# Earth as inertial reference frame: finite radius effects

In a recent question I tried to clarify under what conditions the Earth can be considered as an inertial reference frame. The opinions, summarized in my own answer to the cited question, are that

• Apart from its rotation, the Earth is in a state of free fall and therefore is an inertial reference frame, as only relative accelerations between the bodies can be measured.
• The Earth's rotation is slow, and results only in small non-inertial corrections, as compared to typical accelerations observed in our everyday life. (One can easily account quantitatively for these corerctions or the errors resulting from their omission - see fictitious forces.)

However, this answer seems to suggest that free fall conclusion is only approximately accurate for a finite-size object, and there are also corrections due to this finite size. Hence the answer given previously is possibly incomplete. Specifically,

• are these corrections distinct from the Earth's rotation (or do they express the same thing really)?
• can they be compensated by the Earth's rotation?
• what are the limitations on the Earth's size/radius for making this corrections small (e.g., in comparison to the gravity force experienced at the surface)?

Are these corrections are distinct from the Earth's rotation (or do they express the same thing really)?

The answer by Luboš Motl was describing the contribution of spacetime curvature as deviations from flat spacetime. In flat spacetime, you can always find a global inertial reference frame, and due to the "automatically guaranteed" fact he mentioned we know that in curved spacetime you can always find a local inertial frame where deviations from inertial are second order in space and time.

Flatness means that gravity is uniform or absent. Spacetime curvature means that gravity is non-uniform, which is tidal gravity. The further away you go in curved spacetime the more non-uniform gravity is. This causes spatially separated geodesics to accelerate relative to each other more than nearby geodesics.

Luboš Motl made one mistake in his answer, and that was to claim that the greatest contributions come from the moon and the sun. The largest source of non-uniform gravity near the earth is the earth itself. Due to the curvature and the finite size of the earth, geodesics on opposite sides of the earth accelerate towards each other at 2 g, which swamps any contribution from the moon or sun. It is this non-uniformity in gravity, this curvature of spacetime, that keeps the surface of the earth from expanding although it is accelerating outward at 1 g.

As your region of interest covers a larger and larger area, the deviations from flatness increase. So in a small lab, we can treat free-falling objects as inertial. If they start out at rest with each other then they stay approximately the same distance apart. Over a large region, such as the whole earth, that doesn't work. Free-falling objects initially at rest with each other do not stay the same distance apart.

Can they be compensated by the Earth's rotation?

No. They have nothing to do with the earth's rotation. The effects described above by me and in the other answer by Luboš Motl are for a non-rotating object. Earth's rotation produces an additional but very small effect. I have neglected that here.

What are the limitations on the Earth's size/radius for making these corrections small (e.g., in comparison to the gravity force experienced at the surface)?

Basically, for the curvature corrections to be small you need to have a small enough region of space and time so that objects at rest in free-fall do not appreciably change their distances with each other. If your room is small enough that everything falls in the same direction then you should be fine. If you have a lab that spans a continent then objects falling on one side of the lab will be accelerating slightly towards objects falling on the other side of the lab.

• Going back to my question about when the Earth can be treated as an inertial reference frame - we also need to be working with small enough length and time scales? Commented Sep 13, 2021 at 14:24
• Yes. That is correct.
– Dale
Commented Sep 13, 2021 at 14:35

Are these corrections are distinct from the Earth's rotation (or do they express the same thing really)? Can they be compensated by the Earth's rotation?

The are distinct and they cannot.

As a clear example, consider tides of our oceans. At one place on the Earth's equator the tides are rising, and at another they are falling. In rotating frame the centrifugal force is the same along whole equator, so there is no way to attribute this effect to rotation.

In Luboš answer he claims, that center of the Earth is moving along the geodesic (i.e. is inertial), but the further away you are from the center, the bigger noninertial effects you will experience. This effects are due to the fact, that two parallel geodesics will not remain parallel for long - there is a geodesic deviation. So if we have two freely falling particles in the vicinity of Earth that are initially parallel (in four-dimensional view, 3D parallelism does not suffice), they will start to move away from/closer to each other, just like in figure 4 in this wiki page. The formula for tidal effects is given in the page, together with some values.

But the thing is, we are mixing two frameworks together. In GR, one could hardly consider Earth inertial frame, since everything on the surface of the Earth is accelerated upward with pretty significant acceleration. Every engineer accounts for this when he constructs car or building, so the only way in which to interpret the question whether Earth is inertial frame is in the framework of Newtonian gravity. There, the tidal forces are not due to the failure of our frame to be inertial, but due to the gravitational interaction between bodies. From this point of view, the Earth can be considered inertial.

But we know, it rotates around its own axis and around Sun. Rotation around axis makes surface noninertial as you correctly stated in your answer. If we fix the rotation of our frame by the distant stars, then we get rid of this and rotation around Sun will produce the larges noninertiality in our frame.

In Newtonian framework, one does not consider freely falling bodies inertial, but accelerated. However, from equivalence principle we know, that this acceleration of freely falling bodies simulates inertial frame pretty accurately, we just need to ignore gravitational force that governs the free fall. In Newtonian framework, to describe motion of Earth around Sun in the frame of the Earth, we would need to compute gravitational force from the Sun and centrifugal force from noninertiality of our frame and we arrive at conclusion that they cancel each other out. This is mathematically equivalent to asserting, that frame of Earth in vicinity of it is inertial and there is no gravitational field from the Sun. This is in fact what engineers do. They assume Earth is inertial frame and they do not bother with Sun's gravity. So the approximation that uses Earth's frame as inertial frame is much better than what we would have guessed just by looking at the motion of the Earth and seeing that it moves around an ellipse.

That being said, Earth's center does not move on a geodesic, but this is different effect than tidal forces that Luboš describes. This arise due to the fact, that each particle constituting the Earth tries to move on its own geodesic, but intermolecular forces prohibit this and resulting motion one could call a compromise. But this is very small effect, probably not even worth mentioning.