I thought I should wait at least a day to let your answers to the problem come in. Here’s a summary of them, in kilometers per hour except for those marked “mph”:
4.3 mph = 6.92 km/h
750 mph = 1207 km/h
The first is my figure in the Astronomical Companion; the second is John Lash’s figure that got me worried. The others are in chronological order, or at any rate the reverse of the order they appear in the blog “Comments,” which can be confusing, because some get inserted after others on which they are secondary comments – or something like that.
(WordPress offers me some control over this, and maybe I’ll uncheck an option that enables “threaded/nested comments 5 levels deep,” and see what happens.)
The question might be settled by this list of the answers y’all have found, if it were a matter of a vote, or of an average! (That’s an idea. The gravitational constant, the speed of light, perhaps even the value of pi or of e, were not inevitable: they were agreed to by subcommittees of archangels before the Creation.)
The question might also be settled if we could find some authoritative source that gives the number we seek; but I haven’t found one, and was surprised that it isn’t in the index of Allen’s Astrophysical Quantities.
John reasoned as follows: “I use (roughly) the radius of the earth 3886 miles. Barycenter is 1025 miles below the surface. Radius (mean) of barycenter orbit within the mantle: 3886 – 1025 = 2861 miles. Circumference: 17,976 miles, distance traveled by the barycenter in a month. Divided by 27.56 days for anomalistic period = about 104 miles per day, 4.3 mph.”
My answer, when I found his email, was to do the begining of a recalculation, then to confess that I had just arrived home from four nights away, tired and facing the usual other catchings-up, so I would pass the rethinking back to him; then it occurred to me that others might contribute some perspicacious rethinking; as y’all have done.
Now I re-check my own answer, like this:
C is the center of the Earth. The view is from the north, so east is to the left.
M1 and M2 are positions of the Moon at the beginning and end of an hour. The distance to the Moon doesn’t need to be to scale, since all we need is the angle over which it moves in the hour.
The larger curve is the surface of Earth, but it isn’t really needed for our calculation.
The inner curve is part of the equator of an imaginary globe that we could call the “baryglobe.” Or it could be just a line: the orbit of the barycenter, B1 and B2 are the positions of the barycenter at the beginning and end of the hour.
So the radius from C to B1 or B2 is 4728 kilometers. So the circumference of this baryglobe is 2 pi 4728: 29707 kilometers.
In one hour the angle moved by the Moon, and the barycenter under it, is 360 degrees divided by 27.32 days (the sidereal period of the Moon) divided by 24 hours: 0.55 degree. This is indeed the approximate mean hourly motion of the Moon against the starry background.
So the linear distance from B1 to B2 is the circumference of the baryglobe, 29707, multiplied by 0.55/360: 45.39 kilometers. This then is the linear distance the barycenter travels in an hour. Or is it?
No, it is the distance it would travel in a stationary Earth. But the Earth is rotating in the same eastward direction, much faster, at (approximately) 360/24 = 15 degrees an hour; or more accurately, using the length of the sidereal day, 360/23.93 = 15.04 degrees. So any physical point, any particle, on the equator of the baryglobe, such as a particle at B1, travels in one hour 29707 *15.04/360 = 1241.09 kilometers eastward, to point B3.
But that particle that has rushed away eastward is no longer the barycenter. The particle that now hosts the barycenter is back at B2.
So the distance the barycenter has traveled through the rock in one hour is 1241.09-45.39 = 1195.7 kilometers.
Better still, we should write this as 45.39-1241.09 = -1195.7. In other words, the barycenter has traveled backward, westward, through the rock.
This makes sense, doesn’t it? In our sky, the Moon, though it is orbiting eastward, appears to move westward toward its setting, because we on the surface are moving eastward faster. If you are at point E1 on the surface, looking straight upward at the Moon, then at the end of one hour you are not at E2 but at E3, looking back westward at the Moon. The sub-lunar point, the point under the Moon, travels westward geographically. So the barycenter, which is vertically under the sub-lunar point, must also travel westward, somewhat slower.
My logic is about the same as those that others used, with varying orders of the steps. Our answers are, as several noted, still only approximations, neglecting further refinements. The velocity will vary with time because of factors that vary: for instance the radius to the barycenter, being a fraction of the distance to the Moon, will vary with the Moon’s varying distance. Also varying with that distance is the Moon’s speed. When the Moon, instead of being over the equator, is at a high declination – it can be more than 28° north or south – it is over a “thinner” part of Earth, because of Earth’s slightly flattened shape; and then the radius of the baryglobe is smaller.
And there’s a variety of other reasons for getting different answers. You may have used different “months” (the synodic or the anomalistic, instead of the sidereal) to get the angular average distance the Moon moves per hour. You may have rounded numbers differently, used more digits of pi (though I don’t think that would make much differenmce). You may, as I was afraid I had done, omitted a step, such as dividing days to get hours, or a factor, such as the rotation of the Earth.
I was going to write a little program which would, for different underlying figures, display the different answers. But I’m not doing so, because all this is enough and also because I mashed my index finger in the heavy steel door to the recycling yard and that’s hampering my use of the keyboard. Rest or roll in peace, barycenter!