I understand that you weigh less at the equator due to the increased centrifugal force. From my understanding, the faster you circle the Earth, the less your effective normal force you would feel, up to the point where you are orbiting the planet and the force becomes zero.

The Earth rotates about 1000mph near the equator, faster than an average commercial plane. It seems to me then that a commercial plane travelling west near the equator would cancel some of the centrifugal force of the Earth's rotation and become heavier compared to an eastward flying plane. Would it?

  • 2
    $\begingroup$ How would you weigh it? An aeroplane in steady flight has no weight. If it did, it would fall out of the sky. I love the fact that your link says that 'pounds' and 'grams' and not interchangeable. $\endgroup$ Aug 21, 2021 at 18:47
  • 19
    $\begingroup$ @WeatherVane maybe you could weight it by measuring the air lift necessary to keep it flying. It is not in free fall like the ISS, after all $\endgroup$
    – Prallax
    Aug 21, 2021 at 18:53
  • 1
    $\begingroup$ @Wolphramjonny Except that $m$ changes constantly, because fuel is burnt, and total $g$ depends on coordinates, direction and speed. There's really no reason why $mg$ should stay constant. $\endgroup$ Aug 22, 2021 at 12:10
  • 1
    $\begingroup$ @WeatherVane Re I love the fact that your link says that 'pounds' and 'grams' and not interchangeable. That is incorrect. Pounds are a US customary unit of mass, defined as exactly 453.59237 grams, so obviously, pounds are interchangeable with grams. The scientist who wrote that page (US engineers knows better, we're the ones who are forced to work with US customary units) was perhaps thinking of the pound-force, defined as exactly 4.4482216152605 newtons, which obviously are not interchangeable with grams. $\endgroup$ Aug 22, 2021 at 18:17
  • 5
    $\begingroup$ @WeatherVane Pounds are convertible to grams (or kilograms). Being well aware of the Mars Climate Orbiter fiasco (this is my field of work), I have long been an advocate of using metric units in all flight and ground software. I'm not the only one; I'm but a peon. For a short while this was mandated from high levels in NASA. But then the hardware people fought back. They don't want to go metric. Metric units are communistic??? So I'm on a project where the flight and ground software have to use customary units. $\endgroup$ Aug 22, 2021 at 18:34

3 Answers 3


What you describe is called the Eötvös effect, named after the hungarian physicist Loránd Eötvös

Eötvös devised ways to perform high accuracy measurements of gravity.

For charting gravity over larger areas devices were developed that could be operated on a moving ship. Eötvös noticed that the measurement from the moving ship needed a correction for the velocity of the ship.

(The wikipedia article was started/written by me. It was last attended to by me in 2008)

I prefer to think of the Eötvös effect in terms of the way it will affect the operation of an airship. For level flight the buoyancy of an airship is trimmed to achieve neutral buoyancy.

If an airship is flying eastward, trimmed to neutral buoyancy, and it makes a U-turn, the airship must be retrimmed.

Copied from the Wikipedia article:

(In the following the velocity term notated with $u$ is for velocity in purely westward/eastward direction.)

$ a_u $ is the total centripetal acceleration when moving along the surface of the Earth.
$ a_s $ is the centripetal acceleration when stationary with respect to the Earth.
$ \Omega $ is the angular velocity of the Earth: one revolution per Sidereal day.
$ \omega_r $ is the angular velocity of the airship relative to the angular velocity of the Earth.
$ \left(\Omega + \omega_r\right) $ is the total angular velocity of the airship.
$ u = \omega_r R $ is the airship's velocity (velocity relative to the Earth).
$ R$ is the Earth's radius.

$$ \begin{align} a_r & = a_u - a_s \\ & = \left(\Omega + \omega_r\right)^2 R - \Omega^2 R \\ & = \Omega^2 R + 2 \Omega \omega_r R + \omega_r^2 R - \Omega^2 R \\ & = 2 \Omega \omega_r R + \omega_r^2 R \\ & = 2 \Omega u + \frac{u^2}{R} \\ \end{align}$$

The above derivation shows that the change in required lift to maintain neutral buoyancy is proportional to the velocity relative to the Earth.

It's interesting to note that the Eötvös effect and the Earth rotation effect that is taken into account in Meteorology are physical counterparts.

The component of the Earth rotation effect that is perpendicular to the local surface is referred to as the Eötvös effect. In meteorology/oceanography the component parallel to the local surface is referred to as the Coriolis effect.

In meteorology/oceanography: a mass of fluid (air/water) that is moving in eastward/westward direction tends to drift away from the latitude line that it is moving along. When moving eastward the fluid mass tends to drift to the outside, away from the axis of rotation. When moving westward the fluid mass tends to drift to the inside of the latitude line that it is moving along, coming closer to the axis of rotation.

(On a sphere: an object initially moving along a latitude line, and then proceeding straight ahead will proceed along a great circle. On the other hand: fluid mass moving in westward direction tends to move to the inside of the latitude line that it is moving along, which is not straight ahead.)

[later addition]
In the derivation above a term $2u\Omega$ arises. This term has the same form as the coriolis term in the equation of motion when using a rotating ccordinate system.

