Is the force of gravity less on an airliner at cruise speed and altitude? I'm not talking about a special reduced gravity flight with a parabolic flight path, just a typical long distance flight.

Seems to me there should be less as passengers are further away from the centre of the earth, and also possibly a tiny effect from the speed the plane as it travels round the curve of the earth (altitude remaining constant) but does anybody actually know how much less?

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    $\begingroup$ Welcome to Aviation.SE! This is really more of a Physics.SE type of question, since the answer is purely in the realm of physics and not aviation per-se. Yes, because you are farther from the center of the earth flying at 40,000' above sea level than you are on the ground, the effect of gravity will be minutely less. But for the distances involved, we're talking imperceptably slight, so much so that in practical terms, it's all the same. But not exactly the same. Physics guys will be a better source of the formulas, math, and numbers you're looking for. $\endgroup$
    – Ralph J
    Mar 21, 2016 at 14:07
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    $\begingroup$ I'm voting to close this question as off-topic because it belongs on Physics.SE instead -- not about aviation, within the scope defined in the help center (aviation.stackexchange.com/help). $\endgroup$
    – Ralph J
    Mar 21, 2016 at 14:09
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    $\begingroup$ Why would you retype it? Computers have had "Copy-n-Paste" for decades. $\endgroup$
    – abelenky
    Mar 21, 2016 at 14:11
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    $\begingroup$ I'd be more inclined to move it if we didn't already have 2 good answers. If y'all really want us to, though, go for it. $\endgroup$
    – egid
    Mar 21, 2016 at 15:16
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    $\begingroup$ @egid Physics mod here: FWIW on Physics we'd want to see the asker of a question like this show some effort to work out the answer themselves, or at least research it a bit. Topic-wise, it's definitely within our scope, but if this question had been posted as-is over there, it probably would not have been received so well. Just in case you do consider migrating it. (Also, there is some ambiguity in how the question is posed, as shown by the discrepancy between sweber's and the other answers.) $\endgroup$
    – David Z
    Mar 22, 2016 at 14:25

7 Answers 7


There is less gravitational force, but by how much? An insignificant amount. The gravitational force of attraction between two objects is given by,

$\displaystyle F_{\mathrm g} = \frac{G m_{1} m_{2}}{R^2}$,


$G$ is the graviational constant,

$R$ is the distance between the object's centers, and

$m_{1}$ and $m_{2}$ are the masses of the objects.

Instead of finding the variation in force between the aircraft and earth, it would be be better to find the variation in the acceleration due to gravity, $g$ (as $F_{\mathrm g} = m_{\mathrm a} g$, with $m_{\mathrm a}$ being the mass of the airliner)

We have, on earth's surface,

$\displaystyle g = \frac{G m_{\mathrm e}}{R_{\mathrm e}^2}$


$m_{\mathrm e}$ is the mass of the earth, and

$R_{\mathrm e}$ is the radius of the earth.

For the aircraft at an altitude $h$ above the surface of the earth, this becomes,

$\displaystyle g_{h} = \frac{G m_{\mathrm e}}{\left(R_{\mathrm e} + h\right)^2}$

Taking ratio, we get,

$\displaystyle \frac{g_{h}}{g} = \left(1 + \frac{h}{R_{e}}\right)^{-2}$

Plugging in numbers, we get, for an airliner cruising at 12 km,

$g_{h} = 9.773\ \mathrm{m\ s^{-2}}$,

or about 0.37 % less compared to the sea level value. This is quite small and would not be noticable to all but the sensitive instruments.

