INWIT™ — HOW MANY EARTHS CAN THE SUN CONTAIN?


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How many Earths can the Sun contain?

by Vincent Mallette
Copyright © 1999 Inwit Publishing, Inc.

[written for juveniles]

If you go purely by volume, the standard figures1 give about 1,307,000 earths in the sun. Say about 1,300,000.2 But this number does not take packing into account. If you have a barrel full of baseballs, you know there is space — air — between the balls, no matter how carefully you pack them in. You could in fact pour water into the barrel, without disturbing the balls, once they were packed in. This would actually be a pretty good way of assessing the space in the interstices3 between the balls. You know how many baseballs you put in, you know that the official baseball has a circumference of between 9 and 9 1/2 inches,4 you know the seldom-used formula, volume of a sphere equals the circumference cubed divided by 6 pi squared,5 yielding a baseballís volume as about 12.8 cubic inches.6 Lastly, you know the volume of the barrel, 7 and how much water you put in when it was packed with baseballs. From this you can easily calculate the percentage of the barrelís volume that was filled with "solid baseball." You should get a number somewhere around 74%. There is a purely mathematical, I should say geometrical, way to get this number. The closest packing occurs when every baseball is in contact with 12 other baseballs (hard to visualize, I admit; imagine the baseballs arranged in tetrahedrons — unit cells, they call them in crystallography). A mathematical analysis of this situation shows that each baseball in the most closely packed arrangement "reserves for itself" a volume of d3/Ö 2 (where d is the diameter of a baseball). The "solid baseball" matter, whose volume is p d3/6, occupies [(p d3/6) / (d3/Ö 2)] = 0.740480489693 of its "unit cell."8

So we see at last that the sun — our "barrel" — can hold about 74% of 1,300,000 earths, or about 962,000 copies of our home planet, packed in like sidereal sardines.

And for the sockdolager here...how many baseballs can you pack in the Sun? With interstices, even? The answer is...

4,986,867,311,173,653,589,793,238,462,643


1 The sunís volume, based on a radius of 6.966 ◊ 108 meters, is about 1.416 ◊ 1027 cubic meters. Thatís about 1Ĺ octillion cubic meters. You see how convenient scientific, or exponential, notation is. The octillion number, written out, would be 1,416,000,000,000,000,000,000,000,000 cubic meters. The earthís volume, based on a radius of 6.371 ◊ 106 meters, is about 1.083 ◊ 1021 cubic meters, or about one sextillion cubic meters. I leave it for you to figure out how many zeros thereíd be, if it were written out.

2 Depends on choice of mean radius of earth; books differ. The earth, as you may know, is about 43 kilometers thicker through the equator than through the poles. Newton calculated this number fairly accurately without going to Lapland or the Sahara. He imagined, in effect, that there was a water hose running from a pole to the equator. At what levels would the water stand, comparatively, at each location?

3 A good word to know for your SAT. Somewhat surprisingly, it is related to the word solstice.

4 Sports Rules Encyclopedia, Second Edition — ed. by Jess R. White (Champaign, Illinois: Leisure Press, 1990, p. 29

5 See if you can derive this from the more common formulas,

V = 4/3 p r3 or V = p d3/6, where r and d are radius and diameter of a sphere.

6 Iíll stick with inches here, since the American baseball is defined in inches. But for reference, 12.8 cubic inches is almost exactly 1/5000 of a cubic meter. This answers the frequently asked question, How many crushed baseballs would it take to fill a dugout a meter on a side — 5000. Try this on your friends. And by the way, your barrel probably had a capacity of 31 1/2 gallons. Unless it was used for beer, in which case it was 31 gallons. Of course, if it were used for cranberries, by law it was 87 dry quarts. Thatís only for cranberries, believe it or not.

7 Here is a cookbook formula for the volume of a barrel: V = [approx.] 0.26 L(2D2 + d2), where L is the length of the barrel, d is the diameter of its "head," and D is the greatest diameter

8 This is based on the analysis in Gordon Ferrie Hullís article in The American Physics Teacher, Feb. 1937, p. 23.


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