The question has been asked (e.g., here and here), but I would like to get a more definitive and mathematically formal answer.

The Earth rotates around its axis, around the Sun, and participates in larger scale motions as a part of the Solar system. Yet, we often can get by treating it as an inertial reference frame (e.g., when constructing furniture, cars and buildings). In some cases we do need to take account for the effect of its rotation - e.g., in weather prediction one takes account for the Coriolis force, but we still consider the Solar system as the inertial reference frame.

We do that because:

  • accelerations that we deal with (notably $g$) are much greater than the accelerations due to the other motions that it is involved in?
  • we can neglect the non-inertial forces because all the objects around experience the same accelerations due to these forces?
  • something else?

I am looking for a mathematically motivated answer. I also suggest delineating between what is specific to Earth (accidental) and what would apply to all (or most) planets/stellar bodies.

I took the liberty to summarize the opinions expressed so far in my own answer. Yet, there remains non-inertial effects not covered by free fall and the Earth's rotation - those related to the Earth's finite size and responsible for the tidal forces (more specific question is here). Thus, this question still needs a canonical answer.

  • 1
    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – SuperCiocia
    Sep 9, 2021 at 20:46

9 Answers 9


If you need to throw a ball down a field, then corrections for the rotation amount to about 0.0001g acceleration. Small enough to be ignored for the 3 seconds the ball is in the air.

If you're shooting a projectile into another state, then that apparent acceleration is going to add up to many meters of deflection against the predictions when using an inertial frame for calculations.

Most human-scale experiences don't have sufficient distance or velocity for the error in assuming an inertial frame to become apparent. So the extra complexity of calculating the changes due to the earth's movement is unnecessary. The same way that for many benchtop experiments we might be able to ignore air resistance and the non-uniformity of the gravitational field.

There's no hard rule about this. You just choose the simplest model that is sufficient for your purpose. Want a pipe that takes water from the top of a 10m building to the ground? Probably can ignore earth rotation for how pressures in the pipe develop. Need to model how air circulates around the planet? You can't take that shortcut.

doesn't it mean that the accelerations due to any motion that Earth is involved it are much smaller than the accelerations that we encounter in everyday life? Is this an accident or a result of some properties of the universe?

The accelerations due to earth motion are smaller than the accelerations that you notice in everyday life. It's not an accident, it's just that the earth rotates fairly slowly on the human scale.

Go get a merry-go-round, hook up a motor to it to make it rotate once every 24 hours. Stand on it and you won't be able to tell much is going on. Accelerations due to earth's orbit and due to galactic rotation take even longer and will be even smaller.

The Coriolis corrections are proportional to the rotation rate of the frame ($\omega$). One revolution a day just isn't very large.

  • 1
    $\begingroup$ +1 but doesn't it mean that the accelerations due to any motion that Earth is involved it are much smaller than the accelerations that we encounter in everyday life? Is this an accident or a result of some properties of the universe? Can we do similar claims for other planets or stars? $\endgroup$
    – Roger V.
    Sep 9, 2021 at 18:44
  • 4
    $\begingroup$ @RogerVadim But, in a different sense, it is an accident of how our human brains perceive things, of the way we adapted to function at spatial and temporal scales of our everyday experience. E.g. on everyday time scales, a glacier is basically solid rock, but on decades-long time scales, the behavior of a glacier is perhaps better modeled as a non-Newtonian fluid. Everything is a model, and every model ignores something (like, say, the inertial forces); the only question is whether or not the things you ignored have a significant effect on the usefulness of your calculations (predictions). $\endgroup$ Sep 9, 2021 at 20:22
  • 1
    $\begingroup$ Nice link: ourplnt.com/relative-rotation-speeds-of-the-planets - the speed of a person standing at the equator of Jupiter is almost 30 times that of earth. Would be interesting if we would notice anything different (except that there's only thin air to stand on :) ). $\endgroup$
    – AnoE
    Sep 10, 2021 at 7:21
  • 1
    $\begingroup$ @RogerVadim - Not quite. E.g., if you choose to ignore air drag, and get a number, then do a bunch of experiments to check it, and decide, somewhat arbitrarily that the margin of error is good enough for what you need the calculation for, there's nothing fundamentally objective about that cutoff point; the question is will other people find it useful as well. Now, you can make your choice more formal and more justified (both mathematically and experimentally), but that doesn't make it "objective reality", only a better representation (model) of reality - which is all physics deals in. $\endgroup$ Sep 10, 2021 at 15:25
  • 1
    $\begingroup$ You wrote "ignoring things is based in strict mathematical criteria". By my understanding this is not really accurate, but one might say a couple of things in its place, such as "ignoring things is based on scientific criteria" or "ignoring things is based on statistical criteria". $\endgroup$
    – Lee Mosher
    Sep 10, 2021 at 22:06

It is necessary to distinguish two types of motions that the Earth is involved in:

  1. The rotation about its own axis
  2. The motion of the Earth as a whole around the Sun, the center of the Galaxy, etc.

