Adam Hibberd
I have had many queries concerning the calculation of the likelihood of 3I/ATLAS's orbital plane lying within 5° of that of the ecliptic of the Solar System.
This calculation appears in the paper 'Is the Interstellar Object 3I/ATLAS Alien Technology?' which can be found here.
The calculation exploits a simple equation based on the ratio of two areas on a spherical surface.
If you imagine the plane of 3I/ATLAS's orbit has a vector perpendicular to it, and the inclination of its orbit therefore is the angle between this vector (more precisely known as the 'angular momentum' vector) and the axis representing the ecliptic north pole.
The probability therefore that the inclination of the orbit lies within 5° of this ecliptic north pole, is the area, A, of a circular cap surrounding this North pole (extending 5 degrees either side of this pole) divided by the total area of the sphere, S.
Now by Archimedes 'Hat Box' theorem we have A=2πr2( 1 - cos(5°)), and we have the total area of a sphere, S = 4πr2. The ratio A/S =( 1 - cos(5°))/2 = 2e-3 or ~ 0.2 %. This latter figure is quoted in the paper. In fact to be exact we need to multiply this by 2, since we might be within 5° of the south polar axis also.
There is an alternative logic, however, which has been put to me which comes up with a much higher probability. This I believe to have a flawed logic. I shall elaborate below.
The counter-argument goes like this. Let's say we represent the ISOs as darts and assume they arrive, isotropically (meaning from all directions with equal likelihood) at the Solar System which is represented again as a perfect sphere.
The idea is that the number of darts, L, arriving within 5° latitude of the ecliptic plane should then be divided by the total number of darts, N, and thus the probability of the plane having an inclination < 5° is simply the ratio of these two numbers, i.e. L/N.
What one find when one does this calculation is that you get a MUCH HIGHER PROBABILITY, than that derived in the paper. In other words L/N >> A/S.
So why the disagreement?
The reason is because there is a fallacy in the argument of the latter scenario. Let's say that a dart arrives at 4° latitude above the ecliptic, does it necessarily follow that the inclination of its orbit is also exactly 4°?
It turns out: not at all! If the velocity when it hits the sphere is vertical (so parallel to the ecliptic north) then, its inclination will be exactly 90°, even though it strikes the sphere at only 4° latitude!
Thus by adding up all the darts arriving at less than 5° latitude we shall arrive at a spurious answer since we shall be unwittingly including in our calculations, darts that have a much higher orbital inclination than 5°.
Please also refer to my personal blog, which has the same post.