# How are Earth's rotational changes due to large earthquakes calculated?

According to NASA (and many other sources), Japan Quake May Have Shortened Earth Days, Moved Axis and in the answers to the question here What geophysical events can (temporarily) increase the Earth's rate of rotation?, that major earthquakes result in a slight shift in the length of the day, from NASA:

calculations indicate that by changing the distribution of Earth's mass, the Japanese earthquake should have caused Earth to rotate a bit faster, shortening the length of the day by about 1.8 microseconds (a microsecond is one millionth of a second).

Past great earthquakes, such as the 8.8 in Chile in 2010 and the 9.1 Sumatra in 2004 have caused similar rotational 'wobbles'. The NASA article states that the earthquake's magnitude, location and dynamics result in the orbital 'wobbles'.

How are Earth's rotational changes due to large earthquake calculated, and especially distinguished from oceanic and atmospheric effects?

Changes in the Earth's rate of rotation is measured by the International Earth Rotation and Reference Systems Service (IERS) using observational techniques. They publish daily the rate and orientation of the Earth's rotation, and publish near-term forecasts (of great practical use if you are launching satellites into orbit.) Mass movements in the atmosphere due to weather can change the rotation. I know papers written by Richard Gross at JPL have discussed methods used to examine rotation/earthquake interactions - you can try looking there.

Earthquakes, specially those at subduction zones, result in redistribution of mass (as one plate slides on top of the other). This change in distribution of mass causes changes in gravitational field that can be measured by satellites such as GRACE.

Now angular momentum is conserved. If we know the change in mass then we can calculate the change in moment of inertia and thus the change in the angular velocity.

At a given location, measure the time between when Sirius (or any star) is due south today and due south tomorrow. This is the sidereal day, which is approximately 23h56m long. The more familiar 24h solar day can be computed from the sidereal day.

To be absolutely precise, astronomers measure the due south time of the vernal equinox, not an actual star. A sidereal day is the interval between "successive culminations of the vernal equinox".

The earth's solar day is increasing at a near-constant rate due to (lunar) tidal friction. When scientists see a change that's not consistent with this near-constant increase, they know it's due to some other event.

You might also want to ask in http://astronomy.stackexchange.com

• I do not know enough to know if this is possible, but am skeptical, perhaps you could expand on this?
– Neo
Oct 15 '14 at 18:21
• astronomy.stackexchange.com/questions/7683/… may or may not help.
– user967
Oct 15 '14 at 23:05