Do flights affect tides? 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 &amp; 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?
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 &amp; $r$ 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 $0.099$ $mm$.