1 - L^2}} - \omega L ,\ \frac{dN}{dt} = - r \frac{LN}{\sqrt{1 - L^2}}$$
    and from the first two,
    $$ \frac{d^2 L}{dt^2} + 2r \frac{L}{\sqrt{1-L^2}} \frac{dL}{dt} - (r^2 - \omega^2) L = 0 $$
    I'm stumped.

    However, I notice some relatively simple solutions in this case.

    One of them is Greg traveling on the equator. From the Moon at his zenith to the Moon at his nadir, he travels in the Moon's direction for a time of ## \pi/(\omega - r) ##, and while from the Moon at his nadir to the Moon at his zenith, he travels in the reverse direction for a time of ## \pi/(\omega + r) ##. He thus goes back and forth, with each cycle advancing him an angle of ## (2\pi r^2)/(\omega^2 - r^2) ##.

    Another is Greg going around either pole in synchrony with the Moon: ## L = 0 ,\ M = - r/\omega ,\ N = \pm \sqrt{\omega^2 - r^2}/\omega ##.

    Beyond that, I've had a little bit of success in finding closed-form solutions, but they are rather complicated.
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