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I believe the ISS uses control moment gyroscopes (CMGs) to always have the same side facing the earth.

How far from the earth does an object the size of the ISS need to be tidally locked with the earth like the moon?
What about objects of other sizes? (Small/big artificial satellites or an asteroid)

Speedphoenix
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    The ISS is not akin to a natural satellite. It has a very irregular shape, might change in shape (e.g. when orienting solar panels, or manipulation of external instruments) and is subject to atmospheric drag.

    Furthermore, if considering higher altitudes, tidal locking works by deformation of the body due to tidal force, it is quite possible that this effect would be negligible for a body as small as the ISS.

     – M'vy Jun 05 '19 at 08:42
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    Closer is better for tidal locking of small bodies, because of the gravity gradient. – CourageousPotato Jun 05 '19 at 13:15
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    Do you mean gravitational tidally locking? – Seth Kurkowski Jun 07 '19 at 00:51
  • @uhoh As I don't have 15 reputation on this community votes I cast aren't shown yet, but I did up vote your answer. That was most likely someone else. I wish they would provide a better answer if they down voted though. – Speedphoenix Jun 07 '19 at 13:12
  • @Speedphoenix okay thanks for the speedy reply! – uhoh Jun 07 '19 at 13:13

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I don't think it is possible at any altitude because the ISS has the wrong moment of inertia (mass distribution) for this.

Moments of inertia

As far as I understand it, to use gravity-gradient stabilization efficiently, you need to have the largest moment of inertia in the direction of the gradient; the zenith-nadir or vertical direction. You also need a source of damping force to remove energy in the oscillations that can build up over time due to any other torques. See for example this answer. See also this answer and the extensive series of comments be low it.

From Gravity Gradient Stabilization of Earth Satellites by R. E. Fischell as discussed in this answer (click for full size):

gravity gradient capture damping used in gravity gradient stabilization

In the case of a long thin rod, there is a cancellation of terms and angular acceleration is independent of spacecraft length or mass. From this answer:

For some math, see (Help with my tensor tension; how to derive and calculate this rigid body gravity gradient torque?) and especially @Litho's answer. For a thin rod of mass $m$ and length $l$ in circular orbit, with a perpendicular moment of inertia $\frac{1}{12}ml^2$, rotating in-plane around the short axis, the torque (to first order) is given by

$$L_G = -\frac{GM_Eml^2}{8R_C^3}\sin 2\theta,$$

and the instantaneous angular acceleration is simply

$$\ddot{\theta} = -\frac{3GM_E}{2R_C^3}\sin 2\theta.$$

That's a pretty amazing result! With $GM_E$ of about 3.986E+14 m^3/s^2 and an altitude of 400 km, $\ddot{\theta}$ can be as large as 0.4 degrees per minute^2 at 45 degrees, and that is independent of length!

Aerodynamic drag

Let's compare that to an order of magnitude estimate of the angular acceleration due to aerodynamic drag on the ISS' solar panels.

The moment of inertia of ISS can be approximated with $\frac{1}{12}ml^2$ using a mass of about 400,000 kg and a length of about 70 meters I get a moment of inertia of about 1.6E+08 kg m^2.

I estimate a solar panel area of all the panels on one end of the ISS to be about 800 square meters. Using a simple model for drag force

$$F_D = \frac{1}{2} \rho v^2 C_D A$$

and high solar activity density at 400 km of about 5E-11 kg/m^3 and a worst case configuration, I get a drag force of about 1.2 Newtons at about 30 meters from the ISS' center of mass.

Using $\ddot{\theta} = 1.2 N \times 30 m / 1.6 \times 10^8 kg \ m^2$ I get an angular acceleration due to an unfortunate drag scenario of only about 0.05 degrees per minute squared, way below the torque that can be generated by the gravity gradient once the ISS starts tilting towards vertical orientation.

Conclusion

The ISS can't use gravity gradient stabilization to keep it in it's current orientation - long direction pointed forward. However it could certainly be used if you wanted to orient it nose-down. In that case the stabilization is much larger than even an unfortunate drag scenario, and it's probable that by articulating the solar panels properly you can use drag as the damping force to maintain stabilization.

From here:

International Space Station after undocking of STS-132

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uhoh
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  • @Speedphoenix This is very mis-leading/wrong. You are looking at the tidal torque, but it's how the tidal torque self-damps that's significant in determining tidal lock. The ISS, as any other body that experiences tidal forces, will eventually become tidally locked, if not subject to other forces. The damping force has to do with how the body absorbs force as is distorts. I have none of the numbers for this but I would guess the ISS is pretty subject to damping (especially as there is water on board). How low before atmospheric forces dominate, I don't know but it almost certainly will happen. – ANone Jun 10 '19 at 09:55
  • @ANone there's no such thing as "self-damping torque" but certainly any realistic structure will depart from strict rigid body motion so it can exhibit damping. There's also ohmic losses produced by eddy currents induced by movement in the Earth's magnetic field, though those would be pretty slow. – uhoh Jun 10 '19 at 10:05
  • modulo wording it very much is. It relates to something referred to as the "dissipation function of the satellite", or Q. there are papers about it (http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.890.4049&rep=rep1&type=pdf) as stands in your answer you only talk of the complete rigid-body version. this has a torque but it doesn't damp the rotation, only cause it to oscillate, which doesn't result in tidal lock. Your answer is incorrect. – ANone Jun 10 '19 at 14:44
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    @ANone I haven't listed sources of damping in the answer (beyond mentioning aerodynamic drag) because that's outside the scope of the question ask asked. That doesn't make the answer incorrect. So until you point out something specific that is incorrect, the idea that the answer doesn't say what you would like it to therefore it is incorrect is a bit of a stretch. If you'd like to write an additional answer to the current question as asked, and also write about sources of damping, go for it! Let's see what your ideal answer looks like! – uhoh Jun 10 '19 at 16:37
  • Its not good enough to be 'wrong'. It completely misses the point, and doesn't answer the question. Reaction control is about stability and orientation, and aero-drag causes instability. Gravity torque doesn't damp this. That It can be a large in some cases make it worse. As for something specifically incorrect: "However it could certainly be used if you wanted to orient it nose-down.", is no where close to the mark at its current altitude (due to aero-forces, which directly relates to the question). But that's not the issue here it's that your answer is misleading and confuses concepts. – ANone Jun 11 '19 at 09:15
  • @ANone it really doesn't. Your sentence "Gravity torque doesn't damp this." doesn't even make sense from a physics point of view. I'm concluding that your criticisms are cantankerous and not base on a genuine appreciation of the topic. – uhoh Jun 11 '19 at 09:19
  • This does miss the point. Stabilisation is a second order effect to gravity torque itself. The ISS would not be stable in any orientation given how low in the atmosphere it is, and if it followed only the equations you give, nothing would be tidally locked. – drjpizzle Jun 12 '19 at 08:19
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    @drjpizzle Assuming the ISS is pointing long-axis down, I'm pretty sure I've demonstrated using correct math that the torque due to the gravity gradient will be much larger than any torque due to differential drag, and the solar panels can be manipulated to generate sufficient damping to stabilize. If you would like to argue otherwise, why not write an answer and show your math? I'd love to be proven wrong, because then I learn something. But just an off-hand comment with the words "second order" but without anything quantitative is completely unconvincing. – uhoh Jun 12 '19 at 09:45
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    @drjpizzle Why not take a minute, work the problem, and show your work in a new answer? That's the best way to support your comment. – uhoh Jun 12 '19 at 09:48