<p>Bigger than an atomic nucleus but smaller than an atom.</p>
<p>The tidal forces scale as <span class="math-container">$MR^{-3}$</span>, where <span class="math-container">$M$</span> is the mass of the object causing the tides and <span class="math-container">$R$</span> would be its separation from the centre of the Earth. Note there will be second order terms when <span class="math-container">$R$</span> is not much bigger than the Earth's radius, that I ignore.</p>
<p>Determining the <em>height</em> of a tide depends on all sorts of geographical factors, but if you were to assume a model of the Earth entirely covered by a fluid, then the tidal force above translates into an equipotential surface that also has an amplitude (i.e. +/- from a sphere) that scales as <span class="math-container">$MR^{-3}$</span> (see <a href="https://physics.stackexchange.com/questions/806704/how-to-calculate-equilibrium-height-of-tidal-bulge">https://physics.stackexchange.com/questions/806704/how-to-calculate-equilibrium-height-of-tidal-bulge</a> for example).</p>
<p><a href="https://en.wikipedia.org/wiki/International_Space_Station" rel="noreferrer">Wikipedia says</a> the mass of the space station is 450 tonnes and it is in an orbit about 420 km above the Earth's surface. i.e. <span class="math-container">$M \sim 4.5 \times 10^5$</span> kg and <span class="math-container">$R \sim 6.8 \times 10^{6}$</span> m.</p>
<p>We can now compare this with the tides raised by the Moon (ignoring the contribution of the Sun), which have a theoretical average peak-to-peak amplitude of 0.54m (<a href="http://www.cambridge.org/resources/0521846560/7708_Tidal%20distortion.pdf" rel="noreferrer">Bradt 2008</a>), where <span class="math-container">$M \sim 7\times 10^{22}$</span> kg and <span class="math-container">$R \sim  3.8 \times 10^8$</span> m. Note that this calculation assumed an entirely fluid Earth, however, it appears to produce a quite accurate number for the <em>average</em> tidal amplitude, which is all we actually need here. Thus the amplitude you are looking for is
<span class="math-container">$$\Delta h \sim 0.54\left(\frac{M_{\rm ISS}}{M_{\rm Moon}}\right) \left(\frac{R_{\rm Moon}}{R_{\rm ISS}}\right)^3 = 6\times 10^{-13}\ {\rm m}\ . $$</span></p>
<p>The deviation of the Earth's centre from the Earth plus space station barycentre (about <span class="math-container">$4\times 10^{-11}$</span> m) is an unrelated quantity which increases the <em>further away</em> the space station is, whereas the tidal effects would decrease as <span class="math-container">$R^{-3}$</span>.</p>

CC BY-SA 4.0
2024-09-07T18:18:19.053Z
Could we use this same relation to estimate tidal effects caused by an airliner (similar mass to ISS but flying some 400km lower) or does it only apply to objects in orbit? Im wondering because the aircraft would continually be pushing the atmosphere in the opposite direction so effects might cancel somewhat?
Seb
CC BY-SA 4.0
2024-09-07T21:35:09.303Z
@Seb pretty much yes. Aircraft will be the same order of magnitude as ISS, but large mountains do create measurable (constant) tides around themselves.
fraxinus
CC BY-SA 4.0
2024-09-08T00:25:56.913Z
I think that by "pretty much yes" he means that the airliners would also cause some tidal effect to the surface of the sea. The fact that they are pushing the atmosphere in the opposite direction is absolutely of no consequence what-so-ever.
Mike Nakis
CC BY-SA 4.0
2024-09-08T00:32:41.087Z
@ProfRob I do not see the average depth of the ocean factoring into your calculations, so I think that what you are assuming is a model of the Earth not just entirely covered by a fluid, but actually, the entire Earth being nothing but a ball of fluid, right?
Mike Nakis
CC BY-SA 4.0
2024-09-08T00:52:01.647Z
[-1] for being unphysical. I think there is a real problem using the scaling for an object much greater than one Earth diameter away for something *much closer* than one Earth diameter. The ISS is only 400 km from the ocean surface. And even if you were to really calculate the amplitude of the theoretical hydrostatic equilibrium surface displacement *properly*, would the ocean swell really move at 7 km/sec across the ocean? I think this is a bad, unphysical approach.
uhoh
CC BY-SA 4.0
2024-09-08T06:48:21.690Z
@uhoh I find your dislike to my rough, order of magnitude calculation quite odd. I won't be refining it to take account of what amounts to a small geometric factor in your first objection. As regards the second, nothing has to move by more than $10^{-12}$ m. Do you think the Moon causes water to race around the Earth at 0.5 km/s?
ProfRob
CC BY-SA 4.0
2024-09-08T07:03:09.633Z
@MikeNakis I am simply scaling the result for the Moon. The calculation I reference does assume a fluid body. However, it does produce around the correct empirical 
 result. Substitute 0.54m for a favourite value of the average tidal amplitude due to the Moon .
ProfRob
CC BY-SA 4.0
2024-09-08T08:13:52.453Z
@ProfRob The speed of sound in water is 1.5 km/s so I DO expect the water to respond to a stimulus that moves 0.5 km/s and I DO NOT expect it to respond to a stimulus that moves at 7 km/s. And it's just common sense that an approximation derived for a >> b should not be used for a << b without at least warning the reader what's been done. If someone did that on a homework set or exam and handed it in without making such a note, would you give them a perfect score?
uhoh
CC BY-SA 4.0
2024-09-09T01:40:00.793Z
@DaveTheWave Yes, I believe *a lot lower* perhaps several orders of magnitude or more - enough orders of magnitude that *even an astronomer* should take note! (I'm not disparaging astronomers - orders of magnitude estimates and scaling laws are standard tools. But there are limits, even in Astronomy)
uhoh
CC BY-SA 4.0
2024-09-07T10:50:01.973Z
2024-09-08T06:58:46.240Z
2024-09-08T06:58:46.240Z
ProfRob
<p>Does the tide rise by the length of a water molecule? By  one Planck length?</p>
<p>I'll settle for the answer to: What's the distance between the center of the earth and the barycenter of the earth / space-station system?</p>
<p>BTW, as I described here at stax, the sun's barycenter motion is fascinating. It looks chaotic, but it's really the superposition of many pretty circles and spirals. <a href="https://astronomy.stackexchange.com/questions/54713/does-the-sun-have-different-unique-orbits-around-barycenters/58206#58206">Pluto pulls the sun about 20 miles off-center</a>.</p>

