Can artificial satellite positions affect tides? on Space Exploration.SE asks if artificial satellites can affect tides, because they are a lot closer than our natural satellite, even though they are much smaller in mass. The top answer to that question is basically "no."

A commenter estimates that Pluto might have a larger effect, but Pluto is such a distant object I didn't think most people could easily relate to that, so I tried running the numbers with something people might be more familiar with, closer to our surface: airplanes.

Let's start by considering a big plane in relatively common use. The Boeing 747 is well known as a popular aircraft type, and it seems that the version of it they're producing now is the Boeing 747-8. This page says it has a maximum takeoff weight of 4.48*105 kg and a maximum landing weight of 3.12*105 kg, with a service ceiling of 1.3*104 m. I'll round that to a weight of 3.5*105 kg mass and 1*104 m height above the surface of the ocean.

Now, the moon. Wikipedia says the moon has a mass of around 7*1022 kg and a distance from the center of the earth on the order of 4*106 km, which is 4*109 m. Earth's radius, about 6.3*103 km, is about a tenth of a percent of that distance and lost in the rounding/simplification of the Moon's orbit to be circular instead of its actual eccentricity.

This means that the moon's mass is (7*1022 kg / 3.5*105 kg =) 2*1017 times that of the 747. The moon's distance from the surface of the ocean is (4*109 m / 1*104m =) 4*105 times that of the 747.

Comments on this answer note that "the amplitude of tides is proportional not to the strength of the gravity force, which is proportional to r2, but to its gradient, which is proportional to r3...a factor of 103 in distance means a factor of 109 in tidal effect." Therefore, the effect that the distance has on the amplitude would be ((4*105)3 =) 6.4*1016.

Thus the overall effect of the 747 on tides should be (6.4*1016 / 2*1017 =) about a third of the effect of the moon. It's a back-of-the-envelope order-of-magnitude calculation, but a much larger result than I expected.

Did I mess up somewhere in the math or facts above? (If not, I'm willing to cut & paste that content into an Answer, but for now it's just showing the work I've done in arriving at this question and attempting to solve it myself.)

Now, there are many flights in the sky at any given moment, all around the world, and to a certain extent any tidal effects of one flight would be cancelled out by the effects of one or more others. However, the flight map is not completely balanced or evenly distributed north-south or east-west.

Do flights affect tides?

  • $\begingroup$ Your heights should probably be from the centre of the earth, not from sea level. But I'm not sure off the top of my head whether the r^3 formulation will hold with something as close to the earth as an aeroplane - I don't recall the derivation, and whether it assumes that the distance from earth to moon is much greater than the diameter of either... remember that tides are a global phenomenon, not a local thing. You can calculate the simple gravitational pull between plane and a parcel of water, but then you're back to r^2, not r^3. $\endgroup$ Commented Jun 13, 2019 at 20:27
  • $\begingroup$ Because the gravitational field of a body is much more inhomogeneuos close to it, and because tidal forces originate from field differences, the tidal effect would be larger if the airplane were flying directly above the water; but it wouldn't be a tide in the classic sense, obviously. I remember a school experiment measuring the gravitational attraction of a lead ball on a near-by mass. In effect, that is a tidal force if we consider the near-by mass part of the earth (most of the rest of the earth was not really attracted by the lead ball). $\endgroup$ Commented Jun 14, 2019 at 16:31

3 Answers 3


Both airborne planes and the Moon affect the tides, but the effect of airborne planes is so small as to be negligible.

  • From your sources the mass of a Boeing 747 ($m_{747}$) is $447.696$ $(10)^3$ $kg$.
  • Likewise, from your sources, the mass of the Moon ($m_{Moon}$ is $7.347$ $(10)^{22}$ $kg$.
  • Typically a 747 would fly at an altitude ($r_{747}$) of around $30$ $000$ $ft$. This is close to $10$ $000$ $m$ ($10^4$ $m$).
  • The distance the Moon is from the Earth ($r_{Moon}$) is $384$ $400$ $km$, which is $384.4$ $(10)^6$ $m$.

Now, Newton's Law of Universal Gravitation states, $F$ $=$ ${G.m_1.m_2}/{r^2}$

Where: $G$ is the gravitational constant, being $6.6743$ $10^{-11}$ $m^3.kg^{-1}.s^{-2}$, $m_1$ and $m_2$ are the masses of the two objects being considered & $r$ is the distance between them.

Because tides are being considered only the mass of water is to be considered, not the mass of the Earth. To simplify things I am assuming the area of water affect by each is the same and hence the mass of water will be the same in each case.

For the Earth and Moon system, the force of gravity will be:

$F_{E-M}$ $=$ $6.6743(10)^{-11}$ $.$ $7.347(10)^{22}$ $.$ $m_{ocean}$ $/$ $[{384.4(10)^6}]^2$ $=$ $3.3186(10)^{-5}$ $.$ $m_{ocean}$

For the Earth Boeing 747 system, the force of gravity will be:

$F_{E-747}$ $=$ $6.6743(10)^{-11}$ $.$ $447.696(10)^{3}$ $.$ $m_{ocean}$ $/$ $[{(10)^4}]^2$ $=$ $2.9881(10)^{-13}$ $.$ $m_{ocean}$

Dividing $F_{E-M}$ by $F_{E-747}$ the gravitational force of the Moon is $111$ $060$ greater than that of the plane.

