This question asks about placing an object at rest in the International Space Station. But the ISS is a large object, large enough that gravity will vary across it and cause tidal forces that can accelerate objects away from where they were left.

How big are the tidal accelerations in the ISS? Are they big enough to noticeably move things to the ceiling and floor, and in from the ends?


2 Answers 2


Tidal forces come from gradients in gravitational fields.

Along the horizontal axis of the station, the distance to Earth stays the same, so there's no gradient and therefore no tidal forces.

To find the gradient along the vertical axis, we can simply use the derivative.

$$\left(\frac{\mu}{r^2}\right)' = -\frac{2\mu}{r^3}$$

At the the orbital radius of the ISS, this works out to a gradient of $2.6 \cdot 10^{-6} s^{-2}$

The ISS is quite flat, so you can only multiply that gradient by the handful of metres between the "floor" and "ceiling" in the modules, so in the order of $\approx10^{-5} m/s^2$

That's still about a magnitude more than the acceleration due to aerodynamic drag on the station, $\approx 10^{-6} m/s²$

It's not the dominating acceleration within the station though. Lost items end up in the air filters.

edit: thanks, uhoh, for reminding me that there's a third axis to consider, the north-south axis. Rigidity here should provide a slight effect, lower than at the vertical axis, but it isn't cancelled out by orbital velocity in the same way the prograde-retrograde axis is. Here's a NASA page mapping out all the 3 axis

  • $\begingroup$ Isn’t there a gradient along the station due to the variation in the direction of “down”? $\endgroup$ Jan 18, 2020 at 23:37
  • $\begingroup$ There would have been, if the station wasn't at orbital velocity. $\endgroup$ Jan 18, 2020 at 23:56
  • $\begingroup$ slightly related: Lowest ISS microgravity $\endgroup$
    – uhoh
    Jan 19, 2020 at 2:18
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    $\begingroup$ There has to be a gradient along the length, unless the station is curved to fit the profile of a 400km-radius circle. If you assume it's straight, and "horizontal", then the ends should be around 170cm further from the Earth than the middle. $\endgroup$
    – hobbs
    Jan 19, 2020 at 19:02
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    $\begingroup$ @hobbs: I think you dropped a few digits there. And forgot to include Earth's radius. Specifically, to get an elevation difference of 170 cm, it looks like you used an orbital radius of 400 meters, not 400 + 6371 kilometers. With the correct radius, I get an elevation difference of about 0.1 mm between the ends of the ISS and the middle. $\endgroup$ Jan 21, 2020 at 9:28

Update: This is not direct answer to the question as asked (thanks to @ruakh for pointing this out in the comments), but rather a supplementary answer that outlines another gravitational factor that would affect motion of an object at rest inside ISS.

According to This NASA article, an object would move inside ISS due to the mutual gravitational force between the object and ISS.

The article suggests that due to general law of mass attraction in absence of other disturbances (such as air flow inside the station) any object inside the ISS, regardless whether it is in relative motion to ISS or without it, will eventually end up at rest touching the wall which is the closest to the common center of mass (of the object and the ISS):

The general law of mass attraction is valid even for the space station itself and causes all masses to be attracted toward the common center of mass; however due to the relative insignificance of the entire mass they are attracted at such an extremely slight acceleration that traveling only one meter takes hours. However, nonsecured objects will finally impact one of the walls of the room either as a result of this or of their other random movement, and either immediately remain on this wall or, if their velocity was sufficiently large, bounce back again and again among the walls of the room depending on the degree of elasticity, floating back and forth until their energy of movement is gradually expended and they also come to rest on one of the walls. Therefore, all objects freely suspended within the space station will land on the walls over time; more specifically, they will approach as close as possible to the common center of mass of the structure.

This phenomenon can extend over hours, sometimes over many days, and even a weak air draft would suffice to interfere with it and/or to tear objects away from the wall, where they are already at rest but only adhering very weakly, and to mix them all up.

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    $\begingroup$ The OP is asking about the effect of differences in the Earth's gravitational field (an object that's slightly further from the Earth than the ISS's center of mass will feel a bit less gravity than the ISS, and therefore drift outward relative to the ISS; and conversely with one that's slightly closer), whereas that article is about the effect of the ISS's own gravitational field (an object in or near the ISS will be pulled toward the ISS's center of mass). But it's good to mention the latter, because the former seems moot if it's dwarfed by the latter (which, dunno if it is). $\endgroup$
    – ruakh
    Jan 19, 2020 at 6:45
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    $\begingroup$ @ruakh thank you for the comment. Would it be safe to assume that there are at least 4 factors that would cause relative motion between a "steady" object inside the ISS and the ISS itself?: 1) difference in radius vector, i.e. distances of the object's and ISS's center of mass/gravity to the center of Earth; 2) mutual gravitational pull between the object and ISS if the object's c.g. isn't in exactly same place as the ISS' c.g.; 3) change in ISS trajectory/velocity due to atmospheric drag whilst the object inside doesn't experience the drag; 4) forces due to circulation of air inside the ISS. $\endgroup$ Jan 19, 2020 at 7:22
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    $\begingroup$ Yes, that's correct. $\endgroup$
    – ruakh
    Jan 19, 2020 at 8:43

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