Very similar question at Astronomy SE: https://astronomy.stackexchange.com/q/6183/4042

Given a typical protoplanetary disk made up of the usual planetary system stuff, dust and gas and whatnot, orbiting a common center: Material coalesces into planets, clearing the soupy disk in favor of a handful of massive objects.

And this is where my question begins: the conventional wisdom of orbital mechanics is that the smaller the orbit, the faster the motion:

Orbital speed

The orbital speed is given by:

$$ v = \sqrt{\frac{GM}{r}} $$

so in the diagram $v_1 > v_2$ i.e. the inner body is moving faster.

So say you've got an asteroid belt that gathers itself up into a proto-Mars. In this case our two bodies aggregate into a single body. If we zoom in to the centre of mass of the two bodies their relative velocities look like:

Centre of mass frame

So the two bodies are revolving clockwise about their centre of mass i.e. in a retrograde direction to the accretion disk as a whole. Doesn't this mean the average angular momentum of the Mars-forming belt will be retrograde with respect to the orbit, given that the inside of the belt moves faster than the outside of it? Why then is the prevailing rotational direction of the planets prograde?

I realize that that assumes circular orbits for all the particles involved, which obviously is unlikely. I suppose in a system of elliptical orbits, a particle or object is probably more likely to be captured by a coalescing mass if it is on the climbing (and therefore slowing) portion of its orbit, which would tend toward an overall prograde contribution of momentum. But is that effect sufficient to explain the overwhelming prevalence of prograde rotation in our solar system?

  • 1
    $\begingroup$ This question is rather confused, and because of that its rather confusing how to answer it. I don't know where to begin. $\endgroup$ Commented Jan 3, 2015 at 10:25
  • $\begingroup$ @DavidHammen comments on the specific aspects of the question which are confused would allow the asker to edit it appropriately, which would probably grease the gears a bit. $\endgroup$
    – Asher
    Commented Jan 3, 2015 at 10:42
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    $\begingroup$ That something inward of Venus is moving faster (a whole lot faster) than something outward of Jupiter has nothing to do with "prograde" versus "retrograde". What matters is the direction in which the specific angular momentum vector $\vec h = \vec r \times \vec v$ points. In the disk, all such vectors point in more or less the same direction. $\endgroup$ Commented Jan 3, 2015 at 11:46
  • $\begingroup$ Asher, I've attempted to clarify your question with a couple of diagrams. If you don't like what I've done please shout and I'll back out the changes. To potential downvoters - please don't downvote if you don't like my edit because that's unfair on Asher. Just shout at me instead :-) $\endgroup$ Commented Jan 3, 2015 at 11:48
  • $\begingroup$ @JohnRennie no, that's great. I'm limited to posting from my phone for the time being, so your edit is a large help. Thanks. $\endgroup$
    – Asher
    Commented Jan 3, 2015 at 11:59

2 Answers 2


In the question above, the given orbital speed of: $$v=\sqrt{\frac{GM}{r}}$$ does not hold. Early gas and dust clouds rotate a star at sub-Keplerian velocities. According to Semenov and Teague: "The departure from the perfect Keplerian rotation is due to the thermal gradient, which makes gas orbital velocities to be slightly sub-Keplerian."

Astronomers disagree about just how much slower the gas travels for various models, but we can get a good idea at least for one system from Hayworth et. al in their paper: "Radiation-pressure-driven sub-Keplerian rotation of the disc around the AGB star L-2 Puppis."

enter image description here

In this case, the gas is moving much more slowly than a Keplerian satellite would.

In the circumstellar disk, solids in the form of tiny particles move at nearly the same slower rate as the gas due to drag. As the solids coalesce into clumps of solids, their ballistic coefficient $C_b$ goes up as expected from: $$C_b= \frac{m}{C_da}$$ where $m$ is the mass of the object, $a$ is the cross-sectional area, and $C_d$ is the drag coefficient. This is a natural result of a shape's mass to cross-sectional area ratio ($m/a$) increasing with size (consider a sphere).

A bigger solid will then have a higher terminal velocity in the same orbit as a smaller solid. As it orbits the star, it will pull in slower, smaller, solid matter pieces due to it's own gravity. The combination of gaseous drag and collisions with slower moving solids will cause the angular momentum of the big piece around the star to decrease. As the angular momentum of the piece decreases, it's orbit will migrate inward. Each orbital pass will have more solid debris inside the orbit of the big piece than outside of the orbit, since the big piece will have already "swept out" debris from its previous passes. The slower moving solids closer to the star will hit the big piece from underneath, causing it to spin in a prograde motion, consistent with the motion of the whole disk.

As the planets lose angular momentum, they create a bow-wave which pushes the gas. This bow-wave adds turbulence and also forces pebbles to land at prograde angles. Johansen and Lacerda performed hydrodynamical simulations in their paper: "Prograde rotation of protoplanets by accretion of pebbles in a gaseous environment". Here is one of their figures.

enter image description here

In the above graph, the x-axis points away from the star, and the y-axis is oriented in the direction of the travel of the disk.

Notice that as the Hill Sphere of a protoplanet increases, it will pull solid particles from increasingly far distances, and they will land at increasingly high velocities, to impart increasing amounts of angular momentum.


  1. This is just one theory of how most of the planets and big asteroids ended up with prograde rotation. As recently as 2019, Visser et. al are running more complex numerical models of pebble accretion landing trajectories due to the protoplanet bow-wave and associated shear forces. Here is an interesting figure from their paper:

enter image description here

  1. I recently posted a similar answer on the astronomy stack exchange here: 1. There are dupes with additional information here: 2 and here: 3.

The diagram above moves from a Heliocentric reference frame to a rotating local reference frame: that means a Coriolis force needs to be added which will push the inner body into a slower orbit that matches the outer body, assuming that it does move outwards towards it.

Think about this on a roundabout while extending your arms and then pulling them in again.


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