If the tides produced by the Moon are up to $10$ $m$, then the maximum height of a tide produced by a 747 would be around $0.099$ $mm$.

  • $\begingroup$ Thanks for your answer! This relies entirely on the assumption that the area of water affect by each is the same and hence the mass of water will be the same in each case. I'm not fully convinced that assumption is accurate. Can you discuss more about it and why you think it should apply in the answer? $\endgroup$
    – WBT
    Commented Jun 15, 2019 at 3:24
  • 2
    $\begingroup$ This answer is incorrect. Tidal forces exerted by a satellite are proportional to the inverse cube of the distance to the center of the planet rather than the inverse square of the distance to the surface of the planet. $\endgroup$ Commented Jun 16, 2019 at 13:12
  • $\begingroup$ @DavidHammen I suggest you write a competing Answer with some additional details, especially to counter the intuition that an effect on water would seem most likely related to distance to that water. $\endgroup$
    – WBT
    Commented Jun 17, 2019 at 14:02
  • $\begingroup$ @WBT What he says is true. Different parts of the ocean area pulled at different rates towards the moon, that is what causes the tidal force. Consider that the moon pulls on the Earth's continental masses at the same rate as it pulls on the adjacent bits of ocean, so why should that cause any acceleration of the water relative to the land? The ocean being liquid, differing forces from the moon in different parts of the ocean are what drive 'circulation' and cause the tides. $\endgroup$
    – kingledion
    Commented Jun 26, 2019 at 14:42
  • $\begingroup$ @kingledion Maybe you should write a competing answer then. On the SE network, is it better to have answers posted as such than as comments on the question or a different answer, so they can be discussed and voted on independently. $\endgroup$
    – WBT
    Commented Jun 26, 2019 at 15:34

A plane might be located 1.3*103 meters away from the surface of the ocean right underneath it, but for most of the ocean, it's located much further away, and at a very different angle.

While it may pull more than the moon on the water it flies over for the short time it flies over it, that doesn't make a tide by any definition. At most a wave. But I think that effect is hardly significant as opposed to other effects creating waves. I'm not that knowledgeable on the maths, though.

  • $\begingroup$ The 10^3 was a typo in the write-up, now corrected to 10^4, not affecting the math. $\endgroup$
    – WBT
    Commented Jun 15, 2019 at 3:11

I think you should calculate the distance between the plane and the oceanic floor (instead the surface), or at least some depth. Unlike plane, the moon has almost the same gravitational influence on the surface of the oceans as in the deepest ocean trenches.

Plus, planes flies very fast and are very small objects. Oceans moves too slow, a wave can't follow each plane. No temporary effect is measurable because instantly dissipated by the waves (think about wind). The gravitational influence of the moon affects a vast part of the ocean, giving it time to react.

Oil platforms have certainly more influences, they are just above the surface, heavier, can be far in the ocean and do not move!

Planes have surely more influence on weather (due to reactors / fuel) than waves.

  • $\begingroup$ Planes don't fly all that fast. The 747-8 mentioned elsewhere has, according to Google, a service speed of 988km/h. But the moon and the sun effectively have a "speed" of 1666km/h, which is the circumference of the Earth divided by 24 hours. Your other points, though, are valid. (not the oil platform one.) $\endgroup$ Commented Jun 13, 2019 at 20:12
  • $\begingroup$ Speed of moon does not matter, because gravity of the moon affect all earth, while gravity of a plane only affect a small part of the ocean. The amplitude of tides is < 20m, when ocean have 3,5km of deep, it's very limited, even for the moon. Maybe because moon move too fast! $\endgroup$
    – Melnofil
    Commented Jun 20, 2019 at 21:02
  • $\begingroup$ The gravity of the moon does not affect all of the earth equally. That's the whole reason we have tides. The speed of the moon's orbit really does matter; it provides the most important input into the periodicity of the whole system. $\endgroup$ Commented Jun 20, 2019 at 22:03
  • $\begingroup$ But… You're talking about a non-Euclian space! It's very hard to calculate tides (due to a massive space object) in a non-Euclian geometry (spherical geometry in you're case). In any Euclian space, the gravity of the Moon affect all the Earth equally (almost same force vector at New York, Tokyo, Paris and Antarctica). $\endgroup$
    – Melnofil
    Commented Jun 21, 2019 at 8:00
  • $\begingroup$ For simplicity, I suggest define X=0 is the center of the target (Earth) and X=1 is the center of the mass (Moon), in this way, the moon have no speed by definition. We have tides because Earth turn in this system (on itself). If a planet have a rotation period synchronized with his moon's orbital period, then this planet have no rotation in this system, thus no (periodical) tides due to this moon (the surface of the planet is permanently deformed by gravity of this moon). $\endgroup$
    – Melnofil
    Commented Jun 21, 2019 at 8:01

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