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    $\begingroup$ That's over a ton for a fully loaded A380 :) $\endgroup$
    – Antzi
    Mar 21, 2016 at 14:33
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    $\begingroup$ But a ton when you weigh 590 tons, is still pretty insignificant... $\endgroup$
    – Jon Story
    Mar 21, 2016 at 14:43
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    $\begingroup$ Most of those $R$'s should be $R^2$ instead. $\endgroup$ Mar 21, 2016 at 16:35
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    $\begingroup$ The difference is big enough to be considered in A/C performance calculations, bigger than other effects that manufacturers spend significant amount of money (like parasitic drag of specific areas). However, is a data, a you can not do a lot to use it... bigger is the effect of air density. Relevant, but only for complex calculations. $\endgroup$ Mar 21, 2016 at 16:50
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    $\begingroup$ If you count centrifugal forces, then you get another 0.11% decrease at 12km and 965km/hr!! By a = v^2/r $\endgroup$
    – user14115
    Mar 21, 2016 at 19:45

Gravity itself

@aeronalias is absolutely right. Given the gravitational acceleration of $g=9.81m/s^2$ on the ground, a perfect spherical earth of radius $R_E=6370km$ with homogenous (at least: radially symmetric) density, one can calculate the gravitational acceleration at an altitude of $h=12km$ by

$$g(h)=g\cdot\frac{R_E^2}{(h+R_E)^2}= 9.773 \rm{m}/s^2$$

Expressed in terms of $g$, the difference is

$$g_\rm{diff} = 0.0368565736 m/s^2 = 0.003757g$$

Centrifugal forces

The question also asks for the centrifugal effect on the aircraft as it travels round the curve of the earth, which has not yet been answered yet. The effect is considered small, but compared to the effect on gravity itself, it isn't always.

I got some heavy objections on my answer and I have to admit, I really don't see their point. Therefore, I've edited this section and hope this helps.

In general, an object moving on a circular path experiences a centrifugal acceleration, pointing away from the center of the circle:


$\omega=\frac{\alpha}{t}$ is the angular speed, i.e. the angle $\alpha$ (in radians) the object travels in a given time $t$ (in seconds).

Now let's consider a "perfect" Earth as described above, plus no wind. A balloon hovering stationary over a point at the equator at 12km altitude will do one revolution ($\alpha=2\pi[=360°]$) in 24 hours. So it is $\omega=\frac{2\pi}{24\cdot60\cdot60s}$. Together with $r=R_e+h$, one gets for the balloon:

$$a_{cb}=0.03374061 m/s² = 0.0034394098 g$$

The circumference of the circle the balloon flies is $2\pi(R_e+h)=40099km$

Now consider an aircraft flying east along the equator at the same altitude at 250m/s (900km/h, 485kt) with respect to the surrounding air. (Keep in mind: no wind). In 24h, this aircraft travels a distance of 21600km, or 0.539 of the circumference. This means the aircraft does 1.539 revolutions of the circle in 24h, which means its angular speed is $\omega=1.539\cdot\frac{2\pi}{24\cdot60\cdot60s}$. Thus, the centrifugal force on the aircraft flying east is

$$a_\rm{ce} = 0.0799053814 m/s^2 = 0.0081452988 g$$

The same way, one can calculate what happens when the aircraft flies west: $\omega=(1-0.539)\cdot\frac{2\pi}{24\cdot60\cdot60s}$

$$a_\rm{cw} = 0.0071833292 m/s^2 = 0.0007322456 g$$


Let's write the values together to compare them. I've also added how much lighter a 100kg (220lb) person would feel due to the effects:

                                             | "weight loss"
g_diff = 0.0368565736 m/s²  = 0.003757 g     | 376gram (0.829lb)
a_cb   = 0.03374061   m/s²  = 0.0034394098 g | 344gram (0.758lb)
a_ce   = 0.0799053814 m/s²  = 0.0081452988 g | 815gram (1.797lb)
a_cw   = 0.0071833292 m/s²  = 0.0007322456 g |  73gram (0.161lb)

Note: The 100kg is what a scale at the North Pole (i.e. without any centrifugal effect) shows. The person already feels 344g lighter on the ground at the equator. The balloon doesn't change this (much). But moving east/west has a larger effect on the weight than gravity alone. A person flying west feels even heavier than on ground!

