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I’m writing a SF story with a question about an element of my plot. I have a Manhattan-sized NiFe asteroid, with little or no spin, in orbit at 2.5 astronomical units from the Sun.

Would it be reasonable to assume that the potato-shaped Ni-Fe asteroid, undisturbed for millennia, would settle into an orbit with its long axis (through the c.g.) parallel to its velocity vector? enter image description here

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    $\begingroup$ Interesting question. Aligning the long axis with the direction (velocity vector) at all times, requires it to have a perfectly fitting spin. This is exactly the case for our Moon, which always has the same face towards Earth since it spins in perfect alignment with it's motion around Earth. See this article or google more: discovermagazine.com/2014/dec/2-ask-discover $\endgroup$ – Steeven Mar 24 '17 at 15:04
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    $\begingroup$ As pointed out by @Steeven, tidal forces can stop the rotation such that always the same side is facing the sun. My impression is that this mechanism would not move the long axis in direction of motion if, e.g., the initial condition was a spinning asteroid with the long axis perpendicular to the orbital plane. It would be perfectly allowed, though. And other mechanism might actually cause this. Although, I can't make one up, right now. $\endgroup$ – mikuszefski Mar 24 '17 at 15:12
  • $\begingroup$ @Steeven The planetoid has no spin, is that realistically impossible? $\endgroup$ – catsteevens Mar 24 '17 at 16:45
  • $\begingroup$ No spin at all might not be very likely. I don't think there is any known astronomical object that is perfectly spinless - chances are small. But lets leave answers about how realistic something is to someone who knows more about space than me $\endgroup$ – Steeven Mar 26 '17 at 17:32
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No. Not only is there no reason to expect this, but it is especially unlikely to occur. For the asteroid to remain "pointing" along it's velocity vector, that means that it actually completes a rotation with the exact same period as it's orbit. This is called "tidal locking" and requires tidal forces to be important. Tidal forces scale like,

$$F_\mathrm{tidal} \propto \frac{L}{r^3},$$

For an object of size $L$ (in this case very very small for an asteroid) and a distance between objects of $r$ (where here, $r = 2.5AU$). Even Mercury's orbit isn't tidally locked (corrected by @rob, thanks!). So the only reason this asteroid would continue to point in the same direction is if it coincidentally happened to have the exact right spin angular momentum.

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  • $\begingroup$ Is it possible for the asteroid to have no rotation and therefore no spin angular momentum? Or is that very unlikely? $\endgroup$ – catsteevens Mar 26 '17 at 0:04
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    $\begingroup$ @catsteevens, also unlikely, but definitely not impossible... it could also just be quite small, such that it rotates very slowly. It would be easier to come up with a no-spin explanation than a tidally-locked explanation, however. $\endgroup$ – DilithiumMatrix Mar 26 '17 at 15:19
  • $\begingroup$ For what it's worth, Mercury's spin and orbit are tidally locked --- but in a 3:2 resonance, rather than a 1:1 resonance. $\endgroup$ – rob Mar 27 '17 at 16:36
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N.B.: I did not read the question carefully enough, and so the answer below answers a different question than the one was asked, namely "is it possible for the long-term spin axis of an asteroid to align with its long axis?" I have left it up for posterity.


No. In fact, it will be likely to settle into a rotation about its principal axis with the largest moment of inertia, i.e. one of the "short axes". Rotation about the "long axis"—i.e., the principal axis with the smallest moment of inertia—is unstable.

The reason for this is the following. The angular momentum vector $\vec{L}$ of a rotating body is constant in space. However, if a body is not rotating about its symmetry axis, the different parts of the body will experience time-dependent centripetal acceleration (due to the precession of the angular velocity vector $\vec{\omega}$ in the body frame.) If the body is not perfectly rigid, it will deform ever so slightly under these forces, and frictional forces due to this time-dependent "kneading" will slowly sap the body's rotational kinetic energy.

Now, the kinetic energy of a body rotating in 3D is $$ K = \frac{L_1^2}{2 I_1} + \frac{L_2^2}{2 I_2} + \frac{L_3^2}{2 I_3} $$ where $L_1, L_2, L_3$ are the components of $\vec{L}$ in the directions of the body's principal axes, and $I_1, I_2, I_3$ are the corresponding moments of inertia. It is not too hard to see (or to prove rigorously using Lagrange multipliers) that for a fixed value of $L^2 = L_1^2 + L_2^2 + L_3^2$, the magnitude of $K$ will be the lowest when $\vec{L}$ points along the principal axis with the highest moment of inertia.

Don't believe me? Well, just ask the folks who launched Explorer I, the first US satellite:

Explorer 1 changed rotation axis after launch. The elongated body of the spacecraft had been designed to spin about its long (least-inertia) axis but refused to do so, and instead started precessing due to energy dissipation from flexible structural elements. Later it was understood that on general grounds, the body ends up in the spin state that minimizes the kinetic rotational energy for a fixed angular momentum (this being the maximal-inertia axis). This motivated the first further development of the Eulerian theory of rigid body dynamics after nearly 200 years—to address this kind of momentum-preserving energy dissipation.

The above argument is paraphrased from Ch. 7 of Kleppner & Kolenkow's Introduction to Mechanics. More detailed information can be found in the following review:

M. Efroimsky, Relaxation of wobbling asteroids and comets—theoretical problems, perspectives of experimental observation. Planetary and Space Science, Volume 49, Issue 9, August 2001, Pages 937–955. Arχiv version.

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