The axial tilts of asteroids seem to vary randomly (let me know if this premise is wrong), while the planets have a strong tendency to rotate in the same way. If planets were formed by colliding asteroids, shouldn't the sum of random tilts result in random planetary rotation? Of course other factors are important, like angle and velocity of impacts, YORP-effect, centrifugal breakup and whatnot, but how could that taken together have any systematic effect on rotation?

Ceres does behave with 4° tilt, but the other of the first discovered asteroids have tilts like 84°, 50°, 42°. Dust particles (and gas molecules if applicable) surely rotate randomly. The Solar nebula had a net spin which gravity and friction has manifested in the orbits of the planets. But shouldn't the netting of rotation be individual for each planet, with uncorrelated tilts, as orbital orientation is for each star?


4 Answers 4


You are right that the tilt of the asteroids are distributed in very random way, and that the rotation of the Solar nebula is a minor contributor to that tilt, and only skews it a little.

However, you are not right that randomness simply adds up. The randomness does in fact cancel out more and more when you combine a large amount of asteroids, until the rotation of the nebula becomes the dominant factor. This is related to the Law of large numbers.

For instance, throw a dice. The outcome is random. Throw 10 dices, calculate their sum, and divide by 10. Not so far from the average any more? You can do the same thing with thousands of dices, or millions of asteroids. When the number of asteroids that form an object is really high, the tilt is not going to be far from the average value, determined by the nebula's rotation.

The same argument goes for inclination, and the fact that even though the planets' orbits are elliptical, they are not as far from circular that a random orbit would be.

  • $\begingroup$ But the law of large numbers adds up to an average. Throwing planets like dice would add up to no planet rotation at all on average. Isn't it strange that the dice shows an even number of dots almost all of the time? If the rotation of the nebula affects the rotation of each planet, but not that of the asteroids, then I need more explanation to understand how that is so. Is there any relationship between the rotation of the Solar nebula, and the rotation of the individual planets formed within it? $\endgroup$
    – LocalFluff
    Commented Feb 3, 2016 at 14:30
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    $\begingroup$ @LocalFluff That the random motion's average is going to be zero is my point! The resulting rotation is due to the only non-random component, the rotation of the solar nebula. $\endgroup$ Commented Feb 3, 2016 at 14:39
  • $\begingroup$ That would be the most sane explanation, but still quite short. How does the rotation of the Solar nebula systematically affect each and every individual planet which is formed in it in the same way? Shouldn't half of the planets have been impacted in a way that tipped them over? $\endgroup$
    – LocalFluff
    Commented Feb 3, 2016 at 14:46
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    $\begingroup$ I hate to nitpick, but your third paragraph mixes up the word dice a lot. "throw a dice" should be "throw a die", and "dices" isn't a word, the proper plural form is "dice". I tried to suggest an edit but it wasn't enough characters to count. $\endgroup$
    – Cody
    Commented Feb 7, 2017 at 17:16
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    $\begingroup$ I have to agree with LocalFluff here. You stopped just short of actually answering the question by describing how the "non-random rotation of the solar nebula" actually causes the planets to rotate as they do. If your argument is that the randomly combining asteroids combine to the average, then on average the asteroids are rotating with the disk and the question then remains how did they come to rotate that way (on average). You just shifted the question to a different realm, but provided no answer. $\endgroup$
    – zephyr
    Commented Feb 7, 2017 at 18:46

Remember that in a protoplanetary disc the rotation velocity $v_r$, which is keplerian, as a distance of star r varies like
$$v_r(r) = \sqrt{\frac{GM}{r}} ~ ~~~~~~~~~~~~ (1)$$ This should serve to illustrate part of the point: at any $r<r_0$ we have $v_r>v_r(r_0)$ and the other way around. Thus viewed from the planet's position gas and dust 'left' of it is systematically flowing faster, and 'right' of it systematically flowing slower than the planet.
Thus, if you would accrete a substantial fraction of your total, final mass and thus angular momentum from this flow, you'd induce a systematic spin automatically.

But when is this relevant?
The region from which a protoplanet or asteroid can accrete is maximally its gravitational sphere of influence, also Hill-sphere with radius $$r_H = r_0 \sqrt[3]{\frac{m_{planet}}{3 m_{star}}} ~~~~~~~~~(2) $$ where $r_0$ as above, is a semi-major axis distance.

Now if this $r_H$ is too small to feel the velocity-gradients in (1), or said differently, if the accreting object is not massive enough for $r_H$ to extend significantly into the protostellar disc, then accretion will accumulate random momenta.
If the protoplanet manages to grow to a substantial Hill-sphere, it starts accreting gas and solids with a huge velocity difference $v_r(r)-v_r(r_0)$, which is always systematic, instead of random.