In fact, the OP is ambiguous in whether it means the reference frame associated to the Earth surface or to its center - in the latter case we are concerned only with the second type of motion.

Free fall
The motion of the Earth as a whole is a free fall in gravitational field. The equivalence principle of the general relativity states that in this case the reference frame can be considered inertial, because all the objects in it experience the same accelerations due to gravity, and only their relative accelerations can be detected.

Earth rotation
A reference frame attached to the Earth surface is non-inertial, and fictitious forces need to be introduced: the centrifugal force, the Coriolis force, and the Euler force. These forces can be neglected, if they are are small, as discussed in the older version of this answer (see below). Moreover, one can argue that, if these were not small, the conditions on Earth would be too unstable to allow for the existence of life.

The two points above essentially correspond to the two bullets in the OP.

I appreciate the help from all who participated in the discussion and helped me to clarify different parts of this answer.

Old version of the answer

The first bullet in the OP is the correct answer:

accelerations that we deal with (notably ) are much greater than the accelerations due to the other motions that it is involved in?

Judging by the comments, many people have the gist of the idea - moreover, it was already mentioned in the answers quoted in the OP. However, the approach of trying to calculate all the possible accelerations - due to the Earth rotation around its axis, rotation around the Sun, motion in respect to the Galaxy - is a hard (if not impossible) way to prove it.

In fact, all these accelerations (and hence the pseudoforces appearing when treating the Earth as an inertial reference frame) must be small compared to typical accelerations that we experience on Earth (which are of the order of $g$) as a condition of stability of our little world.

Indeed, let us consider the acceleration due to the rotation of the Earth at angular speed $\omega$. Assuming for simplicity that we are at the equator, the condition that we can neglect the non-inertial effects is $$a=\omega^2R\ll g.$$ It is significant that characteristic everyday accelerations are of the order of $g$ or smaller, since the condition above becomes the condition that we deal with velocities smaller than the escape velocity: $$\omega^2R=\frac{v^2}{R}\ll \frac{GM}{R^2}.$$ That is, if the fictitious forces in question were comparable to the accelerations that we deal with, and orovided that these fictitious forces are due to the motion in the gravity field, our environment would not hold together.

Acknowledgement: I thank @rob for bringing my attention to this simple fact.

  • 1
    $\begingroup$ Surely if the centripetal force was equivalent to, say, 50% of gravity, we'd have to start including it in our calculations, but it wouldn't be enough to make the planet fall apart because it would still be less than gravity. $\endgroup$
    – user253751
    Sep 10, 2021 at 9:24
  • $\begingroup$ @user253751 I think it would change quite a bit in terms of the geological and atmospheric conditions, and whether life would be possible. But I think it deserves a separate question. $\endgroup$
    – Roger V.
    Sep 10, 2021 at 9:30
  • $\begingroup$ +1: The new version of this answer perfectly captures the fundamental point! :) $\endgroup$
    – user87745
    Sep 20, 2021 at 20:13

The reason why you can ignore the Earth's rotational motion for most day to day purposes is that the motion is circular with a long period, a constant tangential velocity and a large radius. Those points mean that we are close to being inertial- the acceleration we experience is purely centripetal (ie downwards) and small compared to the downward force of gravity. Our tangential speed is effectively constant. The curvature of the path we follow as a consequence of the Earth's rotation is of the order of a few inches per mile (being greatest at the equator and zero at the poles), so it hardly varies from a straight line. That explains why the Earth can be considered close to being inertial for many purposes.

As for when, you must take the acceleration due to the Earth's rotation into account when you need to perform calculations to a high degree of precision or when the effects you are trying to model occur over long distances compared with the curvature of the path we follow.

