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No. I am not asking for the center of Earth's map.

On average, the kg/m3 of Earth's soil is heaver than the kg/m3 of oceans, and the earth does not have a flat surface. And as we have quite some oceans on earth, this means the center of mass of earth is not exactly in 'in its middle'.

enter image description here

I know, the earth's crust is just a fraction of Earth's layers, as you can see in the picture above. But say if the left side of imaginary earth contains a lot of oceans, and the right side none, that that mean that the center of mass of earth is more to the right?

Here, my beautifully drawn image to create an image of my question: enter image description here

Note the water on the 'left' side of Earth, and mountains on the 'Right' side of earth.

So there is no way I can really ask 'coordinates' of the exact center of mass of Earth, but which point on Earth's surface (=includes oceans) is closest to the center (of mass)?

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    $\begingroup$ It might be better to specify in the title that you're looking for center of mass. $\endgroup$
    – Kyle Kanos
    Commented Dec 14, 2016 at 12:23
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    $\begingroup$ When you ask about which point on Earth's surface do you really mean the surface (including top of mountains) or do you mean "point at sea leavel". The polar radius of earth is some 20 km shorter than the equatorial radius, so that point might be close to the poles. $\endgroup$ Commented Dec 14, 2016 at 12:46
  • $\begingroup$ @user1583209 yes, including mountains. So not at sea level (per se). $\endgroup$ Commented Dec 14, 2016 at 12:52
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    $\begingroup$ Earth science SE would be the home for this question. $\endgroup$
    – Mass
    Commented Dec 14, 2016 at 15:42
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    $\begingroup$ I agree. Where is the center of mass relative to what? If the earth is a spinning potato, then by definition the spin axis has to go through the center of mass. But the earth is not a rigid body, as parts of it spin at different rates. So, I am not sure the question has an answer. $\endgroup$ Commented Feb 6, 2019 at 15:35

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I know, the earth's crust is just a fraction of Earth's layers, as you can see in the picture above.

I'm afraid you are still severally overestimating this fraction. Note the text "Not to scale".

The image below is for a carbon purpose, but it shows thicknesses of the layers to scale. You can't even see the crust.

enter image description here

Source: http://phys.org/news/2013-03-deep-carbon-quest-underway-quantity.html

Compare the crust of maximum 50 km in depth to Earth's radius of 6370 km. This is $0.7\;\%$.

If we assume sphere-shape, the volume $V=\frac43 \pi r^3$ of Earth is $V=1.0827\times 10^{12}\;\mathrm{m^3}$. The crust volume is all this subtracted all below the crust: $V=(1.0827-1.0574)\times 10^{12}\;\mathrm{m^3}=0.0253\times 10^{12}\;\mathrm{m^3}$. That is $2.3\%$.

And furthermore, presumably the density difference from ocean to mountain is not enormous.

If you have density values, multiply them onto the volumes and find the masses to compare. But a presumably tiny mass difference in a tiny volume fraction... I doubt there is any practical change of centre of mass.

So there is no way I can really ask 'coordinates' of the exact center of mass of Earth, but which point on Earth's surface (=includes oceans) is closest to the center (of mass)?

Apart from the neglibility, I am not sure which answer you are looking for. Since you say mountains are denser, the centre of mass is displaced slightly towards the more mountainrich side. So are you in the same depth in the ocean at one side of Earth as you are in a mountain cave on the other side, of course the mountain cave brings you closest - though negligibly.

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    $\begingroup$ Fully agree with your assessment that the effect will be minimal. Given that the earth is not a sphere, I still think that the closest point to the center of mass might very well be near the poles as they are some 20 km closer than the equator. $\endgroup$ Commented Dec 14, 2016 at 13:25
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    $\begingroup$ @user1583209 One of my soapboxes as a physics teacher is about relative scale. Do a toy calculation for yourself to work out how big this effect might be. Something like this: assume you start with a uniform spherical planet of Earth's mass and radius but with a ocean covering the land uniformly at all places. You then scoop up a 10 km block of (dense, perhaps limestone) rock from the crust in one place and transport it to the antipodes (i.e. make a hole comparable to a large ocean trench and pair with with a mountain comparable to Mt. Everest.). How much does the CoM move? $\endgroup$ Commented Dec 14, 2016 at 17:58
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    $\begingroup$ Also it should be pointed out that the density increases as one gets closer to the center of the earth, making anything happening in the crust even less relevant. $\endgroup$ Commented Dec 14, 2016 at 20:14
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    $\begingroup$ This doesn't answer the question. The OP was aware that the effect would be small. $\endgroup$
    – user4552
    Commented Feb 6, 2019 at 15:26
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This question is about the earth's axial precession. Recently it has been changing as measured by aquifer depletion using NASA's GRACE mission.

"A study by Rodell et al. [2009] in northwest India used terrestrial water storage-change observations from GRACE and simulated soil-water variations from a data-integrating hydrological modeling system to show that groundwater is being depleted at a mean rate of 4.0 +/- 1.0 cm yr-1 equivalent height of water (17.7 +/- 4.5 km3 yr-1) over the Indian states of Rajasthan, Punjab and Haryana (including Delhi). During the study period of August 2002 to October 2008, groundwater depletion was equivalent to a net loss of 109 km3 of water, which is double the capacity of India's largest surface-water reservoir" as reported by the NASA Jet Propulsion GRACE and Tellus Gravity Recovery and Climate Experiment on their website

https://grace.jpl.nasa.gov/applications/groundwater/

The scientific term for Earth's combined gravitational center with the moon is called the baycenter. It is what creates our four seasons.

