<p>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.</p>
<p>The area on the surface of a sphere that is swept out by a circular strip at inclination <span class="math-container">$i$</span> and opening angle <span class="math-container">$di$</span> is <span class="math-container">$2\pi \sin i \ di$</span>. The angle <span class="math-container">$i$</span> can be thought of as an orbital inclination angle if it is defined as measured down from the &quot;pole&quot; towards the &quot;equator&quot; or reference plane. Since randomly oriented orbits will have angular momentum vectors that populate the surface area of a sphere uniformly, then
<span class="math-container">$$P(i)\ di \propto \sin i\ di\ .$$</span></p>
<p>The mean (absolute) value of <span class="math-container">$i$</span> is then straightforwardly given by
<span class="math-container">$$\bar{i} =\frac{\int i\sin i\ di}{\int \sin i\ di}\, $$</span>
where the integrals are performed from <span class="math-container">$0$</span> to <span class="math-container">$\pi/2$</span> (for prograde orbits). This yields
<span class="math-container">$$\bar{i} = \left[\sin i- \theta\cos i\right]^{\pi/2}_{0} = 1\ .$$</span>
i.e. The average inclination angle is 1 radian or 57 degrees.</p>
<p>Obviously you could equivalently define an &quot;inclination&quot; 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 &quot;inclination&quot; is 33 degrees.</p>
<p>Edit: Other interesting factoids - the <em>median</em> inclination is 60 degrees (see Connor's answer); the mean <span class="math-container">$\sin i$</span> is <span class="math-container">$\pi/4$</span> (corresponding to an angle of 51.8 degrees), while <em>both</em> the mean and median <span class="math-container">$\cos i$</span> are 0.5, corresponding to an angle of 60 degrees because <span class="math-container">$P(\cos i)$</span> is <em>uniform</em> for randomly oriented orbits.</p>