  • $\begingroup$ I agree that these accelerations are small - by why are they small? Is it a matter of accident that the Earth is ritating too slow? $\endgroup$
    – Roger V.
    Sep 10, 2021 at 4:40
  • $\begingroup$ It is purely a matter of accident. That said, if the particles that coalesced to form the Earth had been rotating about their combined centre of mass at a much higher rotational speed, then their mutual gravitational attraction might not have been sufficient to provide a centripetal force necessary to bring them together. There is a limit to how fast the earth could rotate while still providing a stable basis for the development of life. $\endgroup$ Sep 10, 2021 at 4:49
  • $\begingroup$ Then it is a matter of many accidents, for the Earth also rotates about the Sun, about the center of the Galaxy, etc. I think there's a deeper reason for that, as I pointed in my own answer $\endgroup$
    – Roger V.
    Sep 10, 2021 at 4:57
  • $\begingroup$ @tea-and-cake this could be part of your answer :) Note however, that rotation of the Earth is hardly different in this sense, as soon as we talk not about the Earth itself, but about an object on its surface. Also, most people think that slow rotation of the Earth is just a lucky accident, which I think it is not (see my own answer). $\endgroup$
    – Roger V.
    Sep 10, 2021 at 7:30
  • $\begingroup$ @tea-and-cake Write an answer, since what you are saying goes well beyond the answer above - comments are not for an extended discussion. $\endgroup$
    – Roger V.
    Sep 10, 2021 at 14:30

There are two major classes of reasons why we can ignore a physical effect, including that of non-inertial reference frames:

  • The system contains uncertainties which are sufficiently larger than the effects we seek to ignore.
  • The system is stabilized against the effects we seek to ignore.

In most situations, the former condition is the driving reason we can ignore non-inertial effects. If non-inertial effects will perturb my results by 1cm, and my uncertainties in the equipment will perturb it by 1km, there's no reason to include the non-inertial effects. And, no surprise, as the precision and accuracy of the experiment increase, the ability to ignore these non-inertial effects goes down.

The other common case is a system which is stabilized. This is why we typically get to ignore effects like the Coriolis effect when modeling aircraft. There is typically a pilot who is comparing where they are to where they want to be and issuing corrections. In these cases, we may not even notice that the effect occurred, although you would see the corrections if you look hard enough.

Likewise, there are times where you can model gravitational effects as Newton's law, $\frac{GmM}{r^2}$. Most of the time that is accurate to well under your uncertainties. Sometimes you need to upgrade to a J2 model, which accounts for the non-spherical nature of the Earth. Other times you need to upgrade to a J4 model. Others need EGM2009. It all depends on how your uncertainties and stabilizing capabilities compare to the inaccuracies in the simpler model.

And if it's not clear? Determining which model to use can indeed be an art in some cases. Typically we err on the side of caution, and use a higher order model if we can't prove that its effects would be negligible. And the thresholds are not mathematically simple. Needless to say, I will be far more demanding on the precision of eye surgery done on my eyes than I will be when trying to hit a home run in a casual league.


Since you are asking for a mathematical explanation, here is one, tongue-in-cheek:

  • Let the outcome we estimate from our simplified mental models be $o_{estimated}$.
  • Let the actually observed outcome be $o_{observed}$.
  • Let the difference be $\Delta o = o_{observed} - o_{estimated}$.

Then we can ignore $\Delta o$ as long as $\Delta o \ll o_{observed}$.

The big elephant in this mathematically decorated room is the missing definition of "$\ll$". It is simply a shortcut for "small enough to ignore", which leads to the circular reasoning "you can ignore it as long as you can ignore it" (i.e., as long as ignoring it doesn't hurt you).1 In our daily lives we simplify a lot all the time. The world is enormously complex, but most of it does not concern us.


  • We consider the ground flat.
  • We typically do not consider that the speed of sound is finite.
  • We never consider that the speed of light is finite.
  • We ignore the known fact that Newtonian physics is utterly and fundamentally wrong (your point: Instead, moving objects gain mass, acceleration curves space, we only interact with our immediate environment (no remote interaction) etc.)
  • We treat surfaces like the tabletop I'm sitting at as a rigid, continuous plane that occupies an unambiguous spacetime coordinate. (But we know that it consists of atoms with nuclei and electron shells, and if we look really close we have a hard time saying exactly where it is now, or when.)

All these simplifications can be made because the differences between the simplified predictions and the observations in our daily lives are small.

