# The Speed of the Barycenter Settled?

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”:

1196
4.3 mph = 6.92 km/h
42
1196.4
1196.2
1197
759
750 mph = 1207 km/h
1193
1196
1196.1
1200

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!

## 11 thoughts on “The Speed of the Barycenter Settled?”

1. John Stroughair says:

To me it seems as if the difference between the two approaches is the question: who is the observer. If you imagine the earth is transparent and we can mark the barycentre with a little red dot; we could then watch the motion of the barycentre. Now if I am floating in space above the plane of the Earth Moon system I would see the red dot rotate slowly following the Moon’s orbital path, to me it just wouldn’t matter how fast the Earth is rotating – I would calculate the lower figure. On the other hand if I am sitting on the Earth’s surface, I see the red dot slipping behind me as the Earth rotates – now it matters a great deal how fast the Earth rotates – and I would calculate the higher figure.

2. Paul Davis says:

I find that nealmcb’s response is along the line I was thinking as I browsed your post.

Of course you intended to pose the question as the motion of the “barycenter” in a GEOcentric frame –
– and as a sky observer I am partial to the geocentric frame.

But as a thinker (mind problems) I tend toward the barycentric.

If it’s “baryCENTRIC” then the barycenter ought to be the reference or origin of the system – against which motion is to be described. I picture a point, orbiting the “Sun”*, Kepler-style, and the Earth/Moon doing their dance around that point – the Earth-rock sliding by said point at the clip you’ve discussed. It’s actually the simplest picture for me – and with this picture it is pretty easy to intuit an order-of-magnitude answer. Maybe it’s an “Algebraic vs. RPN” kind of thing.

*Solar-system-barycenter!

But I confess (an aside here) that, to my repeated and continued consternation, pretty much all plots I’ve run across of the whole-solar-system barycenter show a fixed Sun and a motion of barycenter about it. Yet, to me, the point is to portray the orbit of the Sun with respect to barycenter instead; to show the Sun as a planet-among-others – orbiting a barycenter.
It is really eye-opening to consider the angular acceleration modulation exerted on the Sun as it dances that squiggly path.

This is also fruitful in dispelling the all-too-common notion that one thing (the smaller thing) orbits another (the larger) rather than all orbits being a mutual relationship. Even that – The Earth and my feet are in mutual attraction – it’s not something the Earth does to me.

Oh, and my preferred H-HGttG meme is the “S.E.P.” field – I see examples every day.
42 may be the answer to “L,tUaE” – but is not to be confused with being the answer to every problem!
(I mean, “What’s 5 X 7?” “42” “not”)

3. Kenneth A. Heisler, M.D., FACS says:

Hello Guy,
Sorry about that index finger.
To distract you from the throbbing consider this added fillip to the problem:
Does not “C”, the center of the Earth, itself rotate clockwise around the putative barycenter, “B”, in the course of a month (however defined)? And ought this vector to be added to the countervailing “stationary Earth” (eastward) and “stationary Moon” (westward) vectors?

1. Yes, in a barycenter-centered framework, or in a heliocentric framework, the Earth’s center revolves around the barycenter. We were using a geocentric framework. They are all (pace Neal) geometrically equivalent. We choose which framework to use for the simplest description of whatever we are trying to describe. And sometimes we have to choose which finger to type with.

4. nealmcb says:

I’d say you’re both wrong. To a local approximation, the barycenter is moving with zero velocity, and the rock is moving past it at the high rate of speed you cite. And the barycenter is moving at the sun at about 30 km/s.

Why? Motion is relative, but we generally measure it relative to inertial frames of reference. After all, we have learned that the earth moves around the sun, not that the sun moves around the earth. Actually neither frame is inertial, but the frame of the sun is far less accelerated than the frame of the earth. Better to say that we both move around the barycenter of the solar system, or the galaxy, or the local cluster, but they get increasingly less relevant.

So saying that “at that moment you can tell yourself that the barycenter of the Earth-Moon system is a thousand miles under your feet, gliding through the rock at the speed of a fast jet plane” reflects a very strange perspective, which set up this whole disagreement in the wrong way.

You should say “at that moment, you can tell yourself that the barycenter of the Earth-Moon system is a thousand miles under your feet. You’ve just come as close to it as you’ll get today, moving past it at about (400? you work it out…) km/hr.
When you’re on the far side of the earth, you’re 6378 + 4728 km away, moving relative to it in the other direction at about (much slower) km/hr.

1. They are equivalent. But the description you favor is less helpful: it takes a good deal of explanation, and even when understood does not communicate so clear a picture. That’s not to say it’s not interesting. Perhaps it should be couched in terms of “Another way of looking at it is …”

1. Anthony Barreiro says:

There is always another way of looking at it! It makes my head spin. I’m almost ready to go back to the days of literal geocentrism, when God was in His Heaven and all was right with the world. Except for the Inquisition and the witch-burnings, we can do without those.

2. nealmcb says:

That’s what they told Copernicus….

5. Anthony Barreiro says:

This is wonderful, thank you.

You should throw out the 42’s. Pretty much anytime you see 42 as the answer, just throw it out.

I was contemplating the eastward sidereal motion of the Moon in relation to her westward apparent motion yesterday evening, while watching the quarter Moon creep eastward toward Regulus as they both started to set toward the western horizon. I did a little rough math in my head and re-remembered that the Moon moves a bit more than her own width of half a degree in an hour. Eyeballing the angular distance between the Moon and Regulus (I could have used my sextant, but it was inside and I was lazy), I estimated it would be about six hours until the Moon passed Regulus. As I would need to get up and go to work in the morning, I only saw a couple of hours of the slow motion chase, but sure enough, she closed a noticeable distance in two hours. I’ll check again this evening to make sure that the Moon passed Regulus in the wee hours last night.

6. I’m glad there are those that enjoy math to figure stuff out for the rest of us.

1. Anthony Barreiro says:

I had a hard time with math in school. This was in the days before electronic calculators when every calculation was made with pencil and paper and you needed to use tables to look up square roots and logarithms. I was very interested in science, but my handwriting was horrible, and simple math errors like reading “69” as “96”, or adding when I was supposed to subtract, would give me the wrong answer, and it was quite frustrating. I decided I wasn’t any good at math, and that math was hard and boring. I failed differential calculus and gave up. Now that I don’t *need* to use math, I enjoy figuring out how to set up an equation and calculating the answer to a question. In the past few years I’ve learned enough algebra and trigonometry to muddle through questions of interest, and I hope to get through calculus before I die.