I asked the same question on Physics StackExchange site. I now know that there are several types of "center" and I can tell, I'm interested in the CM-CF offset (CM = geocenter, CF = center of figure).

Wu, Ray, and van Dam, "Geocenter motion and its geodetic and geophysical implications," Journal of Geodynamics 58 (2012): 44-61.

Based on Fig. 2 in the paper above, I calculated that the distance is 9 meters. However this 9 meters seems too small to me. In addition it's about geocenter motion, but on he vertical axis, it says "meters" and not "meters/year" or "meters / 1000 years" or something similar. On Fig. 5, it's clear that Δt = 1 month. But I don't understand what's Δt for Fig. 2. If there is at all...

I also found in this paper http://goo.gl/pPWD8l On page 174 it says that this offset is 0.8 km. However, it's 1995 data while the other article was published in 2012.

So I'm really confused now, I think I misunderstood something. I'm not a geophysicist, please help me out.

  • $\begingroup$ If the Earth were a pure ellipsoid, these should coincide, so the difference would be due to mountains, continents, etc? $\endgroup$
    – user967
    Commented Mar 15, 2016 at 15:17
  • $\begingroup$ If the Earth were a pure ellipsoid, CM and CF should not necessarily coincide, they could differ due to in homogeneous density, but yes the main difference would be due to mountains and continents. $\endgroup$
    – balping
    Commented Mar 15, 2016 at 21:55
  • $\begingroup$ Considering the center of mass of just Earth is often pointless. The fact is that when talking about the center of mass, we need to consider the barycenter (center of mass of two of more bodies that are orbiting each other) of the Earth-Moon system. In our case, the larger body, Earth, simply "wobbles" slightly, instead of appearing to orbit a common center of mass. The barycenter of the Earth–Moon system on average is located 4671 km from the center of Earth (inside Earth, as the planet has a radius of around 6378 km). $\endgroup$
    – arkaia
    Commented Mar 18, 2016 at 1:02
  • $\begingroup$ I'm asking about the Earth only system, without the Moon. Just read into the first paper I linked above, you can see that the CM of the Earth-only system is actually used in some cases $\endgroup$
    – balping
    Commented Mar 18, 2016 at 20:47

2 Answers 2


There is no inconsistency between the 2012 paper and the older reference(s).

Fig. 2 arbitrarily sets all the components (x,y,z) of all the displacement vectors of all the centers to zero at the beginning of the time period (about 123,000 years ago).

Furthermore the 2012 article states:

The horizontal movement of topography associated with plate motion will then change the CF. Over time, these slow phenomena can create large offsets, up to hundreds of meters, between CM and CF. When the evolution of the internal load is reasonably inferred from other sources, then these processes may be able to explain a significant bulk of the current static offset between CM and CF.

Planetary Sciences, table 1.2, says the distance between CF and CM is 0.80 km, citing to Yoder 1995.

See also Evidence for convection in planetary interiors from first-order topography The Moon March 1973, Volume 7, pages 172-180

It has been known for some time that a center of mass-center of figure offset exists for the Earth (e.g. Jeffreys, 1962). From a spherical harmonic analysis of the Earth's topography with the oceans replaced by an equivalent mass of rock, Balmino et al. (1972) find the center of mass displaced by 1.1 km from the center of figure in the direction 145°W long and 42 ° S lat.

And from the NASA memorandum Reports of planetary geology program - 1981:

For Earth, with its combination of sialic ([denisty]~2.7) and simatic ([density] 3.0-3.3) crusts, the CF is displaced toward 41.6°N, 34.7°E, in the direction of the Eurasian shield, and close to the center of the continental province at 46°N and 27°E (Bills and Ferrari, 1978).

See also Geocenter - Degree 1 :

The spherical harmonic coefficients of degree 1 represent the distance between the center of mass of the Earth and its 'center of figure', which in practice is the center of a set of tracking stations on the surface of the Earth. Because of their physical meaning, time changes in degree 1 coefficients can be expressed in several equivalent forms (1) as distances in mm along the Z (along the axis of rotation), X and Y axes; (2) as (fully normalized) coefficients of the geopotential; (3) as the changes in mass (per unit area) that would give rise to the geopotential coefficients, expressed either in kg/m^2 or cm of equivalent water thickness. The relation between these forms can be found in Swenson et al (2008), equations 5 and 4...

And Planetary Surface Processes (2011):

In fact, the Earth itself has a substantial center of mass – center of figure offset if the water filling the ocean basins is neglected. The floor of the Pacific Ocean is about 5 km below sea level, whereas the opposite hemisphere is dominated by the continental ... The waterless Earth's center of figure is thus offset from its center of mass by about 2.5 km at the present time.

  • $\begingroup$ Thanks very much for your detailed work! I have one question: Planetary Surface Processes (2011) mentions 2.5 km for the solid surface. It's much bigger than 0.8 km by Yoder 1995. I think it's because the second one is about CF of the surface including oceans. Am I correct? I didn't find an exact definition in that book (since google only shows some pages). $\endgroup$
    – balping
    Commented Mar 25, 2016 at 15:25
  • 1
    $\begingroup$ @balping I think the center of figure excluding water is offset from the center of mass excluding water by 2.5 km, but like you say, Yoder doesn't explain what the 0.8km value represents. There are discussion center of mass-center of figure offset going back more than 100 years, for example books.google.com/… I'll put a new bounty on your question and hopefully someone can give a better answer. $\endgroup$
    – DavePhD
    Commented Mar 25, 2016 at 15:45

The CM-CF offset for Earth is 2.1 km, that is that Earth's center of mass as determined by an analysis of the Earth's gravitational potential energy that is represented as a finite sum of spherical harmonics. The rotational geometric centre of figure is obtained by analysis of the hypsometry or topography and bathymetry and calculating the geometric centre of the planet. The offset arises because mass inside the Earth is not homogeneously distributed (e.g., mantle plumes, subducting lithosphere, dynamically displace lithosphere because of glacial isostatic rebound etc..). The later distorts the Earth's equilibrium shape away from a simple spheroid. Good books on the subject are: Physics of Earth by Stacey and Davis Cambridge Press Geodynamics by Turcott and Schubert Cambridge Press

For details of how those parameters are calculated Warning: Mathematical sophistication required is high.

Wieczorek, M. A. (2007). Gravity and topography of the terrestrial planets. In Treatise on Geophysics (Vol. 10, pp. 165–206). Oxford, UK: Elsevier. https://doi.org/10.1016/B978-044452748-6.00156-5


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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