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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?

Qmechanic
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cool joey
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2 Answers2

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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.

tmwilson26
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    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. – Luboš Motl Nov 03 '15 at 15:18
  • 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. – tmwilson26 Nov 03 '15 at 15:20
  • 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. – Luboš Motl Nov 03 '15 at 15:20
  • First, I would be interested in the classroom that leaves children to believe that the value of $g$ is $10, {\rm m/s}^2$. They may be told to calculate with 10 for the sake of numerical simplicity but even 10-year-old kids are told at least the value $9.81,{\rm m/s}^2$. But second, more importantly, you don't really address the problem. No practical person ever talks about the centrifugal force from Earth's spin or the general non-inertiality of the Earth-bound frames even if he's an accurate engineer. And the question "why" is really the reason why the OP asked the question. – Luboš Motl Nov 03 '15 at 15:26
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    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. – Luboš Motl Nov 03 '15 at 15:29
  • At the root of this possibility is the equivalence principle: centrifugal and related fictitious forces have the indistinguishable effect on objects from the gravitational fields. The equivalence principle is also the reason why the freely falling frames in a gravitational field may be considered inertial as well, at least at length scales much shorter than the distance from the source of gravity and time scales before we crash to the planetary etc. surface. ;-) It's actually sensible to consider freely falling frames inertial - it's natural from the GR viewpoint and many things simplify. – Luboš Motl Nov 03 '15 at 15:31
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    Maybe you should put these comments into your own answer, as it is difficult to respond to all of this in a single comment. – tmwilson26 Nov 03 '15 at 15:33
  • I'll just respond to a few points: yes $10 \text{ m/s}^2$ is not accurate, but $9.81 \text{ m/s}^2$ is also not correct for every place on earth, and is even about 0.3% off at the equator, which you have indicated is a considerable effect. Your other points are good points, but are all based on making approximations or "folding" their effects into the gravitational constant. I stated in my first sentence that you'll have to take these effects into account depending on the nature of the problem. Most times they are not considered, because $9.81 \text{ m/s}^2$ will work just fine. – tmwilson26 Nov 03 '15 at 15:41
  • Hi, I totally agree that the deviations from 9.81 are significant, too, and comparably large to the centrifugal acceleration. In engineering applications, it's obvious that we can't neglect either. However, people understand that while the sea level across the globe has the same potential, the acceleration - a derivative of the potential - isn't uniform at the sea level. So the universal 9.80665 is just an idealization. It also decreases at higher altitudes etc. However, these are the "easily understandable" deviations from the simplified value of $g$. The OP asked about a "harder one". – Luboš Motl Nov 03 '15 at 16:20
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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).

CR Drost
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