TL;DR Small objects, roughly below the size of asteroids, accrete random momenta pushes. Massive objects, protoplanetary and above, accrete systematic velocity differences, thus giving them a net angular momentum.

  • 1
    $\begingroup$ Can you be assured that the protoplanetary disk is Keplerian? Do you have a source? As LocalFluff points out, that will result in differential rotation of the disk (faster the closer you are) that should result in the rotations being oppositely aligned from the disk revolution. The disk is an extended object with a lot of competing forces besides a central gravity force and I think saying it is Keplerian is a very rough approximation at best. $\endgroup$
    – zephyr
    Commented Feb 7, 2017 at 18:48
  • $\begingroup$ I can certainly agree that by the time the disk is established, these other forces should be negligible and it will be very closely Keplerian, but by that point, the protoplanets likely already have their final spin directions (barring any major collisions). $\endgroup$
    – zephyr
    Commented Feb 7, 2017 at 18:53
  • $\begingroup$ @zephyr: Absolutely wrong about the timescales. Why would that be? The Keplerian disc establishes itself on a free-fall timescale together with the central star. From then on, between birth of the planets and the dissipation of the disc at an age of 1-10 Myrs it is near-Keplerian. I agree that the disc is not perfectly Keplerian, as there are pressure gradients in the game, but those account for a few percent of sub-keplerianity. For the planetary angular momentum, you have to regard the relative momentum, so LocalFluff's argument is wrong. $\endgroup$ Commented Feb 7, 2017 at 19:25

Conservation of angular momentum. The spin of the protoplanetary disc will be randomly determined when it initially forms, but then it becomes the dominant factor. The matter in the disc is then orbiting the centre of mass in the same direction even as it groups into asteroids and then protoplanets. Even though objects have their own individual spin, they all have the larger effect of the disc influencing them. So all the planets rotate the same direction, except Uranus and Venus. I think the hypothesis for those is still protoplanetary collision which has knocked Uranus on it's side and Venus right over.

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    $\begingroup$ Shouldn't the tendency be to rotate in the opposite direction since the inner part of the disc (and planet) orbits faster than the outer? $\endgroup$
    – LocalFluff
    Commented Feb 7, 2017 at 8:44

Conservation of angular momentum largely preserves angular momentum when gaseous planetary nebulas condense to form planets in spite of friction and collisions.This is illustrated below.

The angular momentum of bodies in our solar system are given in http://www.zipcon.net/~swhite/docs/astronomy/Angular_Momentum.html

They are not constant but the gaseous planets are of the same order of magnitude. Orbital Angular Momentum Body orbital radius(km) orbital period(days) mass(kg) L

Mercury 58.e6 87.97 3.30e23 9.1e38

Venus 108.e6 224.70 4.87e24 1.8e40

Earth 150.e6 365.26 5.97e24 2.7e40

Mars 228.e6 686.98 6.42e23 3.5e39

Jupiter 778.e6 4332.71 1.90e27 1.9e43

Saturn 1429.e6 10759.50 5.68e26 7.8e42

Uranus 2871.e6 30685.00 8.68e25 1.7e42

Neptune 4504.e6 60190.00 1.02e26 2.5e42

They are of order e^43. (Mars has less angular momentum. Some may have been distributed to the asteroid belt.)

Each outer planet seems to carry the same angular momentum!

I originally thought that Surya Siddhanta used constancy of angular momentum but it is even simpler. It is simply a snow plough theory which makes larger orbits collect more particles. See "How did the authors of Surya Siddhanta find the diameters of other planets in the solar system?"

I am giving this table to illustrate the constancy of angular momentum even in our solar system presumed to have condensed out of the primordial solar nebula, a fact ancients could have used to determine planetary diameters. Constancy of angular momentum requires planets to spin and orbit around the Sun (or the center of mass).

If there was an angular momentum to begin with is understood. Any large mass of gas or nebula will form eddies eventually by turbulence with rotations in opposite directions as rotations arise naturally (by fluid instability). If each part condenses to a star (and solar system) planetary systems will occur.

Our solar system may have been formed with another mechanism which is a passing star that imparted angular momentum to the original solar nebula.

Very large scale bodies also condense to galaxies (say) and must have black holes at their centers to trap angular momentum. Angular momentum can't be destroyed.

I wish to add this, the rotational angular momentum of all bodies.

Rotational Angular Momentum, L

Body/ mass kg/ radius(km) rotational period (days)/ L

Sun /695000 /24.6 /1.99e30 /1.1e42

Earth/6378 /0.99 /5.97e24 /7.1e33

Jupiter /71492 /0.41 /1.90e27 /6.9e38

Note that the Sun's rotational angular momentum is is also e^42. The spin angular momenta of all planets are small compared to orbital angular momenta.

The outer planets and the Sun have the same angular momenta!

Some kind of equipartition of angular momenta at work?


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