One of the many skills maths students develop in high school is the ability to change the position of a graph. I sometimes need to remind my students that this is not just “busy” work. It is a skill used frequently in technical fields such as science and engineering. What follows is an example from celestial mechanics.
The Two-Body Problem
Calculating orbits and their characteristics is usually a computer-intensive exercise. For example, how far from the earth is an orbiting satellite at a particular time. However, a good first approximation of this is to pretend we are in the unrealistic universe where only two point masses exist. In this universe, the shape of orbits can be perfectly modelled with equations called conic sections. Why they are called conic sections is a bit beyond the scope of this post, but please free free to look that up.
The orbital shape I want to describe here is the ellipse. The equation of an ellipse which is centered at the origin is:\[\frac{x^{2}}{a^{2}} +\frac{y^{2}}{b^{2}} =1\] where 2a is the length of the ellipse in the x direction and 2b is the length of the ellipse in the y direction. If a > b, the line along the a intercepts is called the major axis of the ellipse and the line along the b intercepts is called the minor axis:
This is a possible shape of an orbit about one of the masses in our perfect two-mass universe. In reality, orbits of say satellites around the earth, approximate this shape but other factors like other objects in the solar system and the earth not being a point mass with an uneven distribution of its mass, make the orbital shape slightly different than an ellipse. Not only that, in our perfect universe, the orbit path is exactly contained in a plane and can be completely drawn on flat paper. In reality, orbits also go below and above the average plane of the orbit.
Now one of the masses, say m₂, is on the ellipse. The other mass, m₁, is not at the center of the ellipse, the origin in the above figure. So where is it?
An ellipse has two special points associated with it call focal points. These points have the property that any point on the ellipse has a total distance between it and the focal points is a constant. This constant is 2a if the longest dimension of the ellipse is along the x-axis. Otherwise, this constant length is 2b.
The other mass, m₁, is at one of these focal points. The coordinates of these points can be found using the Pythagorus theorem. Looking at the blue triangle above, you can see that \[c=\sqrt{a^{2} -b^{2}}\]
An Example
Suppose we want to analyze the relationship between two masses that are in orbit. In orbital mechanics, the shape of an orbit is given by its eccentricity, e. An elliptical orbit has an eccentricity between 0 and 1. By the way, a perfectly circular orbit has an eccentricity of 0. So let’s say we have an orbit with the major axis length 2a = 10 and minor axis length 2b = 6. This is an orbit with an eccentricity of 0.8. If we want to know the position of m₂ with respect to m₁, we need a coordinate system. A convenient one would be a Cartesian system aligning the major axis of the orbit along the x-axis and centering the ellipse at the origin. This way we can immediately write down the equation of the orbit:\[\frac{x^{2}}{25} +\frac{y^{2}}{9} =1\]
as a = 5 and b = 3. This means that the focal points are \[c=\sqrt{25 -9} =4\]from the center:
Two of the more basic pieces of information we would like to know (especially if this is a satellite orbiting the earth) is what is the distance r between m₁ and m₂ and what is the direction of m₂ with respect to m₁. We can define the direction as the angle ???? of the line connecting the two masses with the positive x -axis. This is actually the reference used in much of celestial mechanics. This angle is called the true anomaly.
Well this is nice, but since we want to find relationships of m₂ with respect to m₁, it would be even more convenient to put the origin of our coordinate system at m₁. How do we do that?
Transformations to the Rescue
Somewhere in your year 10 or 11 maths (grades 10 or 11 in the States), you took the equation of a standard parabola, y = x², and replaced the x with x–h to get y = (x–h)². This moved the parabola h units to the left or right depending on the sign of h.
It turns out that in any relationship between x and y, replacing the x with x–h has the exact same effect on its graph. So in our example, if we move the ellipse 4 units to the left, m₁ would be on the origin of our coordinate system.
We can now calculate r and ???? a bit more easily than before the transformation. Let’s look at where m₂ is in the above figure. Here, m₂ is at (-4,3). The following process would work with any point, but at this point, the numbers are “nicer”.
The figure below shows the answers. The distance r can be found using the Pythagorus theorem but once you see that the two sides are 3 and 4, then this is the standard 3-4-5 right triangle:
In this perfect universe, other orbital shapes are possible: circles, parabolas, and a hyperbolic. These have their own standard equations but they all can be transformed to move them to any place you wish on the coordinate system to make your calculations easier.