CC BY-SA 4.0
2023-01-28T03:12:02.057Z
*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!
uhoh
CC BY-SA 4.0
2023-01-28T18:48:51.323Z
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'?
terry-s
CC BY-SA 4.0
2023-01-28T19:53:41.993Z
@terry-s because the probability of an axis pointing into an area on a sphere would be proportional to that area.
ProfRob
CC BY-SA 4.0
2023-01-28T22:25:23.513Z
@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!
uhoh
CC BY-SA 4.0
2023-01-30T02:47:20.383Z
My comment wasn't to question the meaning of uniform prob. density, it was about the nature of the assumption linking such a uniform PDF with a particular parameter of an orbit, and what qualifies an orbit as randomly oriented. Why choose the set of ang.mom. vectors with spherically uniform distribution? Why not choose instead a sph. uniformly distribd set (A) of initial points, and the set of orbital orientations defined by a sph. uniform distribn of initial velocities starting from each initial point of the set (A)?
terry-s
CC BY-SA 4.0
2023-01-30T21:48:39.737Z
@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 https://astronomy.stackexchange.com/questions/35471/why-are-asteroids-with-zero-orbital-inclination-rare/40518#40518 .
Connor Garcia
CC BY-SA 4.0
2023-01-31T10:33:02.210Z
@connor garcia - thanks, but it seems this forum is not well adapted to questions raised so far: Satisfying answers require demonstrative proofs, which are hard to present and have not been presented here. For example, nothing shows that the distribution yielded by the 'ang.momentum' model is equivalent (or not) to the one yielded by the 'Monte Carlo' model, nor that either is equivalent (or not) to a uniform 0..180 distribution. Also needed wd be a basis to distinguish between physical causation and statistical effects as reason for low numbers of  low-inclination orbits.
terry-s
CC BY-SA 4.0
2023-01-31T12:23:49.753Z
@terry-s randomly oriented orbts do not have inclinations that are uniformly distributed between 0 and 180 degrees, That is just spherical geometry. Nothing in this question asks about a physical causation and indeed the question is merely one of geometry not astrophysics, and a routine one at that where the answer is not debated. Connor has used a Monte Carlo simulation to do the analytic integral I show - they are not different "models".
ProfRob
CC BY-SA 4.0
2023-01-31T17:07:01.960Z
@terry-s In this case, the OP simply requested a "derivation" of the mean inclination for random orbits, which both ProfRob and I gave (though our results differ).  A great new question might be something like: "Why represent random orbital inclinations as uniformly distributed points on a sphere?"  You could even ask for "demonstrative proofs"!
Connor Garcia
CC BY-SA 4.0
2023-01-31T18:51:23.840Z
@ConnorGarcia In what way do they differ? Your mean inclination of 57 degrees is 1 radian.
ProfRob
CC BY-SA 4.0
2023-01-31T19:13:39.050Z
@ProfRob  Whoops, I didn't read your recent edits and misinterpreted your answer.  You are right, our answers are the same.
Connor Garcia
CC BY-SA 4.0
2023-01-31T22:24:50.453Z
@ProfRob - If you read my comments again you'd see that I didn't suggest that randomly oriented orbits have inclinations ranging uniformly 0..180 deg. Nor did I ask for a physical causation. My comments were concerned to question the basis for concluding  that orbits with spherically uniformly distributed angular momentum vectors are randomly oriented.  I suggested another possible way of connecting uniform probability densities with orbital parameters, not to say that this alternative was certainly right, but to point out the need to justify the particular connection chosen.
terry-s
CC BY-SA 4.0
2023-01-27T22:45:36.133Z
2023-01-31T23:16:24.920Z
2023-01-31T23:16:24.920Z
ProfRob
<p>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.</p>
<p>Uniformly random distributed points on a sphere (representing random orbits) will be symmetrically distributed below and above the equator, so <strong>the average inclination angle is 90 degrees</strong>, or a polar orbit.</p>
<p>If we restrict the question to prograde only orbits, it can be interpreted as: What <a href="https://en.wikipedia.org/wiki/Solid_angle" rel="nofollow noreferrer">solid angle</a> corresponds to a spherical cap area of half a hemisphere? The unit sphere has an area of <span class="math-container">$4\pi$</span>, so the area of half a hemisphere is <span class="math-container">$\pi$</span>.  The formula of the area of a spherical cap on the unit sphere is <span class="math-container">$2\pi(1-cos(\theta))$</span>, where <span class="math-container">$\theta$</span> is the solid angle. Set <span class="math-container">$$\pi=2\pi(1-cos(\theta))$$</span>
and solve, to get <span class="math-container">$\theta=\pi/3$</span> or <strong>60 degrees, which is your answer above</strong>.  Of course, this is the expected <strong>median</strong> inclination of prograde orbits.  To get the <strong>mean</strong>, I ran some monte carlo runs with 10k points and got about <strong>57 degrees</strong>.  To me, the median better describes the distribution.</p>

CC BY-SA 4.0
2023-01-30T18:40:44.040Z
2023-01-30T20:42:26.810Z
2023-01-30T20:42:26.810Z
Connor Garcia
<p>I know the answer is <span class="math-container">$60^o$</span>. But I can't seem to be able to derive the result.</p>
<p>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=<span class="math-container">$\cos(i)$</span>, I get answer closer to 30. So I'm not sure.</p>

CC BY-SA 4.0
2023-01-27T21:58:45.720Z
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.
uhoh
CC BY-SA 4.0
2023-01-27T21:58:56.263Z
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 $\theta$.
uhoh
CC BY-SA 4.0
2023-01-27T22:40:41.983Z
pdf = probability density function.  ie $d/d\theta P(X<\theta)$
James K
CC BY-SA 4.0
2023-01-28T18:44:53.493Z
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?
terry-s
CC BY-SA 4.0
2023-01-28T22:26:14.147Z
@terry-s see [comment](https://astronomy.stackexchange.com/questions/51734/what-is-the-mean-inclination-for-randomly-oriented-orbits-how-to-derive-it?noredirect=1#comment116237_51735)
uhoh
CC BY-SA 4.0
2023-01-28T22:41:56.870Z
@terry-s yes, that's why the probability  $P(\theta) \propto \sin \theta$ . You can easily show that this uniformly populates a sphere with points and that is what is meant by randomly oriented.
ProfRob
CC BY-SA 4.0
2023-01-31T16:41:49.137Z
Actually the first thing that popped up also in my mind when reading the question was Bertrand's Paradox and how obvious a sensible choice of "random orbits" is.
asdfex
<orbital-mechanics>
What is the mean inclination for randomly oriented orbits? How to derive it?
Indigo1729
# What is the mean inclination for randomly oriented orbits? How to derive it?