Maybe another table, showing the weight of the person:

                                             kg      lb
1. Man at north pole                       100.00  220.46
2. Man at equator                           99.66  219.70
3. Man at equator, in balloon               99.28  218.88
4. Man at equator, in aircraft flying east  98.81  217.84
5. Man at equator, in aircraft flying west  99.55  219.47 <- More than 3.

The numbers shown are only valid at the equator and for flights east / west. In other cases, it becomes a little more complex.

EDIT: Being curious on how this depends on latitude, I created this plot about the absolute acceleration an aircraft experiences.

enter image description here

The radius in the equation of the centrifugal force is the distance of the aircraft to the axis of the Earth. It is clear that it decreases when moving away from the equator, and so does the acceleration.

The speed of the aircraft flying west will cancel out the speed of the earth at about 57° N / S, i.e. there is no centrifugal force. At larger latitude, the aircraft will fly in the opposite direction around the axis of the earth, building up a centrifugal force again.
Near the poles, both aircraft become centrifuges (theoretically). E.g. flying a circle of 500m radius gives an acceleration of 12.7g. This is why the data rises to infinity there.

(When doing the math, one has to keep in mind that gravity always points to the center of the earth, while the centrifugal force points away from the axis. You can't just add them)

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    $\begingroup$ This is wrong. You are computing the velocity's wrong. You can't use the ground velocity as reference because it too is an accelerated frame. Velocity is relative to the center of mass and the equation are totally invariant of direction. $\endgroup$
    – user14115
    Mar 21, 2016 at 21:33
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    $\begingroup$ @aidan.plenert.macdonald: I don't understand your argument. A resting observer in space sees that the earth has a tangential speed of 464m/s at the equator. An aircraft flying east with 250m/s has a total tangential speed of 714m/s, and an aircraft flying west has a total tangential speed of 214m/s. What's wrong with this? (a tangential speed of 250m/s at 12km does not equal 250m/s on ground, but this effect is really negligible) $\endgroup$
    – sweber
    Mar 21, 2016 at 21:51
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    $\begingroup$ If the adversary changes the rotation of one of the inner shells, then under your theory, this only has an effect if I arbitrarily choose my reference to be that shell!!! This is ridiculous! Physics shouldn't depend on my choice of reference! Thus, the only thing is can actually depend on is the center of mass (which is invariant of my choice of reference), the distance between us, and my velocity direction. Thus no-change of velocity of the earth surface can effect my motion. $\endgroup$
    – user14115
    Mar 21, 2016 at 22:04
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    $\begingroup$ @DavidRicherby: Second paragraph of question: and also possibly a tiny effect from the speed the plane as it travels round the curve of the earth (altitude remaining constant) but does anybody actually know how much less? $\endgroup$
    – sweber
    Mar 22, 2016 at 9:41
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    $\begingroup$ just to clear things up (hopefully): for practical calculations, all rotational velocity should be measured relative to the background of stars. in this case, it is reasonable to measure relative to the surface of the earth since we are seeking a delta, rather than absolute value for centrifugal force. more detailed discusion here: en.wikipedia.org/wiki/Absolute_rotation $\endgroup$ Mar 22, 2016 at 18:21

You are correct that the force of gravity is slightly less the further you get from the earth. Airlines typically cruise around 30,000 - 35,000 feet. We can use as a proxy measurement the force of gravity on Mt. Everest, which is 29,000 ft.

The force of gravity on Everest is about 0.434% less than the standard 9.8N/kg. This means that one pound at Sea Level would weigh about 0.995 lbs. at 29,000 ft. Or, a typical 180 lb. human would weigh 179.1 lbs.

I don't consider the speed of the aircraft going around the earth to be significant. Any centripetal force would be extremely tiny.

