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HOW DID KEPLER DO IT (Part II) Popular books on astronomy shy away from using mathematics because the publishers say every equation that appears in the book halves the number of readers. For example, they say that Copernicus stated that the planets revolve around the Sun in circles, but Johannes Kepler proved that the orbits of the planets are not circles but ellipses. Full stop! They don't even tell their readers what an ellipse is, the reader having only a vague idea that it is some sort of oval. Well, an ellipse is a very well- defined curve. Whereas a circle is the path of a point which moves so that its distance from a fixed point remains constant, this constant being the length of the radius of the circle; an ellipse is the path of a point which moves so that the sum of the distances from two fixed points remains constant, so that in Figure 1. a+b = c+d = e+f = g+h. The two fixed points are the foci F1 and F2. The Sun occupies one of the foci. The eccentricity of the ellipse is equal to F1 F2 ÷ AP where A and P are the extremities of the major axis of the ellipse. The further the foci are apart the GREATER IS THE ECCENTRICITY OF THE ELLIPSE. The ellipse also has the property that the radius vector sweeps out equal areas in equal times (Fig II.). The areas of triangles a, b and c, swept out in equal times t, are equal. Now, how did Kepler work out the areas of triangles, having the Sun at the vertex and the base in the sky? He had the advantage of being able to use the very careful measurements that Tycho Brahe had made over a period of 20 years, especially of the positions of Mars. In Figure III M and N represent two positions of the planet separated by say, 24 hours, and S is the position of the Sun. The positions of M and N could be very accurately determined against the background of the stars. 24 hours are represented here by t. The area swept out by the radius vector SM i.e. the straight line from the Sun to the planet, when it moves to the position N, is the area of the triangle SMN which can easily be calculated: Since the angle MSN is very small, being only seconds of arc, the angles SMN and SNM can be considered to be right angled = 90°. The area of a triangle is equal to ½ base times vertical height, This equals . Now multiply by 1 which is equal to . So area of triangle SMN is equal to ½ SM x MN x = ½ MN2 x = ½ MN2 ÷ = ½ MN2 ÷ sin MSN. This shows the power of mathematics, because the length SM was impossible to measure. Kepler could use thousands of positions of Mars for M and N and also the sizes of angle MSN. The problem was to calculate the sines of the very small angles, but Kepler was an outstanding mathematician and he knew how to do the calculations. The important discovery that Kepler made, was that as Mars moved from aphelion to perihelion, the distances M-N grew steadily larger for equal lapses of time, but the areas of the triangles remained constant. The distances SM and SN had thus to be getting less and less, i.e. the planet was steadily getting nearer and nearer to the Sun. This would not be the case if the planet's orbit was a circle. Kepler found that the distance of Mars from the Sun varied from 1,38 AU to 1,66 AU, with a mean of 1,52, as we saw in the previous article. The eccentricity of an ellipse is equal to: . Jan Eben van Zyl |
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