6
$\begingroup$

ABBREVIATED QUESTION/PROBLEM:

The direction of earth's gravity vector does not point towards the center of the earth due to centrifugal force. Theoretically, it should point away by .1 degree. But it doesn't.

The difference between geodetic and geocentric latitude gives us the actual difference between the direction of gravity (geodetic) and the center of the earth (geocentric). Converting between geodetic to geocentric reveals a much greater difference than .1 degree.

Due to centrifugal force, earth's gravity should point away from the center of the earth by approximately a tenth of a degree.

The difference between the direction of gravity and the center of the earth is given by vertical deviation (astronomic latitude).

Astronomic latitude is given by $DEFLEC99$ for all points on earth. $DEFLEC99$ gives astronomic latitude as a difference in latitude from geodetic latitude. However, the difference is mostly negligible ($1$ to $45$ arcseconds).

This means that geodetic and astronomic latitude are practically equal.

We can therefore calculate the actual difference between astronomic latitude (direction of gravity) and geocentric latitude (center of the earth) by subtracting geodetic latitude from geocentric latitude.

As an example, let's take the following:

Geodetic / Astronomic latitude - $49.26083$

Geocentric latitude - $49.07043926$

Difference in latitudes - $0.1903937385$

Now let's see what the difference should be according to the strength of centrifugal force:

Centrifugal force = $(2\pi/86164.091)^2 * (\cos[49.07043926]) * 6365904$ (radius at this latitude) $= 0.02217660948$

Gravity = $GM/R^2 - 3(J2)(GMa^2/R^4)(3/2)(\cos^2[49.07043926]-1) = 9.847454494$

To find the theoretical astronomical latitude, then, we would have to determine the $x$ and $y$ components of the vector sum of centrifugal force and gravity, which is the following:

x component = $\cos[49.07043926] * 9.847454494$

y component = $\sin[49.07043926] * 9.847454494 - 0.02217660948$ (centrifugal force)

new angle = $\tan^{-1} (y component / x component) = 49.16806807$

difference between the latitude above and the geocentric latitude

$= 49.16806807 - 49.07043926 = 0.09762881339$

Compare this figure to the first figures:

Geodetic / Astronomic latitude - $49.260833$

Geocentric latitude - $49.07043926$

Difference in latitudes - $0.1903937385$

Calculated Astronomic latitude - $49.2187618253$

Geocentric latitude - $49.07043926$

Difference in latitudes - $0.09762881339$

Actual difference and calculated difference are off by a good $.1$ degree.

Keep in mind that this is not an isolated case particular to this latitude. The inequality persists at latitudes ranging from $20$ to $50$ degrees geodetic.

Is there a component of gravity I'm not solving for?

Explanation of different latitudes:

Geocentric: Imagine a line drawn from a point on the earth's surface to the center of the earth. The angle between this line and the line parallel to the equator is geocentric latitude.

Geodetic: Imagine a point on the earth's ellipsoid surface. The angle between the line perpendicular to the surface at this point and a line parallel to the equator is geodetic latitude.

Astronomic latitude: The angle between the direction of gravity (plumb line) and a line parallel to the equator. This angle is usually described as a deviation from geodetic latitude (a plus or minus value ranging from $1$ to $44$ arcseconds).

Because the difference between geodetic and astronomic latitude is so small, I tend to refer to them as equal for practical purposes.

SOURCES CITED:

https://physics.stackexchange.com/a/141981/311236 - comment explains the J2 spherical component and provides an equation for it

http://download.csr.utexas.edu/pub/slr/degree_2/C20_Long_Term.txt - NASA text file with up to date figures for C20. C20 currently equals the value at the bottom of the second column

https://grace.jpl.nasa.gov/data/get-data/oblateness/ - NASA website explains that J2 = -C20*(sqrt(5))

https://www.oc.nps.edu/oc2902w/coord/coordcvt.pdf - provides equation for converting from geodetic to geocentric latitude. you don't have to calculate for "Rn" in this equation if "h" is equal to 0. "h" is ellipsoidal height. ellipsoidal height is not altitude above sea level. accuracy of ellipsoidal height is negligible towards the end result, so it's best left to 0 for practicality.