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    $\begingroup$ I don't consider the speed of the aircraft going around the earth to be significant. Any centripetal force would be extremely tiny. Nope, this effect can be significantly larger! $\endgroup$
    – sweber
    Mar 21, 2016 at 19:40
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    $\begingroup$ @abelenky: Did it. (see my answer) $\endgroup$
    – sweber
    Mar 21, 2016 at 19:42
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    $\begingroup$ Standard G is 9.80665, minus G @ Everest of 9.7639, for a delta-G of 0.04275 m/s2. (And that is 0.434% of 9.8m/s2) $\endgroup$
    – abelenky
    Mar 21, 2016 at 21:29
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    $\begingroup$ @abelenky - The first source you quoted was Yahoo Answers, which is not exactly a reputable source. Your second source (New Scientist) quotes: "Mount Nevado Huascarán in Peru has the lowest gravitational acceleration, at 9.7639 m/s2". Mount Nevado Huascarán has an elevation of 22,205, well below that of Mt Everest (or typical airliner cruising altitude). $\endgroup$
    – Johnny
    Mar 22, 2016 at 0:55
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    $\begingroup$ @DavidRicherby: but the questioner says, "I'm not talking about a special reduced gravity flight with a parabolic flight path". To me this makes it clear that they are asking about perceived acceleration within a frame of reference fixed to the plane, not about the gravitational force on the plane observed in a frame of reference fixed to the earth. Because the latter is not "reduced" in a parabolic flight. Whether they should have called this "force of gravity" or not is a quibble you should raise on the question, not on answers. $\endgroup$ Mar 22, 2016 at 10:30

Here's the Fermi Estimate version, in case the math in Aeroalias's answer is difficult to follow:

It's been said that the ISS experiences 0.9 G (90% of standard gravity at sea level), which is of course canceled out by their orbital velocity so the astronauts inside feel like they're in 0G.

Airplanes are said to fly a mile high, and the ISS is over 100 miles high--these are not particularly accurate numbers, but they're good enough for order-of-magnitude estimations.

Therefore, without any complicated math, we would expect an airplane to experience 99.9% of standard gravity. As Aeroalias's answer works out to 99.63%, this is a pretty good estimate.

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    $\begingroup$ @RonBeyer: If the ISS would not experience gravity, it would not orbit the earth. The centripetal force just cancles it out. $\endgroup$
    – sweber
    Mar 21, 2016 at 18:26
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    $\begingroup$ @RonBeyer You can think of the ISS and the astronauts as falling at the same speed. But there is surely still gravity- it is gravity that causes them to fall. $\endgroup$
    – jejorda2
    Mar 21, 2016 at 18:28
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    $\begingroup$ "Airplanes are said to fly a mile high" - Uh... NO. That would be about 5,000 feet. Planes actually cruise around 30,000 - 35,000 ft, or about 6 to 7 miles $\endgroup$
    – abelenky
    Mar 21, 2016 at 19:57
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    $\begingroup$ @abelenky At DEN, planes taxi at a mile (and a hundred fifty feet) high. $\endgroup$ Mar 21, 2016 at 21:46
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    $\begingroup$ @sweber: For an object in circular orbit, gravity is the centripetal force. It doesn't cancel itself out! (However, in a frame that moves with the spacecraft there's a centrifugal force that cancels out gravity). $\endgroup$ Mar 22, 2016 at 11:58

Gravity decreases with altitude

See this table providing gravity at different altitudes:

enter image description here
Gravity field at different altitudes (source: The Engineering Toolbox)

At a height of 10 km, gravity is 9.776 against 9.807 at sea level. That's a variation of 0.32%, which I consider as significant from an aircraft design angle, as it allows to reduce the fuel consumption in a larger proportion.

Can we sense the gravity difference at the altitude of 10 km?

Such difference cannot be sensed by a human, a metering instrument is required. Sensing 0.3% difference just requires a scale. If you "weight" a mass of 100 kg, then at a height of 10 km, the scale will just display 99.7 kg.

Note: A gravity value of 9.78 m/s² already exists on Earth, e.g. in Mexico city and in Singapore, due to gravity anomalies.

How fast gravity decreases?

