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rloginunix
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Chasing the Lunar Shadow
« on: Jun 22nd, 2015, 9:06am » |
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Chasing the Lunar Shadow
Can a human on foot keep up with even briefly match the speed of the Lunar shadow observed at Earth's equator during a full Solar eclipse?
Keep it simple by assuming that:
- Sun, Earth and Moon are coplanar;
- the Earth's axis of rotation is perpendicular to its orbital plane (no 23.5-degree tilt);
- the Lunar orbit is circular;
- the speed of light is infinitely large compared to other speeds (and distances) involved;
- the needed radii and distances are given (look them up on wiki, for example);
- the shadow falls on a plane (for a bit more analytic solution);
(justify your answer)
[e]
Improved the wording (in italic) as per rmsgrey's suggestion.
[/e]
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| « Last Edit: Jun 23rd, 2015, 9:29am by rloginunix » |
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towr
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Re: Chasing the Lunar Shadow
« Reply #1 on: Jun 22nd, 2015, 11:22am » |
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Do you mean for the full period that there is a solar eclipse anywhere on the planet? Or, for a (possibly brief) while during that period? (And perhaps, if the latter, for how long.)
The latter seems doable: the moon is ~400 times closer than the sun, apparent motion of the sun is ~30 times greater (so we can practically consider the moon to hang still overhead), and keeping up with the sun would take around 1700 km/h, so ~ 4km/h should be enough to keep up with the moon's shadow while the sun and moon are roughly overhead.
But you need to go faster and faster to keep you, the moon and the sun on one line as time progresses. (And the curvature of the Earth makes the problem even worse)
[e]I confused lunar orbital velocity with apparent velocity, so that doesn't quite work. It may fly around the earth in 28 days or so, but of course in the meanwhile the earth turns quite a few times.[/e]
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| « Last Edit: Jun 22nd, 2015, 10:11pm by towr » |
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rloginunix
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Re: Chasing the Lunar Shadow
« Reply #2 on: Jun 22nd, 2015, 1:22pm » |
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May be I should have framed the question more directly (sorry about that) - what is the speed of the Lunar shadow as it runs over an equatorial region of Earth during the full Solar eclipse? That is the emphasis of this (estimate) problem.
Whether a human can keep up with that speed or not will become apparent once the above question is answered. So the meaning is the second one - just for a while for a brief moment (it's not the endurance, it's the ability).
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rmsgrey
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Re: Chasing the Lunar Shadow
« Reply #3 on: Jun 23rd, 2015, 8:33am » |
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No matter how fast the shadow moves, it will be possible to stay within it for a non-zero time (the fact the duration of totality is non-zero means you can do so without moving)
I think you mean to ask whether it's possible to match the shadow's speed, even briefly.
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rloginunix
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Re: Chasing the Lunar Shadow
« Reply #4 on: Jun 23rd, 2015, 9:12am » |
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Yes, rmsgrey, your formulation is much more successful (thanks) - that is what I meant but failed to convey.
Is it possible for a human on foot to match the speed of the Lunar shadow ...
towr, you've hit the nail right on its head.
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towr
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Re: Chasing the Lunar Shadow
« Reply #5 on: Jun 24th, 2015, 11:44am » |
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I've done some calculations and numerical approximations and even at the slowest, I get something from 1600 to 2000 km/h. So even in the best case you can't run fast enough to stay in line with the sun and the moon.
I've also found this article, which contains the umbral speeds for two lunar eclipse They bottom out at around 0.6 km/s, i.e. over 2160km/h
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rloginunix
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Re: Chasing the Lunar Shadow
« Reply #6 on: Jun 25th, 2015, 9:08am » |
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That's right.
From Wiki:
Dem = 385 000 km, the (mean) distance between the centers of Earth and Moon.
Tm = 27.322 d, Moon's around-the-Earth revolution time in Earth days.
Ree = 6,378.137 km, Earth's equatorial radius.