in this equation, "e" does not refer to euler's number. It is instead equal to the square root of (1-(b^2/a^2))

where:

a = 6378137 m

b = 6356752.3 m

https://www.ngs.noaa.gov/cgi-bin/GEOID_STUFF/deflec99_prompt.prl - DEFLEC99. you can plug in the example latitude and longitude: 49° 15′ 39″ N, 123° 6′ 50″ W

$\endgroup$
6
  • $\begingroup$ I wondered about the origin of the name 'astronomic latitude'. Celcius, when working as astronomer, did a standard procedure: to compare vertical position of stars to local level surface, in order to arrive at a value for the latitude of the Uppsala observatory. $\endgroup$
    – Cleonis
    Aug 13 at 13:38
  • $\begingroup$ It's a fitting name for sure. I added some useful sources to the bottom of my question, by the way. $\endgroup$
    – Cypress
    Aug 13 at 14:01
  • 2
    $\begingroup$ This question is not concise enough. What's the problem? I mean I calibrated SRTM (Mankind's 1st global map in one single coordinate system), and NASADEM...and I can't really follow it. If you tightened it up, we'll put it into ECEF and be done. It might be of ellipsoid vs geoid origin, but again: pls make it more concise. $\endgroup$
    – JEB
    Aug 13 at 14:15
  • $\begingroup$ I've edited my post to include a more concise summary of my question. $\endgroup$
    – Cypress
    Aug 14 at 5:51
  • $\begingroup$ I found another resource for conversion between geodetic and geocentric latitude This resource defines the flattening coefficient 'f', and then gives the conversion in terms of that 'f'. And yeah, at 45 degrees latitude the difference between geodetic and geocentric latitude is about 0.2 degrees. Interesting. I did not expect that. The difference is a factor of 2. That could be just a coincidence, but I suspect it's not. Anyway: both values are good, presumably there is a mathematical reason that they don't coincide. $\endgroup$
    – Cleonis
    Aug 14 at 8:09
3
$\begingroup$

You are comparing two angles to the second decimal place. To achieve a degree of accuracy to support this comparison your inputs need to be accurate to within one part in ten thousand.

In you calculation of centrifugal force you assume a sidereal day is exactly 24 hours. This is inaccurate by about 4 minutes, or one part in 360.

$\endgroup$
3
  • $\begingroup$ I ran the calculation again using the time for an accurate sidereal day. The difference is .0008 of a degree: from 0.1483225653 to 0.149137695. I hope you weren't expecting much of a difference. The thing with changes in gravity is that the forces and masses involved are so large that even the smallest changes are hard to come by, which is why the discrepancy in question is so significant. $\endgroup$
    – Cypress
    Aug 13 at 8:16
  • 5
    $\begingroup$ @Cypress Before you can draw any conclusions you need to check that all your other inputs also have the required accuracy. Where does your radius value come from when you calculate centrifugal force ? Have you taken into account the actual shape of the Earth, which is not a sphere (as in Ben51's answer below) ? Are the DEFLEC99 values given for a specific location or are there averaged over a line of latitude ? $\endgroup$
    – gandalf61
    Aug 13 at 8:43
  • $\begingroup$ The radius value comes from this handy website: planetcalc.com/7721. It provides the radius of earth at a given latitude specific to .01 of a second. A variable radius already assumes the earth is not a perfect sphere. The DEFLEC99 values are specific to latitude and longitude. $\endgroup$
    – Cypress
    Aug 13 at 8:55
2
$\begingroup$

Part of the discrepancy is likely due to the fact that the vector of pure gravity (not including centrifugal force) does not point directly toward the center of the earth. You can think of the earth as a sphere, approximately spherically symmetric, which produces a gravitational component directly toward its center, plus a “jacket” which is thickest at the equator and thins to zero at the poles. The gravitational component due to the jacket is not, in general, directly toward the center of the earth.

