Johannes Kepler helps to answer the question of how many earths can fit into Jupiter

Orange stack_Nick Lomb

Close-packed spheres (oranges) in a regular sequence. Picture Nick Lomb

Jupiter is the largest planet in the solar system and much larger than the Earth. One way of illustrating the size difference is to ask how many earths could fit inside Jupiter. Taking the question to mean how many times the volume of Jupiter is larger than that of Earth, the answer can be can be easily found to be 1300 by taking the ratio of the cubes of the diameters of the two planets. What though if we take the question more literally to mean that how many earths could be stacked inside Jupiter? What then is the answer?

Here we run into the well-known stacking problem, which is the consideration of the most efficient way of stacking spheres into a container. Amazingly, this problem was first considered by none other than the German mathematician, who is famous in the history of astronomy, Johannes Kepler.


The portrait of a man, believed to be Johannes Kepler, painted around 1611 by Hans von Aachen, courtesy Wikimedia Commons

In 1611 while living in Prague as Imperial Mathematician to Emperor Rudolf II Kepler wrote an essay called, in Latin, “Strena seu de Nive sexangula” or, in English, “New-Year’s gift concerning six-cornered snow”. In this essay instead of looking at the movements of planets, Kepler looks at the structure of snowflakes and in particular their sixfold symmetry. Though, of course, without modern knowledge of crystal structures he could not explain this symmetry he did explore some interesting mathematical concepts along the way.

Inspired by seeds in a pomegranate Kepler considered the most efficient way of packing spheres into a container. He conjectured that the best way was the simplest – putting each sphere into the hollow formed in between four spheres below, a method that is used by every greengrocer to stack oranges and similar fruit.

Surprisingly, Kepler’s conjecture was only proved 1998 by American mathematician Thomas Hales in a 250-page proof accompanied by 3 gigabytes of computer programs. The proof was so complex that a panel of 12 referees took four years to report back that they are 99% certain of its correctness.

Taking Kepler’s conjecture as proved, how many earths can we stack, in this most efficient way, into Jupiter? A quantity called density is defined as the fraction of the available space occupied by the spheres. The higher the density, the more closely packed is the stacking. For this most efficient stacking method the density is ?/?18 = 0.7405. A derivation of this formula is below.

We can now state that the number of earths that we could fit into Jupiter is the number of times Jupiter’s volume is larger than that of the Earth multiplied by the packing density, that is, 1300*0.7405 = 963. Thus the literal and exact answer to the question “How many earths can fit into Jupiter?” is 963.

Derivation of the formula for density


Stacking circles in the most efficient way. Drawing courtesy Wikipedia

Let us first consider the 2-dimensional version of the stacking problem and let us take an array of n by n circles of unit diameter. Their total area will be n*n*?/4.

Looking at the above image the length of n circles is n.

To work out the height of n circles note that the centres of two circles in one row form an equilateral triangle with the centre of a circle above and in between them. Thus the vertical distance between the centres is sin 60° = ?3/2 and the height of n circles is n*?3/2.

The total available area is length times height or n*n*?3/2.

Density for this 2-dimensional case is thus area of circles/available area = ?/4 divided by ?3/2 after the n*n factor cancels out. Density = ?/?12 = 0.907

For three dimensions the derivation is similar. We take an array of n by n by n spheres of unit diameter. Their volume is n*n*n*4/3*?/8 = n*n*n* ?/6.

The length and width of the spheres are both n.

To work out the height of n spheres note that the centres of four spheres in one row form a pyramid of unit length edges with the centre of a sphere above and in between them. Simple trigonometry gives the height of the pyramid as 1/?2.

The total available volume is length times width times height or n*n*n*1/?2.

Density for this 3-dimensional case is volume of spheres/available volume = ?/6 divided by 1/?2 after the n*n*n factor cancels out. Density = ?/?18 as used in the calculation above.


Leave a Reply

Your email address will not be published. Required fields are marked *