For the order of magnitude estimate within the listed assumptions Dem/c is about 1.284 s - can be thrown out. Hence, approximately the speed of the Lunar shadow relative to temporarily stationary Earth can be equated to the linear orbital velocity of the Moon Vm =  m Dem = (2  /Tm)*Dem where Tm = 27.322*60*60*24 s, Vm is about 1.026 km/s.
But the linear velocity of an equatorial point on (a now rotating) Earth is Ve = eRee = (2 /86400)*6,378.137 which is about 0.464 km/s. Since both rotate in the same direction, according to the Galilean rule of summation of non-relativistic velocities x = Vm - Ve = 0.562 km/s or about Mach 1.6.
From Wiki, the fastest Olympians cover 100 m in just south of 10 s - an order of magnitude slower.
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towr
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Re: Chasing the Lunar Shadow
« Reply #7 on: Jun 25th, 2015, 10:23am » |
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I don't think that's a correct approach. And I think a good indicator of that is that the shadow is moving in the opposite direction from what you seem to suggest.
Sun and moon move east to west, but the lunar shadow moves west to east, because the sun moves faster than the moon.
Which also means that if only the moon moved a bit faster, you might be able to keep up with the shadow. In fact, the mistake I made earlier that gave a doable speed, was assuming the moon was in geosynchronous orbit, i.e. a 24-hour orbit, which is equivalent to around 1667km/h, or 463 m/s, projected on the ground (along moon/earth axis)
Another thing, you can't just subtract the earths surface velocity from the moon's orbital velocity, because a flat earth assumption is really bad here. The moon's orbital velocity when projected to earth is only 17 m/s, and so the apparent velocity is 463 - 17 = 446 m/s (and in the opposite direction)
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| « Last Edit: Jun 25th, 2015, 10:41am by towr » |
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rloginunix
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Re: Chasing the Lunar Shadow
« Reply #8 on: Jun 25th, 2015, 8:20pm » |
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I see the source of confusion. When dealing with relative velocities references should be carefully spelled out. In my case we are in the frame where Sun is at rest. Since the Earth-Moon distance is way smaller than Sun-Earth and the duration of the event is so short we can ignore the yearly motion of Earth's center around Sun and only consider Earth's rotation about its axis and Moon orbiting it. Earth of course rotating much faster. Looking at Earth-Moon pair from either celestial pole we will see that Moon orbits Earth in the same direction as Earth rotates about its axis. I think it's counterclockwise if looking from the North and clockwise if looking from the South pole. In either case the shadow will run in the same direction as Earth rotates about its axis and the subtraction of velocities should work in this case.
When we say that Moon and Sun run East to West it means that we are in a different frame - the one where Earth is at rest:
Earth frame drawing
The drawing is horribly out of scale of course but the shadow runs West to East since Sun now runs faster than Moon. At some initial moment Sun is at S, Moon - at M, SMP is perpendicular to Earth's surface. After 1 second Sun is at S', Moon - at M'. Agreed - an approximation - but take some small section of Earth's surface near the shadow's border as flat here. Then the shadow's velocity relative to some (now stationary) Earth's equatorial region is PP', its direction was established already. From similar right triangles:
PP'/L = SS'/SO
In reality L << Ds - Earth-Sun distance, hence approximately SO = Ds
SS' = s * Ds
PP' = L * s
To find L use similar right triangles again:
SS'/MM' = SO/MO
MM' = m * Dem
SO = Ds (approx)
MO = Dem - Re - L
Put it together:
(s * Ds)/(m * Dem) = Ds/(Dem - Re - L)
Solve for L:
L = (s - m) * (Dem/s) - Re
Put it into the equation for PP':
PP' = (s - m) * Dem - s * Re
here s = (2 )/Te (Te is one Earth day) and s - m = (2 )/Tm (Tm is one Earth month). In other words the formula for the estimated speed comes out the same. This should corroborate the initial approach.
Granted, lots of approximations and assumptions - but this is just an order of magnitude estimate, not an exact solution.
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