(Asked by: username_0 on 2023-01-27)

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.

**Tags:** orbital-mechanics
> **Comments:**
> **username_1**: 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.  
> **username_1**: 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 $\theta$.  
> **username_2**: pdf = probability density function.  ie $d/d\theta P(X<\theta)$  
> **username_3**: 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?  
> **username_1**: @terry-s see [comment](https://astronomy.stackexchange.com/questions/51734/what-is-the-mean-inclination-for-randomly-oriented-orbits-how-to-derive-it?noredirect=1#comment116237_51735)  
> **username_4**: @terry-s yes, that's why the probability  $P(\theta) \propto \sin \theta$ . You can easily show that this uniformly populates a sphere with points and that is what is meant by randomly oriented.  
> **username_5**: Actually the first thing that popped up also in my mind when reading the question was Bertrand's Paradox and how obvious a sensible choice of "random orbits" is.  

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## 2 answers

### Answer by username_4 on 2023-01-27. Score: 9

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.

> **Comments:**
> **username_1**: *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!  
> **username_3**: 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'?  
> **username_4**: @terry-s because the probability of an axis pointing into an area on a sphere would be proportional to that area.  
> **username_1**: @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!  
> **username_3**: My comment wasn't to question the meaning of uniform prob. density, it was about the nature of the assumption linking such a uniform PDF with a particular parameter of an orbit, and what qualifies an orbit as randomly oriented. Why choose the set of ang.mom. vectors with spherically uniform distribution? Why not choose instead a sph. uniformly distribd set (A) of initial points, and the set of orbital orientations defined by a sph. uniform distribn of initial velocities starting from each initial point of the set (A)?  
> **username_6**: @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 https://astronomy.stackexchange.com/questions/35471/why-are-asteroids-with-zero-orbital-inclination-rare/40518#40518 .  
> **username_3**: @connor garcia - thanks, but it seems this forum is not well adapted to questions raised so far: Satisfying answers require demonstrative proofs, which are hard to present and have not been presented here. For example, nothing shows that the distribution yielded by the 'ang.momentum' model is equivalent (or not) to the one yielded by the 'Monte Carlo' model, nor that either is equivalent (or not) to a uniform 0..180 distribution. Also needed wd be a basis to distinguish between physical causation and statistical effects as reason for low numbers of  low-inclination orbits.  
> **username_4**: @terry-s randomly oriented orbts do not have inclinations that are uniformly distributed between 0 and 180 degrees, That is just spherical geometry. Nothing in this question asks about a physical causation and indeed the question is merely one of geometry not astrophysics, and a routine one at that where the answer is not debated. Connor has used a Monte Carlo simulation to do the analytic integral I show - they are not different "models".  
> **username_6**: @terry-s In this case, the OP simply requested a "derivation" of the mean inclination for random orbits, which both ProfRob and I gave (though our results differ).  A great new question might be something like: "Why represent random orbital inclinations as uniformly distributed points on a sphere?"  You could even ask for "demonstrative proofs"!  
> **username_4**: @ConnorGarcia In what way do they differ? Your mean inclination of 57 degrees is 1 radian.  
> **username_6**: @ProfRob  Whoops, I didn't read your recent edits and misinterpreted your answer.  You are right, our answers are the same.  
> **username_3**: @ProfRob - If you read my comments again you'd see that I didn't suggest that randomly oriented orbits have inclinations ranging uniformly 0..180 deg. Nor did I ask for a physical causation. My comments were concerned to question the basis for concluding  that orbits with spherically uniformly distributed angular momentum vectors are randomly oriented.  I suggested another possible way of connecting uniform probability densities with orbital parameters, not to say that this alternative was certainly right, but to point out the need to justify the particular connection chosen.  

---

### Answer by username_6 on 2023-01-30. Score: 2

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](https://en.wikipedia.org/wiki/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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