The Moon's position affects tides. So, is there any possibility for man-made satellite positions to also affect tides?


3 Answers 3


Note: this is a 'Fermi estimate' answer, so I'm going to round off and ignore minor effects.

Tides are caused by gravity, and gravity is a really weak force. The mass of the Moon is 7 x 1022 kg, and it causes tides where the difference between water levels is on the order of 10 m.

Satellites are on the order of 103 kg, and they're 103 times closer, so the tidal effect caused by the average satellite is 1013 times smaller, and that's too small to be measurable.

Another factor is the number of satellites: 1000 sats in orbits evenly distributed around Earth will cancel out each other's tidal effects.

  • 31
    $\begingroup$ Another small correction: the amplitude of tides is proportional not to the strength of the gravity force, which is proportional to $r^{-2}$, but to its gradient, which is proportional to $r^{-3}$. So a satellite's tides would be "only" about $10^{10}$ times smaller than the Moon's. $\endgroup$
    – Litho
    Commented Jun 12, 2019 at 7:50
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    $\begingroup$ Good answer, but tides are actually only about 0.5m. Everything bigger than that is caused by the water sloshing around in the ocean bed – or, if you want to put it more respectably, by the rhythm of the tidal pull exciting a resonance in the given body of water. $\endgroup$ Commented Jun 12, 2019 at 10:53
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    $\begingroup$ It might help to be clearer that a fact of $10^3$ in distance means a factor of $10^9$ in tidal effect. $\endgroup$ Commented Jun 12, 2019 at 19:16
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    $\begingroup$ I just did the numbers quickly, but I think Pluto has a substantially larger effect on the tides than a satelite does. $\endgroup$
    – Cort Ammon
    Commented Jun 12, 2019 at 20:33
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    $\begingroup$ Airplanes are both much heavier than satellites (excepting ISS) and a lot closer. So Boeing/Airbus tides should be 100 - 1000 times higher than even the ISS ones. $\endgroup$
    – IMil
    Commented Jun 12, 2019 at 23:31

Everything that Hobbes says in his answer is correct, but doesn't seem to me to give a clear answer, which is YES they do affect tides, though not to a measurable extent.

When they were launched, that had a tiny effect on the earths rotation, though not particularly relevant here.

Every single one has exactly the same effect as the moon does, but to a far smaller extent, due to the tiny mass. The totality of artificial satellites, including the ISS, are NOT uniformly distributed, so they do not entirely cancel each other out.

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    $\begingroup$ It's like asking "Can a pea raise a truck?" That answer is yes, but the net effect is so small as to be considered 0. $\endgroup$
    – Machavity
    Commented Jun 13, 2019 at 14:44

Let's do one small calculation.

The mass of the Moon ~ 7x10^22 kg, roughly 4x10^8 m away.

Mass of a really large satellite (ISS) ~ 5x10^6 kg roughly 4x10^5 m away.

The magnitude of the tide will be proportional to the mass of the orbiting object, and to first order, proportional to the cube of the distance.

For the Moon, M/(r^3) is roughly 0.001

For the ISS, M/(r^3) is roughly 8x10^-11

So, the ISS raises a tide that is around Seven orders (~1x10^7) of magnitude weaker than that of the Moon.

  • $\begingroup$ Proportional to the cube of the distance, as others have pointed out. $\endgroup$
    – TonyK
    Commented Jun 13, 2019 at 21:52
  • $\begingroup$ No, in this case, it is proportional to the inverse square of the distance. $\endgroup$
    – Astrolien
    Commented Jun 14, 2019 at 2:25
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    $\begingroup$ Stating it twice doesn't make it true. $\endgroup$
    – TonyK
    Commented Jun 14, 2019 at 8:39
  • $\begingroup$ Ya, now I understood, will correct it. Thanks! $\endgroup$
    – Astrolien
    Commented Jun 14, 2019 at 13:45
  • $\begingroup$ Surely the magnitiude will be inversely proportional to distance?? $\endgroup$
    – MikeB
    Commented Jun 17, 2019 at 10:43

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