The physical reason for that term $2u\Omega$ arising is the fact that the Earth is rotating. In order to remain co-rotating with the Earth a centripetal force must be provided. Providing this centripetal force goes at the expense of the Earth's inverse-square-law gravitational acceleration. As we know, inertial mass and gravitational mass are equivalent, so what a gravimeter measures is a resultant gravitational acceleration.

With a gravimeter that is stationary with respect to the Earth there is no way to use that gravimeter to assess whether the Earth is rotating. However, if it is made part of the experimental setup that the gravimeter is moved around (with magnitude and direction of the velocity with respect to the Earth recorded) then it is possible indeed to assess whether the Earth is rotating; the Earth rotation can be deduced from the magnitude of the observed Eötvös effect.

The Eötvös effect will manifest itself anyway; obviously it is independent of what coordinate system the experimenter is using.

Repeating the thought demonstration of an airship that is trimmed to neutral buoyancy: if that airship makes a U-turn it has to be re-trimmed. Obviously tthe coordinate system that you are using to describe the motion makes no difference to that.

Sometimes you can see an author stating something along the lines of: "The Coriolis force is only there when you are in a rotating frame of reference." What that author means is: the equation of motion has a coriolis term only when you are using a rotating coordinate system.

In the case of the Earth and the Earth's rotation rate: the equation of motion for a rotating coordinate system has a centrifugal term and a coriolis term. For the specific rotating coordinate system that is co-rotating with the Earth the centrifugal term is equal in magnitude to the required centripetal acceleration, and opposite in direction. So what authors do is that they allow the term for the required centripetal acceleration and the centrifugal term to drop away against each other. The remaining term, the coriolis term, has the same $2u\Omega$ form as the one derived above.

  • $\begingroup$ Excellent. See this funny: pcweenies.com/comic/plagiarism-rules $\endgroup$
    – PatrickT
    Aug 23, 2021 at 7:38
  • $\begingroup$ Seems like overcomplication of $\frac{v^2}{r}$. If planes fly 575mph, and earth surface moving 1000, then eastbound plane goes ~1575 and west goes 425. So the effective accel of gravity is reduced by $1575^2 /r$ or $425^2/r$ which gives about 0.8% of a $g$. Do you know what applications coriolis used for? $\endgroup$
    – Al Brown
    Aug 24, 2021 at 0:45
  • $\begingroup$ What is the effect on the molecular motion of the air? How will this affect the airplane? It will affect the curl of air giving lift according to Kutta–Joukowski theorem. It will oppose the circulation around the wings but very little. $\endgroup$ Sep 3, 2021 at 17:43


Yes what you wrote is correct. Accounting for the rotation and the fact that we face different average jet-streams, the change in gravity flying East vs. West is 0.8%, which should be measurable if you take an OK scale. If you weigh 190lbs, then that’s a pound and a half!

Optional consideration: Jet-streams

(If you don’t care about adjustments from average real-world weather trends as jet-streams, jump to the next section.)

Our atmosphere is full of jet-streams and air pockets constantly moving around the earth. They tend to move the same way the earth spins (West to East), but they move East faster than the surface does (in other words, they move West to East even relative to the surface).

People usually care about the effect of these flows on travel times rather than on weight, so they generally care about how the flows move relative to the surface. In our case, we do need to account for the flow of air, but only relative to the center of the earth, ignoring rotation. (Equivalently, we need to consider motion of earth’s surface and airflow relative to the surface)

At first it would seem that the average jet-stream help to an Eastbound plane would equal the average headwind we must overcome flying West. But that doesn’t take into account the effort of pilots to maximize benefit / minimize delay from jet-streams. They use forecasts to predict where the jet-streams will be blowing. Altitude and flight path are choice variables.

From perusing a few articles, I’ll just guess that (relative to the earth’s surface) flying East is aided by a 100mph stream, and flying West is done into a 50mph headwind, and the plane goes 575mph.

These factors are less than the effect of the earth’s rotation at the equator (~1,000mph), but they do function in a contributing way, and increase the difference in weight between flying East vs flying West.

So, speeds relative to the center of the earth, flying East and flying West, if Eastward is positive:

$$v_e= 575 + 1,000 +100=1,675 = 750 m/s $$ $$ v_w= -575 + 1,000 +50= 475 = 210 m/s$$

Both directions are East, meaning Westbound commercial planes at the equator move East, even if they don’t face a jet-stream.

So weather increases weight difference.

End historic jet stream consideration

Details continued


Yes your logic is exactly correct. Very simple problem with one calculation, $\frac{v^2}{r}$.

The reduction in weight comes from centrifugal force:

$$F=\frac{m_pv_p^2}{r}$$, where “$p$“ refers to the plane, $v$ is not relative to the surface, and doesn’t “care” about the rotation of the earth. Going East is a higher $v$ as described. In terms of the effective acceleration of gravity rather than force:

$$g_{eff} = G\tfrac{m_e}{r^2} - \tfrac{v_p^2}{r}$$

Finally, planes have less gravity just from their elevation because they are further from the center of the earth, higher $r$. That reduces effective gravity.


