Copyright ©2006  by Paul Niquette. All rights reserved.


hink of the barycenter as the fulcrum of a lever spanning the space between two orbiting bodies. We will find it convenient to let R represent the total length of that lever.  An elementary relationship will assure balance.  The mass of one body M multiplied by its distance from the fulcrum x must be equal the mass of the other body m multiplied by its distance from the fulcrum R – x.  Let K represent the ratio of their masses M / m.  The location of the barycenter x can be found as follows:
M x = m(R - x) ~~~~~ assuring the balance
Km x = m(R – x) ~~~~ substituting M = Km
K x = R - x ~~~~~~~~~ dividing both sides by m
x = R / (K + 1) ~~~~~~~ solving for x
For the earth-moon barycenter, R = 239,000 miles and K = 81, so x  =  2,914 miles from the earth's center of mass.  The circumference of the earth is about 25,000 miles, so its radius is about 4,000 miles, which puts the earth-moon barycenter about 1,000 miles below the surface.  By the way, in the magazine fold-out, the earth appears to be about 10 inches in diameter, so the moon’s orbit would extend out about 300 inches or some 35 page-widths -- not the 3˝ shown in the sketch.  Drawn to scale, the picture of the earth would be the size of a quarter.   And the moon would be -- well, smaller (see Measuring the Moon).

For the sun-earth-moon barycenter, R = 93,000,000 miles and K = 333,000, so x = 279 miles.  That is indeed quite close to the sun's center of mass.  The apparent scale in the sketch is 80,000 miles to the inch, thus the solution is...

The sun-earth-moon barycenter will be located
3.5 thousandths of an inch from the the sun's center of mass at the centerfold -- about the thickness of one human hair.

...which appears to complete the pun in the title of the puzzle.  At that scale, it would take about 137 page-widths to depict the radius of the earth’s orbit.

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Questions from Barycenterfold Solvers

If the earth moves closer to the sun would that change the location of the barycenter? 

Yes.  In the relationship x = R / (K + 1), the ratio of the two masses K may be assumed constant, which means the location of the barycenter x will change directly with the distance between the two bodies R. As the sun loses its mass through radiation, however, K must not be constant but decreases, and x necessarily becomes gradually larger as the earth moves farther away from the shrinking sun.  How fast?
When Jupiter moves closer to the sun would that change the barycenter?
Again, yes.  Let us be precise, though, by naming the bodies to which the barycenter applies: "the sun-Jupiter barycenter," "the sun-earth barycenter,"...  Analyzing orbital mechanics of more than two celestial bodies can be immensely complex. Barycenterfold applied "the earth-moon barycenter" along with purely circular orbits to simplify the location of the earth-moon center-of-mass within the sun-earth-moon system.  That's OK -- but only when K is large enough (groan).
Does the location of a barycenter depend on the location of the observer?
No.  Although, Barycenterfold depicts distances as measured from earth-to-moon and from sun-to-earth-moon barycenter, the relationship x = R / (K + 1) applies to foreshortened views of the orbiting bodies from anywhere in space.
Is there such a thing as a composite barycenter?  The solar system barycenter, for example?
In principle, yes.  As viewed from far outside the solar system, the expressiong center-of-mass may be preferable, keeping in mind that its location will move about in a complicated pattern.  That's because orbits are not generally circular as if defined by fixed length strings.  By the way, elastic strings would be a misleading model, inasmuch as, unlike gravity, the 'attractive force' increases with distance.
If the earth and moon have their own barycenter would not the location of  that center change at new moon compared to full moon?
No.  The earth-moon barycenter is located about 1,000 miles below the surface of the earth.  Regardless of the phase of the moon (new, half, full), as the earth rotates on its axis each day, the earth-moon barycenter will become positioned 1,000 miles directly underfoot at, say, a given point near the equator.  About 12 hours later, the earth-moon barycenter will have apparently whipped around to a position about 6,000 miles farther away from that point, again directly underfoot.
Are tides and earthquakes influenced by barycenters (earth-moon, sun-earth)?
The location of a barycenter x = R / (K + 1) pertains to an instantaneous relationship between orbiting bodies without explicit regard to the presence of gravitational forces between them, which vary inversely as R2.  Tides, whether in oceans or molten layers underneath, result from external gravitational forces and are provoked by the daily rotation of the earth upon its axis.  Tides, both lunar and solar, do dissipate earthly energy and therefore act as -- well, watch for a new puzzle entitled "Day Brake."

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