Making the assumptions listed below, the greatest contribution to the ground speed of the Moon is due to the rotation of the Earth. The angular velocity of the Earth is:
$\omega_{_{E}} = \frac{V_E}{R_E} = \frac{1,675 }{6,378} \approx 0.26262 $ radians/hr,
where $V_E$ and $R_E$ are the tangential ...
"Hopefully, I haven't made any blunders with my geometry or algebra. ;)" With this stuff, it's so easy to make little blunders, especially sign errors, that are hard to detect. I don't think I accidentally reversed Earth's rotation direction...
The Moon's orbital parameters vary in a rather complicated way. astronomy.stackexchange.com/a/55112/16685 But I wouldn't expect that to have such a huge effect on the ground speed. It looks like Randall is taking the Moon's ~5.14° ecliptic inclination into account, but I don't know what simplifying assumptions he's making.
@PM2Ring I upvoted both your related answers, because the presentation is professional, the research and references are extensive and the SageMath program is simply beautiful. Would have upvoted them more if I could :-) As for my answer I have extended it to cover the more generic and realistic situation that allows for the (varying over an 18 year period) inclination of the Moon's orbital plane relative the Earth's equatorial plane, instantaneous latitude taking the Earth's oblateness into account and varying tangential velocity and radius due to the Moon's elliptical orbit into account.
@PM2Ring I now get a minimum ground speed of -389 m/s and a maximum of -452 m/s. Unfortunately our results still do not overlap, so at least one of us has made an error somewhere. I have included a link to a Mathematica notebook that might make it easier for anyone who would like to check my calculations.
@KDP Hmmm. I agree that the Earth rotation speed should be the dominant term, so the ground speed should be under 465 m/s. So it looks like I've made a dumb mistake somewhere. I can't see anything obvious in my code. So I guess I just need to go through it carefully and check everything. Oh well...
Or maybe not. ;) You're defining the ground speed in terms of the difference in angular speeds. I'm defining it in terms of the magnitude of the difference in Cartesian velocities relative to the geocentre. The Moon's mean orbital speed is ~1022 m/s.
@PM2Ring While I am using the difference in angular velocities, I am multiplying the difference by the Earth's radius which gives the linear Cartesian velocity, since $v = \omega r$. In the more detailed analysis I am adding linear velocity vectors to get the resultant linear velocity.
@JohnHoltz I did some research and the shadow of a solar eclipse does indeed move in the opposite direction to the ground track or apparent motion of the Moon in the sky and I think I l know why. Might deserve its own question :-)
@KDP Yes, but that's the velocity of a point orbiting with the Moon's angular velocity, but at a distance of R'. That point stays colinear with the geocentre & the Moon, but it moves much slower than the Moon.
@PM2Ring The fact remains that your first observation that the tangential velocity of the Earth is the dominant factor and the motion of the Moon can only reduce that, is correct, so the ground velocity should be below 465 m/s.
BTW, your equation for the axial radius isn't quite right. The standard latitude is known as the geodetic latitude. It's the angle that the normal of the ellipsoid makes with the equatorial plane. But that normal doesn't pass through the origin, except at the equator & poles.
The simplest way to calculate the axial radius is to use the parametric latitude. See en.wikipedia.org/wiki/…
Let phi = geodetic latitude and beta = parametric latitude. Let a = the semi-major axis and b = the semi-minor axis. Then the equation of the ellipse is (x/a)^2 + (y/b)^2 = 1
x = a × cos(beta), y = b × sin(beta). And a × tan(beta) = b × tan(phi)
@PM2Ring You are talking about my equation for R' right? Using the equation for an ellipse I get R' = (a\times b) / \sqrt{1/a^2-(\tan b )^2/b^2}}$. Does that seem right now?
@PM2Ring You are talking about my equation for R' right? Using the equation for an ellipse I get R' = (a × b) / sqrt(1/a^2-(tan b )^2/b^2). Does that seem right now?
Is it not possible to use MathJax in chat or delete/edit previous comments?
Making a right mess of this lol. Checked my calculations again. Now I think it should be R' = a × b / ( cos beta × sqrt( b^2 + a^2 × (tan beta)^2 ) ). That checks out for beta =0 and beta = Pi/2 now.
@KDP Unfortunately, Chat doesn't have MathJax built in. But we have bookmarklets, thanks to robjohn, one of the math.SE moderators. See math.ucla.edu/~robjohn/math/mathjax.html
It can be tricky installing the bookmarks in some browsers, especially mobile. For security reasons, many browsers restrict the location where bookmarklets can be launched from. The general rule is to put them into a Bookmarks toolbar, so that toolbar must be turned on (if your browser has such a toolbar).
The chatjax bookmarklets work well. But unfortunately they can't give you a preview. So for long or complicated stuff, you can use the editor in the Answers section on a main site like Physics.SE or Math.SE. That is, create a fake answer to some question. Just remember to not post it. :)
Anyway, back to ellipses...
We have $x=a\cos\beta$ and $y=b\sin\beta$. Let $e'=b/a$. (Note that $e'=1-f$ and $e^2+e'^2=1$, where $f$ is the flattening and $e$ is the eccentricity of the ellipse).
The axial radius is $x=a\cos\beta$, with $$x^2=a^4\cos^2\phi / (a^2\cos^2\phi+b^2\sin^2\phi)$$ And the radius (your $R'$) is given by $r^2=x^2+y^2$, $$r^2 = \frac{a^4\cos^2\phi + b^4\sin^2\phi} {a^2\cos^2\phi+b^2\sin^2\phi}$$ which is equivalent to the expression you gave for R'. But as you see, $x\ne r\cos\phi$.
But seriously, it's generally a lot easier to do as much of the work as possible with $\beta$. If it was good enough for Legendre and Bessel, it's good enough for me. ;) en.wikipedia.org/wiki/…
Of course, the difference between phi and beta is pretty small. If it weren't, spherical trig navigation wouldn't work. The maximum difference is under 0.1°, which is 60 nautical miles, in terms of meridian arc. Here's a plot:
The green "Quad" curve is a simple quadratic approximation. The red "Sin2" curve (which is obscuring the blue True plot) uses a sine approximation which is very close to the true value, but only requires 1 trig call.
Discussion between PM 2Ring and KDP
Discussion between PM 2Ring and KDP