When that is no longer true we must reduce the degree of simplification in our models. Examples corresponding to the list above:

  • You cannot observe objects at a great distance.
  • You cannot clap in a synchronized fashion in a stadium.
  • The time intervals between eclipses of Io appear to vary with the Earth seasons.
  • The atomic clocks in your GPS satellites are too fast, even though we subject them to the most diligent quality control before launch.
  • When we shoot X-rays at crystals we observe regular patterns indicating that solid matter has an inner structure.

So all the math boils down to "you can simplify with impunity until you can't". This is actually a special case of the very general tenet "you can ignore reality until you can't".

1 This pragmatic attitude that we are forced to assume lest we drown in irrelevant details has been explored by William James in the early 20th century. He developed a truth theory of pragmatism which in typically American fashion shunned metaphysical esoterics and focused instead on usefulness. In lecture 6 he wrote:

Pragmatism, on the other hand, asks its usual question. "Grant an idea or belief to be true," it says, "what concrete difference will its being true make in anyone's actual life? How will the truth be realized? What experiences will be different from those which would obtain if the belief were false? What, in short, is the truth's cash-value in experiential terms?

  • 1
    $\begingroup$ The ratio $\Delta o/o_{observed}$ tells us how good our approximation is - in percentage points, rather than simply ignoring or not ignoring something. Mathematically, instead of ignoring we actually expand in series in this quantity, and trancate at the desired order. Incidentally, the size of this quantity also tells us whether the series is converging. Finally, in practice, we compare $\Delta o$ with a detection threshold or a standard deviation of $o_{observed}$ or noise level, or calculate a p-value that is compared with a pre-defined confidence level. $\endgroup$
    – Roger V.
    Sep 10, 2021 at 14:19
  • $\begingroup$ @RogerVadim Ratio or difference? All the definitions are so vague, it doesn't matter or doesn't even make sense. The clap from the other side of the stadium arrives 0.7 seconds after I see them clap; that is a ratio of what? 0.05 seconds may be acceptable, 0.2 is too long, in absolute terms. $\endgroup$ Sep 10, 2021 at 15:26
  • $\begingroup$ the time of human reaction, that is the detection threshold. $\endgroup$
    – Roger V.
    Sep 10, 2021 at 16:05
  • $\begingroup$ ratio and difference - these are relative and absolute error. $\endgroup$
    – Roger V.
    Sep 10, 2021 at 17:01

Introductory thoughts

What does it mean to treat the Earth as an inertial frame? I would say it means that we pretend the Earth is an infinite plane, moving with a constant velocity in some direction. This accounts for many everyday phenomena, since we are so small we essentially always live in the tangent plane of the Earth, and for those of us who aren't professionals in aerospace, objects we throw in the air tend to come down before the tangent plane of the Earth changes by very much.

This assumption is good so long as the deviation of some motion of interest, is smaller than some threshold of acceptability, from the motion accounting for non-inertial effects. Note that the deviation of the motion will essentially always rise above any threshold given enough time (even if I apply $10^{-12}$ N to you, if I do it continuously over a long enough time you will eventually notice you are shifted relative to where you would be if no force is applied. The buildup of a small effect that always points in the same direction until it becomes a large effect, are sometimes known as secular variation.

We can illustrate these general points with the simple example of uniform acceleration in 1 dimension. Then the trajectory of an "inertial observer" is $x_I (t) = x_0 + v_0 t$, and an accelerated observer which "matches" the inertial observer at $t=0$ has a trajectory $x_A (t) = x_0 + v_0 t + \frac{1}{2} a t^2 = x_I(t) + \frac{1}{2} a t^2$. We're interested in the error we make in ignoring the acceleration term, compared to some acceptable level of error in the position $\Delta x$ \begin{equation} \epsilon = \frac{\left| x_I(t) - x_A(t) \right| }{\Delta x} = \frac{a t^2}{2\Delta x} = \left(\frac{t}{t_A} \right)^2 \end{equation} where $t_A = \sqrt{2 \Delta x / a}$ is the time scale over which the non-inertial effects become important.

The error grows with $t$, but only becomes large when $t$ is of order the timescale $t_A$. Given that we have some fixed tolerance in position $\Delta x$, and that a typical experiment happens on some human timescale $t_H$, we can neglect the effect of acceleration if $t_H \ll t_A$, which requires $a$ to be smaller than $\Delta x / t_H^2$. This form of this scaling argument gives us intuition for what to expect in more general cases.