https://spaceplace.nasa.gov/seasons/en/

The question being presented is how much the baycenter has deviated due to human activity which redistributes mass globally across the earth thus changing this gravitational center which determines earth's rotational axis that generate the four seasons. This deviation is termed precession. Fortunately this science has already been done. In a research paper titled Measuring the De Sitter precession with a new Earth’s satellite to the ≃10−5 level: a proposal by Lorenzo Iorio his abstract states:

"The inclination I of an Earth’s satellite in polar orbit undergoes a secular De Sitter precession of −7.6 milliarcseconds per year for a suitable choice of the initial value of its non-circulating node Ω. The competing long-periodic harmonic rates of change of I due to the even and odd zonal harmonics of the geopotential vanish for either a circular or polar orbit, while no secular rates occur at all. This may open up, in principle, the possibility of measuring the geodesic precession in the weak-field limit with an accurately tracked satellite by improving the current bound of 9×10−4 from Lunar Laser Ranging, which, on the other hand, may be even rather optimistic, by one order of magnitude, or, perhaps, even better. The most insidious competing effects are due to the solid and ocean components of the K1 tide since their perturbations have nominal huge amplitudes and the same temporal pattern of the De Sitter signature. They vanish for polar orbits. Departures of ≃10−5∘to10−3∘ from the ideal polar geometry allow to keep the K1 tidal perturbations to a sufficiently small level. Most of the other gravitational and non-gravitational perturbations vanish for the proposed orbital configuration, while the non-vanishing ones either have different temporal signatures with respect to the De Sitter effect or can be modeled with sufficient accuracy. In order to meet the proposed goal, the measurement accuracy of I should be better than ≃35 microarcseconds=0.034 milliarcseconds over, say, 5 year."

This research helped contribute to new investigations using GEM-TI and its predecessors GEM-T3 and GEM-T3S by NASA to calculate a new gravitational model for the earth.

https://doi.org/10.1029/JB093iB06p06169

Abstract

A major new computation of a terrestrial gravitational field model has been performed by the Geodynamics Branch of Goddard Space Flight Center (GSFC). In the development of this new model, designated Goddard Earth Model GEM‐T1, the design decisions of the past have been reassessed in light of the present state of the art in satellite geodesy. With GEM‐T1 a level of internal consistency has been achieved which is superior to any earlier Goddard Earth Model. For the first time a simultaneous solution has been made for spherical harmonic parameters of both invariant and tidal parts of the gravitational field. The solution of this satellite model to degree 36 is a major factor accounting for its improved accuracy. The addition of more precise and previously unused laser data and the introduction of consistent models were also accomplished with GEM‐T1. Another major factor allowing the creation of this model was the redesign and vectorization of our main software tools (GEODYN II and SOLVE) for the GSFC Cyber 205 computer. In particular, the high‐speed advantage (50:1), gained with the new SOLVE program, made possible an optimization of the weighting and parameter estimation scheme used in previous GEM models resulting in significant improvement in GEM‐T1. The solution for the GEM‐T1 model made use of the latest International Association of Geodesy reference constants, including the J2000 Reference System. It provided a simultaneous solution for (1) a gravity model in spherical harmonics complete to degree and order 36; (2) a subset of 66 ocean tidal coefficients for the long‐wavelength components of 12 major tides. This adjustment was made in the presence of 550 other fixed ocean tidal terms representing 32 major and minor tides and the Wahr frequency dependent solid earth tidal model; and (3) 5‐day averaged Earth rotation and polar motion parameters for the 1980 period onward. GEM‐T1 was derived exclusively from satellite tracking data acquired on 17 different satellites whose inclinations ranged from 15° to polar. In all, almost 800,000 observations were used, half of which were from third generation (<5 cm) laser systems. A calibration of the model accuracies has been performed showing GEM‐T1 to be a significant improvement over earlier GSFC “satellite‐only” models based purely on tracking data for both orbital and geoidal modeling applications. For the longest wavelength portion of the geoid (to 8×8), GEM‐T1 is a major advancement over all GEM models, even those containing altimetry and surface gravimetry. The radial accuracy for the anticipated TOPEX/POSEIDON orbit was estimated using the covariances of the GEM‐T1 model. The radial errors were found to be at the 25‐cm rms level as compared to 65 cm found using GEM‐L2. This simulation evaluated only errors arising from geopotential sources. GEM‐L2 was the best available model for TOPEX prior to the work described herein. A major step toward reaching the accuracy of gravity modeling necessary for the TOPEX/POSEIDON mission has been achieved.

Using this information one can calculate the earth's gravitational center using gravity wave mechanics and over time derive a histogram of these measured changes.

Unfortunately at this time no published models citing the change in earth's precession or displaying how much it has deviated due to human activity exist for public use as this information would be vital in the development of ICBM targeting technology. While NASA has researched and modeled these gravitational changes extensively due to issues involving national security this information is not currently available to the general public. Perhaps in the future under the freedom of information act it may be possible to obtain a general estimate. Until then we are obliged to do our own calculations and measurements.

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  • $\begingroup$ Not enough characters/reputation, but barycenter, not *baycenter. $\endgroup$
    – phoog
    Commented May 10, 2022 at 9:58
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The center of mass of Earth, using only irregularities you can measure at the surface, projected on the surface is in Turkey. Some of the calculations are driven by the desire to correlate this center with Biblical interpretations, so I wouldn't give them much credibility.

https://en.wikipedia.org/wiki/Geographical_centre_of_Earth

However, the real International Earth Frame Reference System has a model that includes the Earth Observational Parameters that provide you with the location of the poles on the surface. It is based on the observation at many different astronomical stations of very distant radio sources. It does not define where the center of mass is located.

You can read more here:

http://itrf.ensg.ign.fr/general.php

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  • $\begingroup$ The OP is asking where the earth's center of mass is. This answer talks about two things, neither of which answers that question. $\endgroup$
    – user4552
    Commented Feb 6, 2019 at 15:27

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