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Earth rotates about its axis and also revolves around the Sun at the same time. So why Earth is considered as an inertial frame in Newtonian Physics. So technically, I'm effectively asking why the Earth-centered, Earth-fixed (ECEF) frame is considered an inertial frame?

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marked as duplicate by ACuriousMind, Timaeus, Gert, Kyle Kanos, Martin Nov 4 '15 at 15:03

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    $\begingroup$ "In Newtonian physics" no one considers Earth to be an inertial frame. You need to be more specific. $\endgroup$ – ACuriousMind Nov 3 '15 at 15:42
  • $\begingroup$ The opposite question: physics.stackexchange.com/q/13324/2451 $\endgroup$ – Qmechanic Nov 3 '15 at 18:36
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    $\begingroup$ It's treated as an inertial frame in elementary, pre-calculus physics texts because the students don't have the necessary math background to do anything but. There is no reason to invoke fictitious forces (or general relativity) when solving a block sliding down a ramp problem. It's treated as an inertial frame in many areas of the physical sciences and engineering because the errors that result from doing so are demonstrably ignorable. It's not treated as an inertial frame when the errors that result from doing so are demonstrably significant. $\endgroup$ – David Hammen Nov 3 '15 at 19:06
  • $\begingroup$ Possible duplicate of If the solar system is a non-inertial frame, why can Newton's Laws predict motion? $\endgroup$ – Timaeus Nov 3 '15 at 19:17
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You are right that it should not be considered an inertial frame for many types of problems. This is how you end up with fictitious forces to account for (such as the Coriolis effect). However, this only has practical effect for larger scale problems. For the types of problems generally considered in physics class, the inertial frame approximation will work fine. One way to look at it is that in the rotating earth case, the acceleration that you feel from circular motion $v^2/R$ will be much less than that of gravity, so you can ignore it. So for blocks falling from small heights, it won't matter very much. As the problem scales up, you'll need to take these effects into account.

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    $\begingroup$ Sorry but it is simply not true that the acceleration from the circular motion $v^2/R$ is negligible relatively to gravity. For example, the Earth's spin gives the speed 40,000 km per 24 hours i.e. 460 m/s to each point on the equator and $460^2/6378000=0.03$ meters per squared second. This is 0.3% of $g$, the gravitational acceleration, so if one has a better accuracy than 0.3%, like if he measures a liter of wine with a better precision than 3 milliliters, and people often do, the centrifugal acceleration simply cannot be neglected. $\endgroup$ – Luboš Motl Nov 3 '15 at 15:18
  • $\begingroup$ I figured he was talking about classroom style physics problems, where often people approximate gravity as being $10 \text{ m/s}^2$ at all places on the earth. At that level of accuracy it won't matter. Also, if you need that level of accuracy to measure wine, use a balance and that won't be problem either. $\endgroup$ – tmwilson26 Nov 3 '15 at 15:20
  • $\begingroup$ Instead, what we implicitly do is that we clump the centrifugal force with the gravitational one and call it "Earth's gravity" even though Earth's gravity is about 0.3% stronger and this part is subtracted by the centrifugal force. $\endgroup$ – Luboš Motl Nov 3 '15 at 15:20
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    $\begingroup$ So the actual reason that allows people to neglect it is that the fictitious forces may be absorbed to the gravitational ones that they do consider. One doesn't need the centrifugal force to be negligible for that. Instead, what one needs to be obeyed is a much weaker condition: the centrifugal force has to be approximately uniform - like the gravity we consider - at the length scales of the problem. And that's true because both the gravity and the centrifugal force of the spin significantly change at the length scale comparable to the Earth's radius. $\endgroup$ – Luboš Motl Nov 3 '15 at 15:29
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    $\begingroup$ Maybe you should put these comments into your own answer, as it is difficult to respond to all of this in a single comment. $\endgroup$ – tmwilson26 Nov 3 '15 at 15:33
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To expand slightly on John Rennie's comment, almost everyone who discusses ECEF also discusses ECI, the "Earth-centered inertial" frame, and talks about how ECEF is not "inertial," in contrast to ECI. I don't know anyone who considers it "inertial" in all cases. Especially if you're dealing with weather and atmospheric physics, you have the Sun heating up air on the equator into an updraft, but this gets transformed by the Coriolis effect into a wind drifting westward relative to the surface: in a more folksy explanation, the inertial tangent frame travels with speed $r \omega$ as $r$ increases; so something moving with speed $v = r \omega$ that rises by a height $h$ into a tangent frame moving with speed $(r + h) \omega$ will appear to be moving backward with speed $h \omega.$ These west-moving equatorial winds are called the "trade winds" and the hot air that rises from that current tends to "fall" in a corresponding "east" wind at the "horse latitudes", in a "tube" of convection known as the "Hadley cell". (Confusingly these winds are called "westerlies" because ships used to mark which way the wind was blowing based on the direction it is coming from.)

All of that is Coriolis stuff; it depends on the Earth being a rotating reference frame, not an inertial one.

If someone is treating the ECEF frame as "inertial" it's perhaps legitimate if they're travelling northward near the equator (no Coriolis force; centrifugal force can be absorbed into gravitational acceleration). But in general ECEF and ECI are used by people talking about satellite navigation, and on those scales the Coriolis force usually peeks its ugly head in. The only thing I could think that would it negligible is if your satellite is orbiting the Earth many times per day, but the GPS satellites, for example, only orbit twice a day and therefore can't neglect such effects (and should use ECI to correct for them).

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