$\endgroup$
1
  • $\begingroup$ In my equation for gravity, I calculate for uneven mass distribution using the J2 factor. Given the latitude and radius at that latitude, the force of J2 is 0.01142331441. The question is what direction does the J2 force vector point towards? Knowing this would be helpful. $\endgroup$
    – Cypress
    Aug 13 at 8:41
2
$\begingroup$

I concur with the anwer by Ben51.

The center of the Earth's gravitational attraction is not located at the Earth's geometric center.

One way to see that qualitatively is the thought experiment of the center of gravitational attraction of a disk as measured outside the disk, but in the plane of the disk. Since gravity falls off with the inverse square of distance the center of gravitational attraction will not be at the geometric center of the disk.


In the case of the Earth's equatorial bulge: For an object located at the Equator: where is the center of gravitational attraction?

The following reasoning provides a way of assessing the location of the Earth's center of gravitational attraction:
Thought demonstration: an object located at the Equator is at a higher geopotential than at the poles.

If you would have a celestial object, with the same size, shape and gravity as the Earth, but nonrotating and frictionless surface, then an object released from the point of highest geopotential will accumulate kinetic energy as it slides to a point of lowest geopotential.

The dynamic equilibrium of the Earth's equatorial bulge is analogous to the dynamic equilibrium of a Mercury mirror.

The dynamics of a Mercury mirror is that of Hooke's law.

In the case of Hookes law: for circular motion around the center of attraction the ratio of kinetic energy and potential energy is 1:1 (virial theorem, applied for the case of Hooke's law)

The kinetic energy of co-rotating with the Earth, along the Equator, corresponds to a geopotential difference of about 10 kilometer. (Allowing for an error of 10% or so)

Earth radius from geometric center to poles is about 20 kilometer less than radius from center to Equator.

(And of course, for an object not located at any point not on the Equator the center of gravitational attraction is not in the Earth's equatorial plane.)

$\endgroup$
1
  • 1
    $\begingroup$ I think you might be interested in my response to Ben51. The J2/C20 factor takes into account mass distribution. I'd like to know your thoughts on the direction of its force. $\endgroup$
    – Cypress
    Aug 13 at 8:50
1
$\begingroup$

There are some minor problems with your calculation of theoretical gravity. The term theoretical gravity refers to the gravity of a rotating oblate spheroid (ellipsoid) in the absence to tidal effects. This gravity is always normal to the surface of the ellipsoid, so its angle is given by geodetic latitude.

Of course, the Earth is not a uniform ellipsoid, and there are many models and datasets that deal with the subsequent deviations. Step one is to choose an ellipsoid, as there are more than one. Some are global, some are best for specific continents. The most common global one is WGS84.

Thanks to variations in Earth's structure, an equipotential surface deviates from the ellipsoid surface. The deviations are called geoid undulations and are captured with a thing called "The Geoid", of which, there are many. EGM96 is popular, but has been superseded by EGM2008. It is spherical polynomial expansion of theoretical mean sea level up to order 2500-ish. Of course, gravity is a vector, so it deviates from geodetic latitude, and various models/datasets capture those two-parameter deviations (such as DEFLECT99).

But we're dealing with theoretical gravity, so there is no need for those models. The gravity at the surface is not uniform. It varies with latitude as:

$$g(\phi)= g_e\Big[\frac{1+k\sin^2{\phi}}{\sqrt{1-\epsilon^2\sin^2{\phi}}}\Big] $$

where $a,b,\epsilon$ are ellipsoid parameters and:

$$ k=\frac{bg_p-ag_e}{ag_e}$$

where $g_e=9.8321849378\,$m/2 and $g_p=9.78\,$m/s are the acceleration due to gravity at the equator and pole, respectively. Also: $\sin{\phi}$ refers to the sine of latitude in radians, as latitude and longitude are formally geodetic coordinates in degrees, not angles, but...you know, they're also angles.

