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I know the answer is $60^o$. But I can't seem to be able to derive the result.

Specifically, how to define the probability density function (PDF) for this problem? Once I have that it should be straightforward. But if I use PDF=$\cos(i)$, I get answer closer to 30. So I'm not sure.

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    $\begingroup$ Interesting question! What does "PDF" mean? Also, in orbital mechanics inclination ranges from 0° to 180°, so due to the symmetry the average is most likely to be 90°. You can ask about the average of the absolute value of inclination which is probably more important for eclipse observations, or specify just the 0° to 90° interval. $\endgroup$
    – uhoh
    Jan 27, 2023 at 21:58
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    $\begingroup$ The way I'd solve the problem would be to associate each orbit with a point on a unit sphere where the angular momentum vector intersects; equatorial orbits are a point at the top, ($\theta = 0$) polar orbits are dots around the equator$\theta = \pi/2$, and then solve for the average $\theta$. $\endgroup$
    – uhoh
    Jan 27, 2023 at 21:58
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    $\begingroup$ pdf = probability density function. ie $d/d\theta P(X<\theta)$ $\endgroup$
    – James K
    Jan 27, 2023 at 22:40
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    $\begingroup$ What defines a randomly oriented orbit? I can see some appeal in @uhoh's idea to take uniformly distributed sampling points on a spherical surface and take each to specify an orbital inclination. In reality, each such point specifies a node as well as an inclination. For any given surface-density of sampling points, there will be far fewer points (near either pole) to provide orbits with a given small inclination than there will be points (near the equator) to provide orbits of a given large large inclination. Can this be called a distribution of randomly oriented orbits? $\endgroup$
    – terry-s
    Jan 28, 2023 at 18:44
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    $\begingroup$ @terry-s see comment $\endgroup$
    – uhoh
    Jan 28, 2023 at 22:26

2 Answers 2

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The orientation of an orbit is uniquely defined (bar the 180 degree ambiguity between prograde and retrograde orbits) by the orientation of the angular momentum vector. A random distribution means there is an equal chance of the angular momentum vector pointing to any unit area on a sphere.

The area on the surface of a sphere that is swept out by a circular strip at inclination $i$ and opening angle $di$ is $2\pi \sin i \ di$. The angle $i$ can be thought of as an orbital inclination angle if it is defined as measured down from the "pole" towards the "equator" or reference plane. Since randomly oriented orbits will have angular momentum vectors that populate the surface area of a sphere uniformly, then $$P(i)\ di \propto \sin i\ di\ .$$

The mean (absolute) value of $i$ is then straightforwardly given by $$\bar{i} =\frac{\int i\sin i\ di}{\int \sin i\ di}\, $$ where the integrals are performed from $0$ to $\pi/2$ (for prograde orbits). This yields $$\bar{i} = \left[\sin i- \theta\cos i\right]^{\pi/2}_{0} = 1\ .$$ i.e. The average inclination angle is 1 radian or 57 degrees.

Obviously you could equivalently define an "inclination" in terms of the angle of the angle between the orbital plane (the plane with which the angular momentum vector is perpendicular) and the reference plane, in which case the mean "inclination" is 33 degrees.

Edit: Other interesting factoids - the median inclination is 60 degrees (see Connor's answer); the mean $\sin i$ is $\pi/4$ (corresponding to an angle of 51.8 degrees), while both the mean and median $\cos i$ are 0.5, corresponding to an angle of 60 degrees because $P(\cos i)$ is uniform for randomly oriented orbits.

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    $\begingroup$ Nice! I started this, but when I saw $$\int_0^{\pi/2}\theta \sin \theta \ d\theta = \int_0^{\pi/2} \sin \theta \ d\theta$$ I got confused & distracted and forgot that I was looking at 1 radian. What a curious equality! $\endgroup$
    – uhoh
    Jan 28, 2023 at 3:12
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    $\begingroup$ How does an area on the surface of a sphere that is swept out by a cone of given opening angle relate to a 'randomly oriented orbit'? $\endgroup$
    – terry-s
    Jan 28, 2023 at 18:48
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    $\begingroup$ @terry-s because the probability of an axis pointing into an area on a sphere would be proportional to that area. $\endgroup$
    – ProfRob
    Jan 28, 2023 at 19:53
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    $\begingroup$ @terry-s I saw your comment on the question as well. If you'd like to see a rigorous mathematical definition of what a uniform probability density function (PDF) means and how the PDF is defined exactly, that might be an excellent new question! $\endgroup$
    – uhoh
    Jan 28, 2023 at 22:25
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    $\begingroup$ @terry-s The choice of spherically random angular momentum vectors is the simplest convenient representative model for this question. If you set up a Monte Carlo run as you suggest with random initial positions and velocities, it would be equivalent, but more complicated. If you assume a uniform distribution from 0 to 180, you end up with way too many orbits near Zero inclination, see astronomy.stackexchange.com/questions/35471/… . $\endgroup$
    – Connor Garcia
    Jan 30, 2023 at 21:48
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The orbital inclination angle is the angle between the rotational axis of the central body and the orbital axis of the orbiting body. Each orbiting body's orbital axis maps to a point on a sphere around the central body. Prograde orbits are from 0 to 90 degrees, and retrograde orbits are from 90 to 180 degrees.

Uniformly random distributed points on a sphere (representing random orbits) will be symmetrically distributed below and above the equator, so the average inclination angle is 90 degrees, or a polar orbit.

If we restrict the question to prograde only orbits, it can be interpreted as: What solid angle corresponds to a spherical cap area of half a hemisphere? The unit sphere has an area of $4\pi$, so the area of half a hemisphere is $\pi$. The formula of the area of a spherical cap on the unit sphere is $2\pi(1-cos(\theta))$, where $\theta$ is the solid angle. Set $$\pi=2\pi(1-cos(\theta))$$ and solve, to get $\theta=\pi/3$ or 60 degrees, which is your answer above. Of course, this is the expected median inclination of prograde orbits. To get the mean, I ran some monte carlo runs with 10k points and got about 57 degrees. To me, the median better describes the distribution.

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