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Log Graphs to Plot Planetary Patterns

This post is inspired by the excellent Professor Stewart’s latest book, Calculating the Cosmos. In it he looks at some of the mathematics behind our astronomical knowledge.

Astronomical investigations

In the late 1760s and early 1770s, 2 astronomers Titius and Bode both noticed something quite strange – there seemed to be a relationship in the distances between the planets. There was no obvious reason as to why there would be – but nevertheless it appeared to be true. Here are the orbital distances from the Sun of the 6 planets known about in the 1760s:

Mercury: 0.39 AU
Venus: 0.72 AU
Earth: 1.00 AU
Mars: 1.52 AU
Jupiter: 5.20 AU
Saturn: 9.54 AU

In astronomy, 1 astronomical unit (AU) is defined as the mean distance from the center of the Earth to the centre of the Sun (149.6 million kilometers).

Now, at first glance there does not appear to be any obvious relationship here – it’s definitely not linear, but how about geometric? Well dividing the term above by the term below we get r values of:

1.8, 1.4, 1.5, 3.4, 1.8

4 of the numbers are broadly similar – and then we have an outlier of 3.4. So either there was no real pattern – or there was an undetected planet somewhere between Mars and Jupiter? And was there another planet beyond Saturn?

Planet X

Mercury: 0.39 AU
Venus: 0.72 AU
Earth: 1.00 AU
Mars: 1.52 AU
Planet X: x AU
Jupiter: 5.20 AU
Saturn: 9.54 AU
Planet Y: y AU

For a geometric sequence we would therefore want x/1.52 = 5.20/x. This gives x = 2.8 AU – so a missing planet should be 2.8 AU away from the Sun. This would give us r values of 1.8, 1.4, 1.5, 1.8, 1.9, 1.8. Let’s take r = 1.8, which would give Planet Y a distance of 17 AU.

So we predict a planet around 2.8 AU from the Sun, and another one around 17 AU from the Sun. In 1781, Uranus was discovered – 19.2 AU from the Sun, and in 1801 Ceres was discovered at 2.8 AU. Ceres is what is now classified as a dwarf planet – the largest object in the asteroid belt between Jupiter and Mars.

Log Plots

Using graphs is a good way to graphically see relationships. Given that we have a geometrical relationship in the form d = ab^n with a and b as constants, we can use the laws of logs to rearrange to give log d = log a + n log b.

Therefore we can plot log d on the y axis, and n on the x axis. If there is a geometrical relationship we will see us a linear relationship on the graph, with log a being the y intercept and the gradient being log b.

(n=1) Mercury: d = 0.39 AU. log d = -0.41
(n=2) Venus: d = 0.72 AU. log d = -0.14
(n=3) Earth: d = 1.00 AU. log d = 0
(n=4) Mars: d = 1.52 AU. log d = 0.18
(n=5) Ceres (dwarf): d = 2.8 AU. log d = 0.45
(n=6) Jupiter: d = 5.20 AU. log d = 0.72
(n=7) Saturn: d = 9.54 AU. log d = 0.98
(n=8) Uranus: d = 19.2 AU. log d = 1.28

 

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We can use Desmos’ regression tool to find a very strong linear correlation – with y intercept as -0.68 and gradient as 0.24.  Given that log a is the y intercept, this gives:

log a  = -0.68

a = 0.21

and given that log b is the gradient this gives:

log b = 0.24

b = 1.74

So our final formula for the relationship for the spacing of the n ordered planets is:

d = ab^n

distance = 0.21 x (1.74)^n.

Testing the formula

So, using this formula we can predict what the next planetary distance would be. When n = 9 we would expect a distance of 30.7 AU.  Indeed we find that Neptune is 30.1 AU – success! How about Pluto?  Given that Pluto has a very eccentric (elliptical) orbit we might not expect this to be as accurate.  When n = 10 we get a prediction of 53.4 AU.  The average AU for Pluto is 39.5 – so our formula does not work well for Pluto.   But looking a little more closely, we notice that Pluto’s distance from the Sun varies wildly – from 29.7 AU to 49.3 AU, so perhaps it is not surprising that this doesn’t follow our formula well.

Other log relationships

Interestingly other distances in the solar system show this same relationship.  Plotting the ordered number of the planets against the log of their orbital period produces a linear graph, as does plotting the ordered moons of Uranus against their log distance from the planet.  Why these relationships exist is still debated.  Perhaps they are a coincidence, perhaps they are a consequence of resonance in orbital periods.   Do some research and see what you find!

This is an example of how an investigation into area optimisation could progress.  The problem is this:

A farmer has 40m of fencing.  What is the maximum area he can enclose?

Case 1:  The rectangle:

Reflection – the rectangle turns out to be a square, with sides 10m by 10m.  Therefore the area enclosed is 100 metres squared.

Case 2:  The circle:

Reflection:  The area enclosed is greater than that of the square – this time we have around 127 metres squared enclosed.

Case 3: The isosceles triangle:

Reflection – our isosceles triangle turns out to be an equilateral triangle, and it only encloses an area of around 77 metres squared.

Case 4, the n sided regular polygon

Reflection:  Given that we found the cases for a 3 sided and 4 sided shape gave us the regular shapes, it made sense to look for the n-sided regular polygon case.  If we try to plot the graph of the area against n we can see that for n ≥3 the graph has no maximum but gets gets closer to an asymptote.  By looking at the limit of this area (using Wolfram Alpha) as n gets large we can see that the limiting case is the circle. This makes sense as regular polygons become closer to circles the more sides they have.

Proof of the limit using L’Hospital’s Rule

  

Here we can prove that the limit is indeed 400/pi by using L’Hospital’s rule.  We have to use it twice and also use a trig identity for sin(2x) – but pleasingly it agrees with Wolfram Alpha.

So, a simple example of how an investigation can develop – from a simple case, getting progressively more complex and finishing with some HL Calculus Option mathematics.

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