At any latitude on the surface, there are two radiuses of curvature (the meridional and normal):

$$ M(\phi)=\frac{(ab)^2}{[(a\cos{\phi})^2+(b\sin{\phi}^2)^2]^{\frac 3 2}}$$

$$ N(\phi)=\frac{a^2}{[(a\cos{\phi})^2+(b\sin{\phi}^2)^2]^{\frac 1 2}}$$

but these aren't the radius to the axis of rotation. For that, transform to Earth Centered Earth Fixed coordinates (ECEF):

$$ x=\cos{\phi}\cos{\lambda}(N(\phi)+h)$$ $$ y=\cos{\phi}\sin{\lambda}(N(\phi)+h)$$ $$ z=\sin{\phi}([1-\epsilon^2]N(\phi)+h)$$

where $h$ is the height above the ellipsoid, so:

$$ \rho = \sqrt{x^2+y^2+z^2}=N(\phi)\sqrt{\cos^2\phi+(1-\epsilon^2)^2\sin\phi}$$

Of course, the longitude $\lambda$ drops out.

$\endgroup$
2
  • $\begingroup$ I ran your equation for gravity and got 9.8022 m/s2. It hardly makes a difference towards the end result, though, about .0004 of a degree. I can edit my post to show the difference if you'd like. However, I think Cleonis is right. A more effectual approach might be to examine the conversion between geodetic to geocentric latitude. This paper includes the equation I used, let me know what you think of it: oc.nps.edu/oc2902w/coord/coordcvt.pdf $\endgroup$
    – Cypress
    Aug 14 at 5:49
  • $\begingroup$ geodetic latitude points along $\vec g$ by definition, so subtracting the centrifugal should give you the solution to Newton's gravity of a stationary ellipsoid straight away. $\endgroup$
    – JEB
    Aug 14 at 13:52
0
$\begingroup$

This is a second answer; this answer has a different focus.

We have for conversion between geodetic latitude and geocentric latitude that at 45 degrees latitude the difference between geodetic and geocentric is about 0.2 degrees, whereas a hypothetical plumb line hangs at an angle of 0.1 degrees.

We are looking at a difference between an angle of 0.1 degrees and an angle of 0.2 degrees, that's two significant figures. Therefore in any exploratory calculation that is done: using three significant figures is sufficient.

Gravity at the poles is about 9.82 $m/s^2$, and effective gravity at the Equator is about 9.78 $m/s^2$. The difference is down in the third figure; it'll be fine to just use 9.80 $m/s^2$

At 45 degrees latitude, how much centripetal acceleration is required to remain co-rotating with the Earth?

Since inertial mass and gravitational mass are equivalent there is no need to incorporate a mass term in the mathematical expression.

Required centripetal acceleration:

$$ a_c = \omega^2 r \qquad (1) $$

The significant figures of the square of the Earth angular velocity: 5.31

The significant figures of the Earth radius at 45 degrees latitude: 4.50

Required centripetal acceleration at 45 degrees latitude: 0.0239 $m/s^2$


For comparison: at the Equator the required centripetal acceleration to remain co-rotating with the Earth is 0.0339 $m/s^2$ Providing this required centripetal acceleration goes at the expense of the inverse-square-law gravitational acceleration, resulting in an effective gravitational acceleration of about 9.78 $m/s^2$.

At 45 degrees latitude we need to decompose the required centripetal acceleration into a component parallel to the local surface, and a component perpendicular to the local surface.

Only the component parallal to the local surface is relevant here; the component perpendicular to the local surface affects the period of a swinging pendulum, but not the angle of a plumb line.

At 45 degrees latitude, required centripetal acceleration in the direction parallel to the local surface:
0.707 * 0.0239 = 0.0169 $m/s^2$

Corresponding angle:
arcsin(0.0169/9.80) $\approx $ 0.1 degree


At all other latitudes the value will be smaller than 0.1
Closer to the Equator the component parallel to the local surface is smaller. Closer to the poles the distance to the center of rotation is smaller.


