Isaac Newton, like many astronomers before and since, was inspired by comets. His theory of gravitation enabled him to compute the orbit of comets as detailed in *The Principia*.

The mathematical formula that Newton developed to calculate the orbit of comets enabled him, and Halley, who was Flamsteed’s successor as Astronomer Royal, to determine if and when a comet would return. They could then try to equate the comets they observed with those in historical records. The margins of error were perilously close, but Halley famously and correctly equated his comet with earlier comets, and predicted that it would return in 75 years’ time, a prediction whose fulfilment caused a popular sensation. There is an excellent account of this in ‘A Brief Account of Halley’s Comet‘ by Evan Andrews.

In this blog I’m going to look at the way that Newton and Halley calculated the orbit of comets.

Newton said that he found this a very difficult problem. By this he meant that finding a method was very difficult. But given the method, how difficult was it to carry it out?

Newton carried out his calculation graphically, working to amazing precision. Halley used Newton’s method, but by calculation, and hence predicted the return of Halley’s comet.

I shall outline the steps in this and the next blog. I should like to hear from anyone who undertakes to carry them out.

The orbit of the comet is a conic, with focus the sun, lying in some plane that passes through the sun. The angle between this plane and the *ecliptic* – that is the plane of the earth’s orbit – might take any value, and the comet might move along this orbit in either direction. The orbit will approximate a parabola. If it is a branch of a hyperbola the comet will never return. If it is an ellipse it will return after a number of years, perhaps after thousands of years, depending on how far out the second focus lies. Of course we are neglecting the effect of the gravitational field of Jupiter and the other planets.

The first step in the calculation is to find the parabolic orbit of best fit.

We consider the projection of the orbit into the plane of the ecliptic. As the comet passes along its orbit, we have a *phantom comet*, its projection, that passes along the projected orbit in the ecliptic. The projection of a parabola is again a parabola. But the focus of the first parabola does not, in general, project onto the focus of the projected parabola.

A straight line from the sun to the phantom comet will sweep out equal areas in equal times, so it moves as if attracted by a centripetal force based at the sun, by Proposition 3, Book 1 of *The Principia*. But this force will not, in general, satisfy the inverse square law, as the sun is not at its focus. I do not know whether anyone has attempted to find a convenient formula for this force.

In Newton’s time the favoured celestial coordinate system was the geocentric ecliptic system, which is exactly what we need for our calculation. This consists of two angles. The *longitude* is the angle, measured anti-clockwise, from a fixed primary direction, to the projection of the line of sight of the celestial body onto the ecliptic. This primary position is taken to be the position of the sun at the vernal equinox, which is now March 20, and was some three days earlier in Newton’s time. One has also to distinguish between the Julian and Gregorian calendars that differ by eleven days, as Newton uses both.

This primary position is referred to by Newton and more widely as `the first point in Aries’. This slow change of direction, rotating through 360 degrees every 26,000 years, as observed by the ancients, and explained and computed (with a significant theoretical slip) by Newton in *The Principia*, is known as *the precession of the equinoxes*. The second angle is the latitude of the body, which gives the angle, North or South, between the line of sight of the comet and the ecliptic. These two angles determine the line of sight of the body, and conversely. Newton gives the ecliptic coordinates and the time of many observations of a number of comets. So we have the required astronomical input.

I shall describe Newton’s algorithm for carrying this out – namely Proposition 41 of Book 3 – graphically, as Newton carried it out. You will get a more accurate result if you follow Halley’s example, and use coordinate geometry; but it may be less fun. In this blog we shall see how to get the astronomical data represented diagrammatically. In the next blog we shall see how to calculate the orbit of the comet from this beginning.

First, draw a circle to represent the orbit of the earth, with centre *S*, the sun. You want to take the circle to be as large as possible, subject to the position of the phantom comet, when the three observations are made, lying within the bounds of the paper. You don’t know how distant the comet is at any time, so you need to use your judgement.

Now take the first point of Aries as being the direction of a line pointing sideways to the right, and transform the times of the observations into angles. So if the time that has elapsed since the vernal equinox (as above) is a fraction *p* of a year, the angular distance between the position of the earth as it moves round the sun and the first point of Aries is 360*p* degrees, anti-clockwise.