CC BY-SA 4.0
2024-09-07T19:14:48.443Z
For terrestrial satellite orbits, it's better to use the EMB (Earth-Moon Barycentre), not the centre of the Earth. See https://i.sstatic.net/QsXQEbSn.png & https://i.sstatic.net/zOdMFc75.png (3 minute timestep, using the Terrestrial Time scale, not UTC).
PM 2Ring
CC BY-SA 4.0
2024-09-08T00:54:10.097Z
I think the currently accepted answer is [way wrong](https://astronomy.stackexchange.com/questions/58497/how-high-does-the-ocean-tide-rise-every-90-minutes-due-to-the-gravitational-pull#comment134791_58498) and should not be accepted.
uhoh
CC BY-SA 4.0
2024-09-08T10:01:49.727Z
@uhoh - You should post an alternative answer, providing a different calculation, then.
Bobson
CC BY-SA 4.0
2024-09-09T01:31:51.003Z
@Bobson And exactly how would you propose that that be done? It's quite a challenging problem *with no easy answers* available! That's exactly why it should not be made to look like it is an easy problem to answer.
uhoh
CC BY-SA 4.0
2024-09-10T09:49:30.650Z
@uhoh - No idea, but it's not my field.  But I'm willing to accept the back-of-a-napkin methodology of a 150k+ rep user as being a good guess at it.  Rep doesn't automatically mean that the user is right on this particular topic, but it does mean that they're more of an expert than I am.  If you have reason to think that the posted answer is way off base, then you should be able to at least post an answer explaining the factors involved and why the assumptions in the current one are invalid, even if the conclusion ends up being "result is impossible to determine".
Bobson
CC BY-SA 4.0
2024-09-10T16:26:39.200Z
@Bobson my comments are already sufficient and complete, and what you propose would not be a proper Stack Exchange answer.
uhoh
<gravity><earth><tidal-forces><iss>
How high does the ocean tide rise every 90 minutes due to the gravitational pull of the space station?
Miss Understands
# How high does the ocean tide rise every 90 minutes due to the gravitational pull of the space station?

(Asked by: username_0 on 2024-09-07)

Does the tide rise by the length of a water molecule? By one Planck length?

I'll settle for the answer to: What's the distance between the center of the earth and the barycenter of the earth / space-station system?

BTW, as I described here at stax, the sun's barycenter motion is fascinating. It looks chaotic, but it's really the superposition of many pretty circles and spirals. [Pluto pulls the sun about 20 miles off-center](https://astronomy.stackexchange.com/questions/54713/does-the-sun-have-different-unique-orbits-around-barycenters/58206#58206).

**Tags:** gravity, earth, tidal-forces, iss

> **Comments:**
> **username_1**: For terrestrial satellite orbits, it's better to use the EMB (Earth-Moon Barycentre), not the centre of the Earth. See https://i.sstatic.net/QsXQEbSn.png & https://i.sstatic.net/zOdMFc75.png (3 minute timestep, using the Terrestrial Time scale, not UTC).  
> **username_2**: I think the currently accepted answer is [way wrong](https://astronomy.stackexchange.com/questions/58497/how-high-does-the-ocean-tide-rise-every-90-minutes-due-to-the-gravitational-pull#comment134791_58498) and should not be accepted.  
> **username_3**: @username_2 - You should post an alternative answer, providing a different calculation, then.  
> **username_2**: @username_3 And exactly how would you propose that that be done? It's quite a challenging problem *with no easy answers* available! That's exactly why it should not be made to look like it is an easy problem to answer.  
> **username_3**: @username_2 - No idea, but it's not my field.  But I'm willing to accept the back-of-a-napkin methodology of a 150k+ rep user as being a good guess at it.  Rep doesn't automatically mean that the user is right on this particular topic, but it does mean that they're more of an expert than I am.  If you have reason to think that the posted answer is way off base, then you should be able to at least post an answer explaining the factors involved and why the assumptions in the current one are invalid, even if the conclusion ends up being "result is impossible to determine".  
> **username_2**: @username_3 my comments are already sufficient and complete, and what you propose would not be a proper Stack Exchange answer.  

