I know that Geodetic Latitude is measured with respect to the normal at a point on the surface of the reference ellipsoid. But what about points above and below the surface? Do they follow a hyperbolic path? (See the graphic I created.) Or do they follow a straight line?

Wikipedia says, "[Ellipsoidal] coordinates are the natural choice in models of the gravity field for a uniform distribution of mass bounded by the reference ellipsoid."

Latitude should follow gravity if possible, shouldn't it?

Reference ellipsoid with lines of latitude

up vote 6 down vote accepted

No, latitude doesn't follow gravity (as @mkennedy notes, it follows the normal to the ellipsoid).

And, no, gravity doesn't follow your hyperbolic curve (nor a straight line).

The simplest model for gravity of the earth which accounts for its ellipsoidal shape and its rotation is "normal gravity". (And the formulas for normal gravity are conveniently expressed in terms of ellipsoidal coordinates.) Unfortunately, the Wikipedia articles on this subject, theoretical gravity and normal gravity formula, are deficient in that the height variation is treated only approximately. (I haven't yet had the energy to fix this!) However, I have written up some detailed notes on normal gravity here.

Here's the figure from those notes showing the field lines (green) and level surfaces (blue) for an exaggerated model of the earth:

field lines and level surfaces for normal gravity

The red curve is the surface of the ellipsoid. Normal gravity is only uniquely defined outside the ellipsoid because the gravity inside the ellipsoid depends on the mass distribution (which is not specified in the derivation of normal gravity). In this figure, normal gravity has been extended inside the ellipsoid assuming that the mass is all concentrated on a disc on the equatorial plane.


By the way, falling bodies don't follow field lines. Because this is a rotating system, Coriolis forces come into play. In addition the bodies interia will cause the body to deviate from a curved field line.


The field lines follow hyperbolas if the ellipsoid is not rotating. Two possible mass distributions which then result in a constant gravitational potential on the reference ellipsoid (i.e., which satisfy the conditions for normal gravity) are:

  • All the mass is sandwiched uniformly between the ellipsoid and a slightly smaller similar ellipsoid. In this case the potential is constant inside the ellipsoid. Such an ellipsoidal shell is called a homoeoid.

  • A massive circular disk of radius E, where E2 = a2 - b2, with mass distribution proportional to 1/sqrt(E2 - R2), for radius R < E. This is the limiting case of the homoeoid.

  • If a < b (the ellipsoid is prolate), the disk is replaced by a massive rod with uniform mass distribution.

Details are given in my notes.


A uniform mass distribution is a possible solution to the problem of normal gravity. This is the so-called Maclaurin spheroid. In this case the flattening is given by the rotation (instead of being independently specified). In this case, the level surfaces inside the ellipsoid are concentric similar ellipsoids and the field lines all terminate at the center of the ellipsoid. (The field outside the ellipsoid is normal gravity, of course.) Here are the level surfaces (blue) and field lines (green) inside the ellipsoid for f = 1/5:

field lines and level surfaces for maclaurin spheroid

  • Within the confines of the reference ellipsoid the field lines (green) are hyperbolic (or nearly so). Hence the hyperbola segment in the graphic in my original question. I did not think there would be such extreme variation outside the reference ellipsoid, however. I will have to read your notes. – posfan12 Aug 24 at 1:36
  • 1
    In a mathematical treatment, "nearly hyperbolic" means "not hyperbolic"! Note the parameters used here: flattening = 1/5 and geostationary orbit = 2.2526 times the equatorial radius. For the earth we would have (approximately) flattening = 1/300, geostationary orbit = 6 times the equatorial radius. – cffk Aug 24 at 3:01
  • If the Earth were a uniform mass, would it make a difference? Or does Normal Gravity already take this into account? – posfan12 Aug 24 at 20:45
  • 1
    The surface of an ellipsoid of uniform density is only a level surface if it's not rotating. This is a special case of normal gravity; but it's not a good model for the earth. Exterior to such a body the field lines are hyperbolic; inside they are not. – cffk Aug 24 at 21:24
  • 1
    Bletch, my last comment is wrong. The surface of a ellipsoid of uniform density is not a level surface. An non-rotating ellipsoidal shell whose density is proportional to the distance from the center of the shell to the tangent plane is a level surface (and the gravity inside such a shell vanishes); see Chasles (1840). – cffk Aug 26 at 14:32

At latitudes nearer the Equator, the inertia produced by Earth's rotation is stronger than at polar latitudes. This counteracts the Earth's gravity to a small degree – up to a maximum of 0.3% at the Equator – reducing the downward acceleration of falling objects.

The difference in gravity at different latitudes is that the Earth's equatorial bulge (itself also caused by inertia) causes objects at the Equator to be farther from the planet's centre than objects at the poles. Because the force due to gravitational attraction between two bodies (the Earth and the object being weighed) varies inversely with the square of the distance between them, an object at the Equator experiences a weaker gravitational pull than an object at the poles.

In combination, the equatorial bulge and the effects of the Earth's inertia mean that sea-level gravitational acceleration increases from about 9.70999 m·s−2 at the Equator to about 9.832 m·s−2 at the poles, so an object will weigh about 0.5% more at the poles than at the Equator.

The same two factors influence the direction of the effective gravity. Anywhere on Earth away from the Equator or poles, effective gravity points not exactly toward the centre of the Earth, but rather perpendicular to the surface of the geoid, which, due to the flattened shape of the Earth, is somewhat toward the opposite pole. About half of the deflection is due to inertia, and the half because the extra mass around the Equator causes a change in the direction of the true gravitational force relative to what it would be on a spherical Earth.


Regarding the points above and below the surface from the observer point of view they follow a straight line.

  • Quote, "Anywhere on Earth away from the Equator or poles, effective gravity points not exactly toward the centre of the Earth, but rather perpendicular to the surface of the geoid..." The hyperbola I drew is perpendicular at the surface. And the Wikipedia article I linked to seems to suggest that gravity follows the curve, not a straight line. (Though Geodetic Latitude as it is used in practice may ignore this.) – posfan12 Aug 22 at 20:27
  • example: When the merry-go-round is not rotating, rolling the ball back-and-forth is simple and straightforward. While the merry-go-round is rotating, however, the ball won’t make to your friend sitting across from you without significant force. Rolled with regular effort, the ball appears to curve, or deflect, to the right. Actually, the ball is travelling in a straight line. Another friend, standing on the ground near the merry-go-round, will be able to tell you this. You and your friends on the merry-go-round are moving out of the path of the ball while it is in the air. – Damini Jain Aug 23 at 3:00
  • that is due too Coriolis effect. – Damini Jain Aug 23 at 3:01

Don't forget that latitude is defined relative to an ellipsoidal surface. A height above, or below, the ellipsoid (HAE) is just offset along that perpendicular-to-surface line.

If you were working with level surfaces instead, the perpendicular to that surface could change as height changed--because the point is now on a different level surface. That difference between the normal to the gravity/level surface and an ellipsoidal surface is called the deflection of the vertical.

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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