Now you can place the earth in its orbit at the times of the three observations, giving positions *T* and *t* and τ, following Newton’s notation. Through each of these points draw a straight line pointing in the direction given as the observed longitude of the comet.

You have now taken the first step, and the first step is often the hardest.

But there are further points to be mentioned. You have to get the astronomical data, and choose your three observations.

As to getting the astronomical data, they are given, for various comets, towards the end of Book 3, the final book of *The Principia*, and you can read them off from the original Latin, which is readily available on the web. Go to *The Principia* as published on The Newton Project and scroll down to page 491. This is the text of the first edition; the third edition has more observations of comets. You need no Latin at all; you can recognise a table of astronomical data.

Above is one of the tables. It is for the comet of 1680, and is known as Kirch’s comet, or Newton’s comet. Halley, following Newton’s approach, calculated the orbit of the comet, and hence its period, and equated the comet with earlier comets, but on this occasion incorrectly. Can you do better? I don’t know to what extent the error was due to the inaccuracy of the observations, or to inaccuracies introduced in the calculation.

The main problem is dealing with the astrological symbols. The 360 degrees of a full circle are divided into multiples of 30 degrees, corresponding to the twelve astrological symbols as follows:

**1** Aries ♈ **2** Taurus ♉ **3** Gemini ♊ **4** Cancer ♋ **5** Leo ♌ **6** Virgo ♍ **7** Libra ♎ **8** Scorpio ♏ **9** Sagittarius ♐ **10** Capricorn ♑ **11** Aquarius ♒ **12** Pisces ♓

So the *n*-th symbol corresponds to 30(*n*-1)^{o}, and a latitude of ♏12^{o} is a latitude of 162^{o}.

You will see that Newton distinguishes between apparent time – that is sundial time (*Tem. appar*) – and true time (*Tem. verum*). This nicety is irrelevant to our purpose here. You will also need the fact that *Borealis* (bor.) is the Latin for ‘North’, as in northern latitudes, and *Australis* is the Latin for ‘South’, as in southern latitudes.

It remains to decide on your choice of observations. Newton’s advice is as follows:

Let three observations be selected at time intervals that are, to a good approximation, equal. But let the interval of time when the comet is moving more slowly be slightly greater than the other interval, so that in fact the ratio of the difference in the lengths of the intervals to the sum of the intervals is the ratio of the sum of the intervals to some six hundred days.

I have truncated Newton’s advice, which then becomes rather technical. I don’t think that the omitted part is very helpful. ‘The comet is moving more slowly’ refers to the observed movement, not the actual movement.

So now we taken three positions of the earth, and three lines of sight of the phantom comet at these points. The fact that these lines intersect the orbit of the phantom comet gives almost no information. But we know the times when it crosses these lines of sight, and the line from the phantom comet to the sun sweeps out equal ares in equal times.

The question of how to compute the orbit of the comet from the information encoded in the diagram, and using the latitude of the comet at the three observations, will be covered in the next blog.

Before we leave the depiction of the astronomical information, let us consider the question of whether we have made a significant over-simplification. We have taken the earth’s orbit round the sun as being circular; but in fact it moves in an ellipse. The eccentricity of the earth’s orbit is 0.0167. This means that the distance between the foci of the earth’s orbit is 0.0167 times the major axis of the earth’s orbit. So while one focus *S* is the centre of the sun, the other focus *H* is still within the body of the sun. The line joining the foci is *the line of apsides*.

On January 2 this line points from the sun to the earth, with *H* further from the earth. Now you have got to work out where the earth will be on this ellipse at any time. This is an example of Kepler’s equation, which has no simple solution. But as the eccentricity is small, an excellent approximation is obtained by supposing that the earth maintains a constant angular velocity about *H*. I think that you will find this correction to be too small to be of practical significance, and Newton’s description of his algorithm seems to imply that he ignored it.

You may recognise the painting that heads this blog as Blake’s famous painting of Newton at work. The next blog will be headed by a more realistic drawing of the same topic knocked up in a few minutes by a less famous artist.

Charles Leedham-Green

C.R.Leedham-Green@qmul.ac.uk

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