---

## 1 answers

### Answer by username_4 on 2024-09-07. Score: 27 | Accepted answer

Bigger than an atomic nucleus but smaller than an atom.

The tidal forces scale as $MR^{-3}$, where $M$ is the mass of the object causing the tides and $R$ would be its separation from the centre of the Earth. Note there will be second order terms when $R$ is not much bigger than the Earth's radius, that I ignore.

Determining the *height* of a tide depends on all sorts of geographical factors, but if you were to assume a model of the Earth entirely covered by a fluid, then the tidal force above translates into an equipotential surface that also has an amplitude (i.e. +/- from a sphere) that scales as $MR^{-3}$ (see <https://physics.stackexchange.com/questions/806704/how-to-calculate-equilibrium-height-of-tidal-bulge> for example).

[Wikipedia says](https://en.wikipedia.org/wiki/International_Space_Station) the mass of the space station is 450 tonnes and it is in an orbit about 420 km above the Earth's surface. i.e. $M \sim 4.5 \times 10^5$ kg and $R \sim 6.8 \times 10^{6}$ m.

We can now compare this with the tides raised by the Moon (ignoring the contribution of the Sun), which have a theoretical average peak-to-peak amplitude of 0.54m ([Bradt 2008](http://www.cambridge.org/resources/0521846560/7708_Tidal%20distortion.pdf)), where $M \sim 7\times 10^{22}$ kg and $R \sim 3.8 \times 10^8$ m. Note that this calculation assumed an entirely fluid Earth, however, it appears to produce a quite accurate number for the *average* tidal amplitude, which is all we actually need here. Thus the amplitude you are looking for is
$$\Delta h \sim 0.54\left(\frac{M\_{\rm ISS}}{M\_{\rm Moon}}\right) \left(\frac{R\_{\rm Moon}}{R\_{\rm ISS}}\right)^3 = 6\times 10^{-13}\ {\rm m}\ . $$

The deviation of the Earth's centre from the Earth plus space station barycentre (about $4\times 10^{-11}$ m) is an unrelated quantity which increases the *further away* the space station is, whereas the tidal effects would decrease as $R^{-3}$.

> **Comments:**
> **username_5**: Could we use this same relation to estimate tidal effects caused by an airliner (similar mass to ISS but flying some 400km lower) or does it only apply to objects in orbit? Im wondering because the aircraft would continually be pushing the atmosphere in the opposite direction so effects might cancel somewhat?  
> **username_6**: @username_5 pretty much yes. Aircraft will be the same order of magnitude as ISS, but large mountains do create measurable (constant) tides around themselves.  
> **username_7**: I think that by "pretty much yes" he means that the airliners would also cause some tidal effect to the surface of the sea. The fact that they are pushing the atmosphere in the opposite direction is absolutely of no consequence what-so-ever.  
> **username_7**: @username_4 I do not see the average depth of the ocean factoring into your calculations, so I think that what you are assuming is a model of the Earth not just entirely covered by a fluid, but actually, the entire Earth being nothing but a ball of fluid, right?  
> **username_2**: [-1] for being unphysical. I think there is a real problem using the scaling for an object much greater than one Earth diameter away for something *much closer* than one Earth diameter. The ISS is only 400 km from the ocean surface. And even if you were to really calculate the amplitude of the theoretical hydrostatic equilibrium surface displacement *properly*, would the ocean swell really move at 7 km/sec across the ocean? I think this is a bad, unphysical approach.  
> **username_4**: @username_2 I find your dislike to my rough, order of magnitude calculation quite odd. I won't be refining it to take account of what amounts to a small geometric factor in your first objection. As regards the second, nothing has to move by more than $10^{-12}$ m. Do you think the Moon causes water to race around the Earth at 0.5 km/s?  
> **username_4**: @MikeNakis I am simply scaling the result for the Moon. The calculation I reference does assume a fluid body. However, it does produce around the correct empirical   result. Substitute 0.54m for a favourite value of the average tidal amplitude due to the Moon .  
> **username_2**: @username_4 The speed of sound in water is 1.5 km/s so I DO expect the water to respond to a stimulus that moves 0.5 km/s and I DO NOT expect it to respond to a stimulus that moves at 7 km/s. And it's just common sense that an approximation derived for a >> b should not be used for a << b without at least warning the reader what's been done. If someone did that on a homework set or exam and handed it in without making such a note, would you give them a perfect score?  
> **username_2**: @DaveTheWave Yes, I believe *a lot lower* perhaps several orders of magnitude or more - enough orders of magnitude that *even an astronomer* should take note! (I'm not disparaging astronomers - orders of magnitude estimates and scaling laws are standard tools. But there are limits, even in Astronomy)  

---