In the question the formula with the J2 term was used to calculate gravity. That is overkill. For the question at hand just looking up the value is sufficient.


We have that when converting between geodetic and geocentric latitude: at 45 degrees the difference is about 0.2 degree. So we have that that value does not coincide with the value of the corresponding plumb line angle, which is about 0.1 degree.

In the exploratory calculation above the focus was on the angle of a hypothetical plumb line. The exploratory calculation is for motion with respect to a sphere, but actual Earth latitude involves using the reference ellipsoid.

The difference between sphere and reference ellipsoid is, I think, where the explanation for the values not coinciding must be sought.

Anyway, given that inertial mass and gravitatational mass are equivalent the hypothetical plumb line angle cannot be measured. Only the local effective gravitational acceleration is accessible to measurement. The hypothetical plumb line angle does not enter any form of calibration, neither in astronomy nor in geophysics, so it's never a problem.



General remarks

Start with looking at how much significant figures you need. If the final value needs to be accurate to three significant figures then any exploratory calculation only needs to carry four significant figures.

In its current form the question has many numbers with 10 or so digits after the decimal point. That makes the question look very crowded.

$\endgroup$
5
  • $\begingroup$ I think it was just difficult for me to present the components of my question in a less overwhelming, more digestible way. Part of including the "overkill" aspects and digits was to emphasize thoroughness. When presenting this kind of discrepancy, people quickly assume you've slacked somewhere. Need to find the right balance of simplicity and precision here. $\endgroup$
    – Cypress
    Aug 14 at 13:26
  • $\begingroup$ Regarding the results: as I've mentioned elsewhere, changes in the direction of gravity are hard to come by due to the magnitude of the numbers involved (millions of meters, mass of earth). Think of it this way: if it takes the combined axial radius and squared velocity of earth to change .1 degrees, what would it take to change another .1? I think it's more than reference ellipsoid because the ellipsoid model was developed to explain current gravity measurements, so any changes to the model would have to yield similar results. Something is very off. $\endgroup$
    – Cypress
    Aug 14 at 13:26
  • $\begingroup$ @Cypress About the two values 0.2 degree and 0.1 degree. I am confident both are good. I repeat what I wrote in this answer: "The hypothetical plumb line angle does not enter any form of calibration, neither in astronomy nor in geophysics, so it's never a problem." You are worried something is off. I say: this has never been noticed before because there isn't any form of calibration that uses it. It lies at an intersection of GIS and physics where no one has gone before. $\endgroup$
    – Cleonis
    Aug 14 at 13:47
  • $\begingroup$ @Cypress My current thinking is as follows. One way to investigate this is to move through a range of rotation rates, from zero rotation rate to, say, double the Earth rotation rate. Each rotation rate will have a corresponding flattening. (flattening of rotating bodies) Each degree of flattening has a corresponding deviation between geodetic and geocentric coordinates. Both can then be plotted: geodetic-vs-geocentric as a function of rotation rate, and centrifugal effect as a function of rotation rate. $\endgroup$
    – Cleonis
    Aug 17 at 21:07
  • $\begingroup$ Defining the deviation accurately over all latitudes would be a useful approach, though it still doesn't answer the discrepancy. I think the problem lies with current gravity measurements and the supposed parameters of earth. Both of these values combined severely restrict the possibilities for a significant change in the direction of gravity. In my response to Thomas, I mentioned how gravity equations define radius as distance to the center of earth, meaning that uneven volume distribution can't be used to explain plumb line deviation. $\endgroup$
    – Cypress
    Aug 17 at 22:50
0
$\begingroup$

This is a third answer, with a focus different from the previous two.