The center of the gravity field is near the center of the Earth. The surface of the Earth is at 6,400 km from the center and the value of gravity is 9.81 m/s² or g.

Each time the distance from the center doubles, the gravity value is divided by 4: At 12,800 km, the value is 1/4 g. This progression is said to be an inverse square law, which looks like this:

enter image description here
Curve of an inverse square law (source)

Many physical quantities are based on this same law (light intensity, sound intensity, radio signal intensity). As you can see after 3 or 4 Earth radius, the variation has slowed down a lot, but it continue to decrease and will never reach zero. it means any object in the universe has an impact on all other objects! (but a small one).

The gravity value decreases when climbing, but also decreases when going underground. Near the center of the Earth the gravity is null (at least that's what we believe, we are not going to be able to check until a very long time, it's easier to explore space than the depths of our planet). This is the complete picture of the gravity:

enter image description here
Gravity field according to the preliminary reference Earth model

Gravity is a puzzling force not yet understood. We know local effects of the gravity, but we ignore the reasons of such effects.

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    $\begingroup$ Measuring gravity from the center of the Earth is a little misleading, as all mass has gravity - if you were in the center you would feel a roughly equivalent pull in all directions, not just "downward". (What "down" even means in the center is up for debate.) If you dug a really deep hole, you'd thus feel the affect of gravity from the ground below you, but also from the sides and all the mass that's above you. Thus, your overall downward pull would actually decrease the deeper you went. A bit pedantic, and not really relevant to aircraft (hopefully), but worth mentioning. $\endgroup$ Mar 22, 2016 at 18:49
  • $\begingroup$ The maximum gravity value is near the center of the Earth (or can be seen as maximum near the center) are you sure that this is the case? Resulting force should zero out as you approach to the center, right? $\endgroup$
    – Sebi
    Mar 22, 2016 at 21:54
  • $\begingroup$ @Sebi, Yeah, that's what my comment above was getting at. It would be true if the Earth were treated as a point-source of gravity, but this is obviously a spherical-cows-in-a-vacuum type of oversimplification. The chart added afterwards clears this up a bit. $\endgroup$ Mar 23, 2016 at 19:11
  • $\begingroup$ @DarrelHoffman: On the other hand isn't our current concept of gravity also oversimplified? Did we see the graviton recently? $\endgroup$
    – mins
    Mar 23, 2016 at 19:28

Most of questions deal with the gravity reduction due increased distance to the Earth but there is also centrifugal force caused by movement. This force allows spaceships to circle the Earth with no power applied and causes weightlessness inside - just increased distance would not be enough for these effects. An airplane is also flying around the Earth, same as a satellite does, just much slower.

This effect described here and may reduce the perceived weight (the aircraft feels lighter and everything inside is lighter) but can also increase it (depending on how the flight direction is related to the rotation of the Earth). The effect is somewhat about 0.3 % of mass at the speeds close to the speed of sound so comparable to the effect from the increased altitude.


Most answers here are from a perspective of "at a height of X, you weigh Y". Lets turn this around.

At the altitude of the ISS, gravity is about 10% less. I doubt you'd notice 10% reduction in weight (double-blind, if such a thing were possible), without using measuring equipment. That's 400km up (250 miles), and well outside our atmosphere.

The only reason people are "weightless" on the ISS, is that they are in orbit.

Things in orbit are always weightless. The scales you are standing on is falling at the same speed you are. This can happen at any altitude - if you kick a ball, it is in orbit - an orbit that intersects with the surface of the earth, so it can't complete the orbit.

  • $\begingroup$ The question is about the effects on an airliner, not on the ISS. $\endgroup$ Mar 21, 2016 at 22:17
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    $\begingroup$ I understand that. My point is that, even at the altitude of the ISS, you'd be hard pressed to notice, so at a plane altitude, which is much less, you definitely wouldn't. $\endgroup$ Mar 21, 2016 at 22:18
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    $\begingroup$ Well, you basically are not answering the question though. $\endgroup$ Mar 21, 2016 at 22:35

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