....is an oblate spheroid because centrifugal force stretches the tropical regions to a point farther from the center than they would be if the planet did not rotate. So we all learned in childhood, and it seems perfectly obvious. However...

I am at $45^\circ$ north latitude. Does that mean

  • An angle with vertex at the center of the earth and one ray pointing toward the equator at the same longitude as mine, and one ray pointing toward me, is $45^\circ$ (that would mean I'm closer to the north pole than to the equator, measured along the surface, as becomes obvious if you think about really extreme oblateness); or
  • The normal to the ground where I stand makes a $45^\circ$ angle with the normal to the ground at the equator at the same latitude (this puts me closer to the equator than to the north pole); or
  • something else?

If for the sake of simplicity we assume the earth is a fluid of uniform density, it seems one's potential energy relative to the center of the earth would be the same at all points on the surface.

  • Would the force of gravity at my location, assuming no rotation, be directly toward the center? Would it be just as strong as if the whole mass of the earth were at the center and my location is just as far from the center as it is now?
  • Would the sum of the force of gravity (toward the center or in whichever direction it is) and the centrifugal force (away from the axis) be normal to the surface at my location?
  • Given all this, how does one find the exact shape?
  • How well does that shape in this idealized problem match that of the actual earth?
  • 4
    $\begingroup$ This belongs on physics, although I would still suggest to edit it to be a single question. $\endgroup$
    – Phira
    Oct 30, 2011 at 18:27
  • $\begingroup$ @Phira If I wondered whether the Cartesian vortex theory or Newton's theory is closer to the truth, that would certainly belong to physics and not to mathematics. But if you accept Newton's physics, all that remains to answer the questions above is mathematics. (BTW, this problem was the "crucial test" of the Cartesian-versus-Newtonian theories. The former predicted an oblong rather than an oblate earth. Geodetic measurements in 1733, paid for by the French taxpayers, showed it was indeed oblate, and the Cartesian theory did not survive that blow.) $\endgroup$
    – Michael Hardy
    Oct 30, 2011 at 18:39
  • 1
    $\begingroup$ @MichaelHardy If you accept all of theoretical physics, then most problems of theoretical physics "are mathematics". But it is much more likely that a physicist knows physics. $\endgroup$
    – Phira
    Oct 30, 2011 at 18:47
  • 1
    $\begingroup$ @MichaelHardy What makes you think that you will get a better answer here than on the physics site? $\endgroup$
    – Phira
    Oct 30, 2011 at 18:48
  • $\begingroup$ Related: physics.stackexchange.com/q/8074/2451 $\endgroup$
    – Qmechanic
    Oct 30, 2011 at 20:21

3 Answers 3


It is not popular to begin an answer with a question but I would do that: How do you know that you are at 45° north latitude?

There are several latitudes defined. If you got the latitude from a map, Google Earth or maps, of read it at your GPS receiver, that its is geographic or geodetic latitude. In such a case 45° is the angle between normal to reference ellipsoid and equatorial plane. The reference ellispoid (for example WGS84 used in the Google and GPS) is a mathematical construction, an easily managed approximation to the actual shape of Earth.

If you got the latitude by your own direct measurement of height of the north pole (in the sky) above horizon or its distance from zenith (supposing you took into account refraction, abberation, nutation etc.), than it is astronomic latitude. In that case 45° is the angle between normal to geiod and the equatorial plane. Geoid is equipotential surface passing through your point of observation.

Finally, there is also geocentric latitude - angle between the equatorial plane and a line passing through your place and Earth's center.

If Earth is a perfect non rotating sphere and distribution of its mass has spherical symmetry than all three latitudes has the same value at every point of globe.

See Coordinate systems by R. Knippers for additional details.

If you require that gravitational force at every point of the globe point towards the Earth's center than mass distribution must have sperical symmetry. Since rotation is axially symmetric you cannot get the required result unless mass distribution is so special (and unphysical) that it compensates for effect of rotation added.

Only if sphape of Earth closely follows equpotential surface (for example is covered by ocean) than neat force (gravity plus centrifugal one) is perpendicular to the surface at every place.


1, At 45 deg (N) latitude you are closer to the North pole, to picture this just draw the Earth as a much more extreme oblate spheroid.

2, The shape of the Earth is set by the outward rotational force exactly balancing the inward gravitational force at every point (except for local geography). So the overall potential is always down (except for local geology)

There is a useful intro to the difference between mean sea level and the Earth's surface at ESRI (makers of the popular GIS/mapping software)

  • $\begingroup$ +1 for answering some of the questions and adding the link. Most of the question remains unanswered. $\endgroup$ Nov 1, 2011 at 15:26
  • $\begingroup$ Your point $1$ does not answer my question about the meaning of latitude, but only says what I already said in the question. $\endgroup$ Sep 6, 2017 at 19:59

I know this thread is old, but wanted to reply to a few of the points that weren't addressed yet.

The most common definition of latitude is the angle between a vector going from earth's center to a point on the surface and the equatorial plane. This is a good choice since it allows use of the "fixed stars" to determine location on earth, as well as orbiting satellites whose dynamics don't depend on whether the earth's shape is spherical or oblate spheroidal - outside of higher order orbital perturbations.

Regarding the gravity question, it does not point toward earth's center at any locations on earth besides the equator and the poles. Everywhere in between, what we measure is actually effective gravitation due to - as you suggest - both the force of gravity, and the centrifugal contribution arising since earth is non-inertial due to its rotation.

This effective gravitation is normal to the surface, but that surface is oblate spheroidal, so the surface normal does not pass through the earth's center except for at the poles and equator, as mentioned above. The magnitude of that effective gravitation is the 9.80 to 9.81m/s/s value that every physics student memorizes. Without the centrifugal contribution it would be closer to 9.83m/s/s. Another related fact is that the direction an object will hang in place (like a plumb bob on a long string) is NOT the same as the path a falling object will follow as it descends. This is due to the Coriolis force that arises for moving bodies in non-inertial frames.

The shape of earth can be calculated using the concept of the effective potential rather than the standard potential energy that most students think of from first semester physics. The whole purpose of effective potential is to create an energy expression that acts in such non-inertial frames as earth. A way to thing about the effective potential is that a rotating earth will tend to "fling" objects outward in proportion to their distance from the rotation axis. As such, it is easier to rise against gravity farther from the rotation axis and therefore the rotation plays into the perception of potential energy in rotating coordinate systems. The same concept is used to find the locations of Lagrange points in 2-body non-inertial gravitational systems.

Regarding how well radar satellite maps match the theoretically predicted shape, it is quite good, but you have to keep in mind that surface topography of earth is far from smooth. So there are places on earth's surface that will differ from the ideal oblate spheroid by miles... and the spheroidal earth itself is only around 14 miles farther across at the equator than at the poles, so to our eye (if we could view earth from space) earth would look pretty much like a perfect sphere in spite of its equatorial "bulge".


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