Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

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?

share|cite|improve this question
Related: and links therein. –  Qmechanic Apr 25 '13 at 19:21

2 Answers 2

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:

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

share|cite|improve this answer
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
@BenCrowell: and yet the question is tagged "Newtonian mechanics"... –  Rody Oldenhuis May 30 '13 at 20:32

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.).

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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