The general formula to express the relation between geocentric latitude and geodetic latitude is as follows: With:
$Lat_c$ => geocentric latitude
$Lat_d$ => geodetic latitude
f => the flattening of the Earth

$$ \tan(Lat_c) = (1 - f)^2 \tan(Lat_d) \qquad (1) $$

In this case we don't need many significant figures; two is enough. For the purpose of this particular comparison we can simplify the relation to the following expression:

$$ Lat_c \approx (1 - f)^2 Lat_d \qquad (2) $$

Now we take a closer look at the $(1 - f)^2$ factor

$$ (1 - f)^2 = 1 - 2f + f^2 \qquad (3) $$

Since the flattening 'f' is very small the square of f is negligable. (comparison: 0.99*0.99=0.9801)

So the following is sufficient:

$$ Lat_c \approx (1 - 2f) \ Lat_d \qquad (4) $$

Crucial for answering the question is the factor '2' that is in (4).


The Earth's flattening

The following two are both proportional to the square of the Earth rotation rate:

A. The magnitude of the Earth's flattening

B. The angle between local plumb line and the direction of the local inverse-square-law gravity

Those two are both proportional to the square of the Earth rotation rate because in both cases the underlying mechanism is identical: providing required centripetal acceleration.


We see that it is in the process of interconversion between geocentric and geodetic latitude that the factor '2' is introduced. This factor 2 enters by way of the squaring in (1)

The flattening 'f' is very small, therefore the effect of the squaring is that it acts as introducing a factor '2'.

$\endgroup$
0
$\begingroup$

The point is that the earth is not a sphere but (in a first order approximation) an ellipsoid, so even without any rotation, the local vertical would not go through the earth's center (unless at the pole or equator). You would have to add this static 'aberration' due to the ellipsoid shape to the centrifugal 'aberration'.

This is is straightforward to show as follows:

the gravitational potential due to a homogeneous static ellipsoid is

$$\Phi(r,\theta)=-\frac{GM}{r}+\frac{2 f GM \overline R^2}{5 r^3}P_2(\cos(\theta)) +O(f^2)$$

where $r$ is the radial distance, $\theta$ the polar angle (90-latitude) of the observation point,$\overline R$ the average radius of the ellipsoid and f its ellipticity (flattening). $P_2(\cos(\theta))$ is the Legendre polynomial

$$P_2(\cos(\theta))=\frac{1}{2}(3 \cos^2(\theta)-1)$$

We can get the acceleration vector from this by taking the gradient

$$\vec g(r,\theta)=(g_r(r,\theta),g_{\theta}(r,\theta))=-\vec \nabla\Phi(r,\theta)=(-\frac{\partial\Phi(r,\theta)}{\partial r}, -\frac{1}{r}\frac{\partial\Phi(r,\theta)}{\partial \theta})$$ ignoring the azimuthal coordinate here due to rotational symmetry. Evaluating this yields

$$g_r(r,\theta)=-\frac{GM}{r^2}\cdot [1+\frac{3}{5}f\frac{\overline R^2}{r^2}\cdot(3 \cos^2(\theta)-1)]$$ $$g_\theta(r,\theta)=\frac{GM}{r^2}\frac{3}{5}\frac{f \overline R^2}{r^2}\cdot \sin(2\theta)$$

Since $g_\theta(r,\theta)\neq 0$, the acceleration vector is in general not going through the center of the ellipsoid but has a component pointing to the equator. Taking an ellipticity $f=1/298$ for the earth, one obtains a maximum deviation from the radial (geocentric) direction at the surface ($r=\overline R$)

$$\alpha_g = \arctan(\frac{g_\theta(R, 45 \deg)}{g_r(R, 45 \deg)})\approx 0.11 \deg$$

This non-radial acceleration caused by the shape of the ellipsoid is pointing towards the equator, as expected (note again that $\theta$ is the co-latitude (90-latitude) here).

If we assume this ellipsoid as perfectly rigid and add a rotation to it, there will be an additional centrifugal force added, leading to the net acceleration

$$\gamma_r=g_r+\omega^2r_x\sin(\theta)$$ $$\gamma_\theta=g_\theta+\omega^2r_x\cos(\theta)$$

where

$$r_x:=r\sin(\theta) = \frac{a\sin(\theta)}{\sqrt{\sin^2(\theta)+\frac{\cos^2(\theta)}{1-f}}}$$

is the projection of the position on the equatorial plane (with $a$ the equatorial radius and $f$ the ellipticity (flattening)). $\omega$ is the rotational frequency of the earth, and $\theta$ the co-latitude.