Rotation of the Earth

Probably the most obvious non-gravitational, non-inertial effect is the rotation of the Earth. I think this point was very well covered by other answers, and I agree the conclusion that the magnitude of things like the Coriolis force are so small that you don't notice them over "human" timescales. The parameter that controls the smallness of rotational effects of the Earth is the angular velocity.


Well, actually that's not quite true since the Earth's motion is very complicated, and in addition to the daily rotation there is also precession and nutation -- while any non-inertial forces from these effects are miniscule by any reasonable defintion, understanding secular effects from precession are crucial for astronomers).

Gravitational and tidal effects from the Earth and moon

Gravity makes this a bit tricky, but I think a uniform gravitational field can be considered a "benign" form of non-inertial-ness, since by the equivalence principle a uniform gravitational field is simply a uniformly accelerating frame of reference. So I think we can say the Earth is "inertial enough," so long as the gravitational field is "close enough" to uniform. The gravitational acceleration $\vec{g}$ for an isolated, spherical body of mass $M$ is \begin{equation} \vec{g} = \frac{GM}{r^2} \hat{e}_r = \frac{GM}{R^2} \left(1 - 2 \frac{\delta r}{R} + \cdots \right) \vec{e}_r \end{equation} where $\hat{e}_r$ is a unit vector pointing from the observer to the center of the Earth, and $R$ is the radius of the Earth. In the second equality, we have written the distance from the Earth to observer as $r=R+\delta r$, where $\delta r$ is the radial distance of the point to the surface of the Earth, and expanded in $\delta r / R$. Tidal forces due to a non-uniformity of Earth's acceleration come from the second term, $\delta r/R$. So long as the height of an object above the Earth's surface is a small fraction of the Earth's radius, we can treat the Earth's gravitational field as uniform. This is a good approximation if you throw an apple 1 meter in the air. It is not a good approximation if you launch a satellite into orbit.

You can try to account for more subtle things like the fact that the Earth is not a perfect sphere. The next best approximation would be to say the Earth is an ellipse; you will get another set of corrections proportional to the small ellipticity of the Earth, $e$. Even more subtle effects come from the local topography of the Earth; these will be suppressed by something like the height of the topography variation (eg the height of a mountain) over the radius of the Earth. All of these effects are in fact noticeable if you perform high precision measurements of the Earth's gravitational field.

We can also look at tidal forces from the moon. Here $\delta R$ is the same, but now $M$ is smaller (it is the mass of the moon instead of the Earth) and $R$ is much bigger (the difference of the moon's center to the Earth's surface), so the tidal force $\sim GM \delta r / R^3$ is much, much smaller than the tidal force due to the Earth itself. We do notice the tides of course; this basically boils down to the facts that we are small compared to the Earth so we do notice small changes, and 12 hours is a substantial fraction of the period of the Earth's rotation so $\omega t$ becomes of order 1. More precisely, what distinguishes the moon's tidal force from the Earth's own tidal force is that $R$ (as I've defined it) is time dependent, and it's often much easier to see a time-varying small effect than a constant one.

Motion through the cosmos

Finally, considering motion of the Earth through the cosmos, as was also discussed in other answers, freely falling motion is inertial. If there are genuinely non-gravitational forces that the Earth / solar system / galaxy experience, these must act on such a long time scale that they will not be significant for Earth-based measurements.

Wrapping up

I think most of the points here would generalize to other planets, or at least gives you a condition for checking if it would generalize. While you need to reason about each effect on a case-by-case basis, there is always (a) some small parameter controlling the non-inertial effect, be it the rotational velocity for Earth's rotation, $\delta R/R$ for tidal effects, the ellipticity, etc, and (b) some time scale that tells us when secular effects become important.

  • 1
    $\begingroup$ FWIW, I give values for the lunar & solar tidal accelerations here. That answer also has a link to a series of articles on measuring those tidal effects on precision pendulums. $\endgroup$
    – PM 2Ring
    Sep 22, 2021 at 5:00

If we could separate the force of gravity between planets, stars and satelites from the same force that each celestial bodies does on loose object on its surface, the effects of being a non-inertial frame would be evident.

For example: the earth is a non inertial frame because is orbiting the sun. So, if a book of 3kg is weighted on a kitchen scale at noon, there is a centripetal force of $$m \omega^2 R = 3*(2\pi \frac{1}{365*24*3600})^2 * 150*10^9 \approx 0.02 N$$

that apparently should be added to the weight. It means 2 grams, what could be detected even by a cheap scale. But there is no difference of the weight of the book along the day. And the reason, according to Newton's theory, is the book be attracted by the Sun with the same force, and the effects cancel out.