The flight effect is lower than the rotation effect. (Flying commercially West at the equator is not enough to make one motionless above the earth.) With the earth’s surface moving 1,000 mph Eastward and planes flying at 575 mph, we can assume air goes with the surface. One speed would be 1,575 and one would be 425 mph. (After adjusting for historic weather, air-stream trends, these values were 1,675 and 475 instead.)

Usung the adjusted speeds in $m/s$ and an $r$ of $6.4E6 ~m$:

$$\frac{v^2}{r} = \frac{(750, ~210)^2}{6,400,000} $$

$$=0.088, ~0.007 m/s^2 = 0.0089 , ~ 0.0007 g$$

Gives a difference in gravity of 0.82%, which would be measurable if you take a good scale. Ignoring jet-stream differences would reduce this from 0.8% to 0.7%, making 0.7% our estimate for 575 mph of ground speed both ways.

  • 2
    $\begingroup$ Would this reduced apparent weight result in less fuel consuption somehow? This seems like it could have measurable consequences. $\endgroup$ Aug 21, 2021 at 19:08
  • 3
    $\begingroup$ The fuel benefit is small, because the main factor is wind resistance when at speed, not weight. But, jets are super inefficient on the ground. They burn a chunk of fuel taxiing. People have proposed towing them around. Of course this weight reduction doesnt apply to taxiing, meaning the impact is even less. But Im sure there’s some implication: pumps maybe. Obviously anything being weighed like food service dispensing. Probably a widely known consideration for designers of planes $\endgroup$
    – Al Brown
    Aug 22, 2021 at 15:53
  • 1
    $\begingroup$ One could estimate an order of magnitude by comparing the speed needed to stay in orbit (between 20k and 30k km/h) and the common airplane speed (~1k km/h). Sure, the orbit is 6600 km and not 6300, let's just ignore that. Also the effect is not linear, but still. $\endgroup$ Aug 23, 2021 at 8:21
  • $\begingroup$ @Peter-ReinstateMonica yeah that works well. Just remember when flying east, groundspeed plus speed of earths surface (1000mph) makes plane go like 2.5-3k km/h not 1k km/h. $~$ 27k km/h vs. 2.7k km/h is 1% after the squaring. And I got 0.89% being more exact. (And 6300 vs 6600 is ~1/20th - that effect isnt actually squared cuz it affects not only g but our mv^2/r too, to the first power there not second, leaving one power of effect) $\endgroup$
    – Al Brown
    Aug 23, 2021 at 9:31
  • $\begingroup$ What you show is the centripetal acceleration expressed as a force while centrifugal acceleration depends upon a cross-product which does NOT depend upon velocity, but on the rotation rate of the reference frame. The Coriolis acceleration does depend upon the velocity of an object in a rotating reference frame, however... $\endgroup$ Sep 3, 2021 at 14:16

The airplanes experience identical centrifugal forces at the equator, regardless of their flight direction.

The equation of motion in a frame fixed to the surface of the Earth is:

$$\vec F -2m\vec{\Omega}\times\vec{v}-m\vec{\Omega}\times(\vec{\Omega}\times\vec{r})=m\vec a$$

If we put that on the equator in standard North-East-Down (NED) coordinates, the centrifugal term becomes a negative force in the down direction:

$$ F_D^{\rm cent} = -m\Omega^2 a \approx -m\frac{g}{289}$$

where $a=6378137\,m$ is the semi-major axis, $g=9.80665\,$m/s$^2$ is the $g$, and $\Omega=2\pi/{\rm day}$ is the angular frequency.

Clearly, the velocity doesn't enter the expression for centrifugal force, so it can't depend on the velocity in the east direction.

For Eastward velocities, $\pm v_E$ (with v_E>0), the cross product in the Coriolis force is also strictly up or down:

$$ F_D^{\rm Cor} = \mp 2mv_E \Omega \approx \mp mv_E\frac g {67238}$$

Hence the east (west) bound plane is pushed up (down), and thus weighs less (more) than a stationary plane. This is known as the Eötvös effect, which is expressed as a reduction (increase) in the gravity for east (west) bound ships...and now of course, planes.

In an inertial frame, there are neither centrifugal no Coriolis forces. The difference in apparent weight is due to different centripetal forces on, which has magnitude for east and west bound planes of:

$$ F_D = -m\frac{(a\Omega + v_E)^2}{a} - m\frac{(a\Omega - v_E)^2}{a}$$ $$ F_D = -\frac{m}{a}\big[((a\Omega)^2+2v_Ea\Omega+v_E^2)- ((a\Omega)^2-2v_Ea\Omega+v_E^2)\big]$$ $$ F_D = -\frac{m}{a}\big[4v_Ea\Omega\big]$$ $$ F_D = -4mv_E\Omega$$

which exactly the expression for the Coriolis force (when comparing $+v_E$ with $-v_E$, the earlier expression is for $\pm v_E$, hence the factor of 2 difference).

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Buzz
    Aug 24, 2021 at 1:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.