The deviation of this net acceleration vector from the direction to the earth's center is

$$\alpha_\gamma = \arctan(\frac{\gamma_\theta}{\gamma_r})$$

Evaluating this for $\theta=45 deg$ gives a value $\alpha_\gamma\approx0.21 \deg$. This is the 'aberration' of the plumb line from the direction to the earth's center for a point on the reference ellipsoid including the centrifugal force. In principle this aberration should be the same as that of the local vertical on the surface of the ellipsoid. The latter can be calculated by taking the gradient of the ellipsoid equation

$$S=\frac{r^2\sin^2(\theta)}{a^2} +\frac{r^2\cos^2(\theta)}{b^2}$$

i.e.

$$\nabla_r S = \frac{2r}{a^2} [\sin^2(\theta)+\frac{\cos^2(\theta)}{(1-f)^2}]$$ $$\nabla_\theta S = \frac{r}{a^2}\sin(2\theta) [\frac{1}{(1-f)^2} -1]$$

from which we obtain the deviation of the normal to the ellipsoid as

$$\alpha_S = \arctan(\frac{\nabla_\theta S}{\nabla_r S})$$

which for $\theta =45 \deg$ and $f=1/298$ evaluates to $\alpha_S=0.19 \deg$ This differs by about $0.02 \deg$ from the direction of gravity. The reason for this is the fact that the earth is not an ideal fluid but has some rigidity. For an ideal fluid the flattening would be larger, and for $f\approx 1/233$ the discrepancy does indeed disappear (see also this reference in this respect).

$\endgroup$
19
  • $\begingroup$ Yeah, it does seem something additive is going on. I visualize the cross section of a sphere, in the process of being flattened to the reference ellipsoid. In spherical state there is a single 45 degrees point; we can plant a flag there, as if planting in soil. Then we start flattening. The flattening is redistributing the volume, and that makes the flag move outwards a bit. The same flattening also changes the angle of the slope. My hunch: for small values of the flattening the two contributions are roughly equal in magnitude $\endgroup$
    – Cleonis
    Aug 15 at 7:14
  • $\begingroup$ Responding to Thomas: I'm not sure if I'm understanding you correctly, but there is no "static aberration". A static earth would not be ellipsoidal, it would presumably be spherical because the ellipsoid shape is a result of rotation. $\endgroup$
    – Cypress
    Aug 15 at 8:26
  • $\begingroup$ Responding to Cleonis: Looking into volume distribution would be useful. In my original equation for gravity, you can see the J2/C20 factor, which accounts for greater mass at the equatorial bulge. I included a useful explanation of J2 at the bottom of my post in sources cited. I believe J2 has significant magnitude. Its direction is equal to pure gravity, though. $\endgroup$
    – Cypress
    Aug 15 at 8:36
  • $\begingroup$ @Cypress What I meant to convey with 'volume redistribution' is this: imagine a cilindrially symmetric hill, and you plant a flag at a particular distance to the center of that hill, about halfway up the slope. Now imagine that hill isn't geologically stable, and there is creep. The entire hillside slumps down a bit, like thick molasses would. That will have two effects: the flagpole, set in the soil, moving together with the soil, moves away from the hill's center a bit. In addition the inclination of the hillside will decrease a bit. Those two effects add up. $\endgroup$
    – Cleonis
    Aug 15 at 9:01
  • $\begingroup$ @Cypress It doesn't really matter how the shape of the earth arises. Just assume that it is an ellipsoid in the first place and that it is so rigid that the rotation does not distort it further. In this case you would have the static 'aberration' of the local vertical + the one from the centrifugal force as well. $\endgroup$
    – Thomas
    Aug 15 at 9:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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