So, on one hand the orbiting celestial bodies are accelerated frames, but on the other hand the expected fictitious forces resulting from it are absent due to the forces of gravity.

The exception is rotation around its own axis. If there is a precise way to know the direction of a straight line from the surface to the center, devices like a plumb line will show a deviation, due to the fictitious centrifugal force.

  • $\begingroup$ True, the Earth is in orbit around the Sun, and a freefall frame is (of course) inertial. OTOH, the book isn't at the centre of the Earth, so it's not quite in orbit around the Sun. But the tidal acceleration from the Sun at the Earth's surface is very small. At its maximum, it's only $\approx5.05×10^{-7} \,\mathrm{m/s^2}$. It is actually possible for an amateur to make a local measurement of the change in gravity due to the tidal influence of the Moon & the Sun, but you need very good equipment. See leapsecond.com/hsn2006 $\endgroup$
    – PM 2Ring
    Sep 11, 2021 at 3:06

Motion near the Earth is basically inertial if an object is not being "pushed" by anything! If you throw a stone it is in inertial motion (apart from air resistance) until it hits the ground.

Motion of the Earth around the Solar System, Galaxy etc. is irrelevant if you are not being pushed, and is completely dominant if you are on the ground!

Artillery is in inertial motion too; the coriolis effect is an artifact only visible to observers on the ground!


enter image description here

The components of the earth rotation at the latitude $\lambda$ are:

$$\mathbf \Omega_E=\begin{bmatrix} 0\\ \cos(\lambda)\, \Omega\\ \sin(\lambda)\,\Omega\\ \end{bmatrix}$$

from here we can obtain the pseudo forces due to the earth rotation

$$ \mathbf F_s=m\,\big(2\,(\mathbf\Omega_E\,\times \mathbf{\dot{R}})+\mathbf\Omega_E\times\,(\mathbf\Omega_E\times \mathbf R)\big) $$

where $$\mathbf R=\begin{bmatrix} x \\ y \\ z\\ \end{bmatrix}\quad,\text{is the postion vector}$$

$$\frac{\mathbf F_s}{m}= \left[ \begin {array}{c} \left( \left( \cos \left( \lambda \right) \right) ^{2}x+ \left( \sin \left( \lambda \right) \right) ^{2}x \right) {\Omega}^{2}+ \left( -2\, \cos \left( \lambda \right) { \dot{z}}+2\, \sin \left( \lambda \right) {\dot{y}} \right) \Omega \\ \left( - \sin \left( \lambda \right) \cos \left( \lambda \right) z+ \left( \sin \left( \lambda \right) \right) ^{2}y \right) {\Omega}^{2}-2\, \sin \left( \lambda \right) \Omega\,{\dot{x}}\\ \left( \left( \cos \left( \lambda \right) \right) ^{2}z- \cos \left( \lambda \right) \sin \left( \lambda \right) y \right) {\Omega}^{2}+2\, \cos \left( \lambda \right) \Omega\,{\dot{x}}\end {array} \right] $$

The EOM's in rotating system free falling

$$\begin{bmatrix} \ddot{x} \\ \ddot{y} \\ \ddot{z} \\ \end{bmatrix}=-\frac{\mathbf F_s}{m}-\begin{bmatrix} 0 \\ 0 \\ g \\ \end{bmatrix}\tag 1\\ g=\frac{G\,M_E}{(R_E+z)^2}=\frac{M_E}{\big(R_E\,(1+\frac{z}{R_E})\big)^2 }\approx \frac{G\,M_E}{R_E^2}$$ where $~R_E~$ is the earth radius , $~M_E~$ earth mass and G the gravitation constant

with $~\Omega=7.2710^{-5}~$[1/s]$~~,\lambda=\frac{40}{180}\,\pi~$

$$\frac{\mathbf F_s}{m}=\left[ \begin {array}{c} 0.000000005285290000\,x- 0.0001113828620\,{ \dot z}+ 0.00009346131845\,{\dot y}\\ - 0.000000002602497284\,z+ 0.000000002183754512\,y- 0.00009346131845\,{ \dot x}\\ 0.000000003101535488\,z- 0.000000002602497284\,y+ 0.0001113828620\,{\dot x}\end {array} \right] $$

thus $\frac{\mathbf F_s}{m}\approx \mathbf 0~$ and equation (1) is the inertial case.


Your Answer

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

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