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In many practical applications, one can consider the Earth-Centered Inertial (ECI) reference frame approximately as an inertial reference system, though strictly speaking, it is non-inertial.

Is there any quotable reference, where this claim is supported by a detailed estimation how small the effects are that one neglects if one considers the Earth's frame as inertial?

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Related: physics.stackexchange.com/q/3193/2451 and links therein. –  Qmechanic Apr 25 '13 at 19:21
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3 Answers 3

I come from the world of precise satellite orbit determination, so my references also come from there. I realize this means most of the equations in my references relate to orbits and not general measurements, however, I do think you'll find them interesting.

I think a good reference is Fundamentals of Astrodynamics and Applications, by D.A. Vallado. The relevant section is Section 3.7 and onward.

Another good reference is Satellite Orbits, by Montebruck and Gill. Relevant sections are in Chapter 3 (could not find a link).

To summarize: quasi-inertial reference frames are frames such that Newton's laws and special relativity can be applied without any significant impact on accuracy. Usually, for applications where this accuracy is not sufficient, accuracy is most easily improved by adding corrective terms to Newton's equations of motion. Corrective terms account for planetary precession & nutation, GR effects, Coriolis effect, tidal effects, etc.

To give an impression of the magnitudes of such corrections; the second paragraph on the wiki is helpful:

ECI coordinate frames are not truly inertial since the Earth itself is accelerating as it travels in its orbit about the Sun. In many cases, it may be assumed that the ECI frame is inertial without adverse effect. However, when computing the gravitational influence of a third body such as the Moon on the dynamics of a spacecraft, the acceleration of the ECI frame must be considered. For example, when computing the acceleration of an Earth-orbiting spacecraft due to the gravitational influence of the Moon, the acceleration of the Earth itself due to the Moon's gravity must be subtracted

The main non-inertial effect to take into account is the acceleration of the ECI frame towards the Moon. If you assume the Earth and Moon are in circular orbit about their barycenter, this acceleration can be estimated;

  1. by Newtonian gravity:

    $$ a_{\mathrm{Earth}} = \frac{\mu_{\mathrm{Moon}}}{r_{\mathrm{E-M}}^2} \approx \frac{4902.8\ \mathrm{km}^3\,\mathrm{s}^{-2}}{(384399\ \mathrm{km})^2} = 33.180\, \mu\mathrm{m\,s}^{-2} $$

  2. from circular motion:

$$ a = \frac{v_{\mathrm{Earth}}^2}{r_{\mathrm{E-M}}} \approx \frac{(1.022\ \mathrm{km\ s}^{-1})^2}{384399\ \mathrm{km}} = 33.180\, \mu\mathrm{m\,s}^{-2} $$

The same can be said for the Sun. Again, assuming circular orbit:

$$ a_{\mathrm{Earth}} = \frac{\mu_{\mathrm{Sun}}}{r_{\mathrm{S-E}}^2} \approx \frac{132712440018\ \mathrm{km}^3\,\mathrm{s}^{-2}}{(149597871\ \mathrm{km})^2} = 5.9301\, \mathrm{mm\,s}^{-2} $$

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The relevant definition of inertial here is the GR definition, not the Newtonian definition. By the GR definition, Lubos Motl's answer is corrrect. The reason the GR definition is the right one is that the OP is asking for locally observable physical effects. –  Ben Crowell May 30 '13 at 17:16
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@BenCrowell: and yet the question is tagged "Newtonian mechanics"... –  Rody Oldenhuis May 30 '13 at 20:32
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In special relativity, it is easy to define an inertial reference frame. However, that is not the universe we live in. General relativity, in particular Einsteins equivalence principal, states that a local inertial reference frame can be defined by a set of free floating particles.

If we consider the earth as a free floating particle, then there is an local inertial reference frame whose center is the center of the earth. However, a local reference frame is defined not only by its center but by its orientation. It might seem that the right choice of orientation is to be fixed with respect to the stars. However, this is wrong. To see this is wrong, imagine a cloud of very light particles with the same shape as the earth, and the same orbit. Since the particles in the cloud would follow the same orbit, the particles at the front of the cloud would always be ahead of the ones at the back. This implies that the inertial reference frame does in fact rotate, at a rate of one rotation a year. (The influence of the moon adds some complications here - to first order, we should also add the effects of the moon)

So an inertial reference frame can be defined as having center at the center of the earth, $x$ coordinate pointing towards the sun, and $z$ coordinate perpendicular to the ecliptic. According to Wikipedia, the ECI reference frame is approximately fixed with respect to the stars, and so is only approximately inertial (even to first order).

On the meaning of inertial in this context: Newton's laws will apply to first order in this reference frame. However, second order corrections may be needed to account for the second order influence of the Moon and Sun's gravity.

The answer of Luboš Motl was correct except for the claim that an Earth centered reference frame whose orientation was fixed with respect to the stars would be inertial.

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The only forces that have a significant impact on the motion of Earth's center-of-mass are gravitational forces. The Earth is "freely falling"; in the language of general relativity, the modern theory of gravity, it is moving along a geodesic.

Because of this fact, it is automatically guaranteed that the spacetime in the vicinity of the Earth's world line is flat to first order; the metric and its first derivatives vanish. (The first derivatives are equivalent to the Christoffel symbol which therefore vanishes at the Earth's center, too.)

This is only modified at the second order: the spacetime curvature (the Riemann tensor) is nonzero near the Earth. Equivalently, the spacetime curvature prevents us from setting the metric tensor equal to the flat spacetime metric at the second order. We may have the metric schematically of the form $$ g_{\mu\nu} (\vec x) \sim \eta_{\mu \nu} + [R_{\alpha\beta\gamma\delta}] [x^\pi x^\rho] $$ So the metric is flat up to corrections that go like $x^2$ where $x$ is the deviation from the Earth's center. These corrections generically manifest themselves as tidal forces; the greatest contributions come from the Moon and the Sun; other planets may matter, too. The non-inertiality of the Solar System as a whole; and the non-inertiality of our local cluster etc. gives increasingly negligible contributions because the tidal forces decrease with the typical distance scale faster than the force itself.

All other deviations from the flat metric are smaller than that. In other words, the tidal forces are the greatest error that you get if you assume that the Earth is an inertial system floating in an empty space; everything else is smaller.

In the text above, I assumed that you use the non-rotating frame of Earth, one that has a fixed orientation relatively to the stars. You may also use a rotating frame that is fixed relatively to the spinning Earth's surface. The inertiality of this rotating frame is obviously violated by the centrifugal and Coriolis forces (and relativistic corrections to them, including frame-dragging etc.).

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ECEF is rotating, so the final two sentences of the final paragraph are the ones that actually answer the OP's question. –  Ben Crowell May 30 '13 at 15:35
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