2014-10-17 17:40:09 8 Comments

Many sources state that the Earth's gravity is stronger at the poles than the equator for two reasons:

- The centrifugal "force" cancels out the gravitational force minimally, more so at the equator than at the poles.
- The poles are closer to the center due to the equatorial bulge, and thus have a stronger gravitational field.

I understood the first point, but not the second one. Shouldn't the gravitational force at the equator be greater as there is more mass pulling the body perpendicular to the tangent (since there is more mass aligned along this axis)?

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## 4 comments

## @Thaina 2016-04-26 04:54:43

The point is if all effect was taken into account. Math would be summed up that effect of more mass under your feet still less than effect of distance from the center of mass

Another view is. At equator there are bulge near you. But from all other side of earth the bulge is far from you. Compare to the pole that all bulge is equally far from you, that account the difference

## @David Hammen 2014-10-18 15:20:52

TL;DR version: There are three reasons. In order of magnitude,

The poles are closer to the center of the Earth due to the equatorial bulge. This strengthens gravitation at the poles and weakens it at the equator.

The equatorial bulge modifies how the Earth the gravitates. This weakens gravitation at the poles and strengthens it at the equator.

The Earth is rotating, so an Earth-bound observer sees a centrifugal force. This has no effect at the poles and weakens gravitation at the equator.

Let's see how the two explanations in the question compare to observation. The following table compares what a spherical gravity model less centrifugal acceleration predicts for gravitational acceleration at sea level at the equator ($g_{\text{eq}}$) and the north pole ($g_{\text{p}}$) versus the values computed using the well-established Somigliana gravity formula $g = g_{\text{eq}}(1+\kappa \sin^2\lambda)/\sqrt{1-e^2\sin^2 \lambda}$.

$\begin{matrix} \text{Quantity} & GM/r^2 & r\omega^2 & \text{Total} & \text{Somigliana} & \text{Error} \\ g_\text{eq} & 9.79828 & -0.03392 & 9.76436 & 9.78033 & -0.01596 \\ g_\text{p} & 9.86431 & 0 & 9.86431 & 9.83219 & \phantom{-}0.03213 \\ g_\text{p} - g_\text{eq} & 0.06604 & \phantom{-}0.03392 & 0.09995 & 0.05186 & \phantom{-}0.04809 \end{matrix}$

This simple model works in a qualitative sense. It shows that gravitation at the north pole is higher than at the equator. Quantitatively, this simple model is not very good. It considerably overstates the difference between gravitation at the north pole versus the equator, almost by a factor of two.

The problem is that this simple model does not account for the gravitational influence of the equatorial bulge. A simple way to think of that bulge is that it adds positive mass at the equator but adds negative mass at the poles, for a zero net change in mass. The negative mass at the pole will reduce gravitation in the vicinity of the pole, while the positive mass at the equator will increase equatorial gravitation. That's exactly what the doctor ordered.

Mathematically, what that moving around of masses does is to create a quadrupole moment in the Earth's gravity field. Without going into the details of spherical harmonics, this adds a term equal to $3 J_2 \frac {GMa^2}{r^4}\left(\frac 3 2 \cos^2 \lambda - 1\right)$ to the gravitational force, where $\lambda$ is the geocentric latitude and $J_2$ is the Earth's second dynamic form. Adding this quadrupole term to the above table yields the following:

$\begin{matrix} \text{Quantity} & GM/r^2 & r\omega^2 & J_2\,\text{term} & \text{Total} & \text{Somigliana} & \text{Error} \\ g_\text{eq} & 9.79828 & -0.03392 & \phantom{-}0.01591 & 9.78027 & 9.78033 & -0.00005 \\ g_\text{p} & 9.86431 & 0 & -0.03225 & 9.83206 & 9.83219 & -0.00013 \\ g_\text{p} - g_\text{eq} & 0.06604 & \phantom{-}0.03392 & -0.04817 & 0.05179 & 0.05186 & -0.00007 \end{matrix}$

This simple addition of the quadrupole now makes for a very nice match.

The numbers I used in the above:

$\mu_E = 398600.0982\,\text{km}^3/\text{s}^2$, the Earth's gravitational parameter less the atmospheric contribution.

$R_\text{eq} = 6378.13672\,\text{km}$, the Earth's equatorial radius (mean tide value).

$1/f = 298.25231$, the Earth's flattening (mean tide value).

$\omega = 7.292115855 \times 10^{-5}\,\text{rad}/\text{s}$, the Earth's rotation rate.

$J_2 = 0.0010826359$, the Earth's second dynamic form factor.

$g_{\text{eq}} = 9.7803267714\,\text{m}/\text{s}^2$, gravitation at sea level at the equator.

$\kappa = 0.00193185138639$, which reflects the observed difference between gravitation at the equator versus the poles.

$e^2 = 0.00669437999013$, the square of the eccentricity of the figure of the Earth.

These values are mostly from Groten, "Fundamental parameters and current (2004) best estimates of the parameters of common relevance to astronomy, geodesy, and geodynamics."

Journal of Geodesy, 77:10-11 724-797 (2004), with the standard gravitational parameter modified to exclude the mass of the atmosphere. The Earth's atmosphere has a gravitational effect on the Moon and on satellites, but not so much on people standing on the surface of the Earth.## @Peter Mortensen 2017-05-07 07:12:36

Re

"The poles are closer to the center of the Earth due to the equatorial bulge. This strengthens gravitation at the poles and weakens it at the equator.": This wouldbe true if the Earth had a uniform mass distribution.not## @David Hammen 2017-05-07 14:26:59

@PeterMortensen - That is incorrect. Even if the Earth had a uniform density, gravitational acceleration at the pole would be greater than that at the equator by a factor of about $1+\frac 1 5 f$, where $f$ is the flattening factor. See Distribution of Gravitational Force on a

non-rotatingoblate spheroid.## @uhoh 2017-09-22 11:42:12

It's really helpful to have all of this in one place; I never really realized the gravity of the situation until going through all of it at once.

## @Qmechanic 2014-10-17 18:15:51

The point is that if we approximate Earth with an oblate ellipsoid, then the surface of Earth is an equipotential surface,$^1$ see e.g. this Phys.SE post.

Now, because the polar radius is smaller than the equatorial radius, the density of equipotential surfaces at the poles must be bigger than at the equator.

Or equivalently, the field strength$^2$ $g$ at the poles must be bigger than at the equator.

--

$^1$ Note that the potential here refers to the combined effect of gravitational and centrifugal forces. If we pour a bit of water on an equipotential surface, there would not be a preferred flow direction.

$^2$ Similarly, the field strength, known as

little$g$, refers to the combined effect of gravitational and centrifugal forces, even if $g$ is often (casually and somewhat misleading) referred to as thegravitationalconstant on the surface of Earth.## @Floris 2014-10-17 18:18:11

Does the argument "you are closer to the center of mass" work?

## @user4552 2014-10-17 18:25:33

Nice. Although the answer never uses the term "centrifugal force," that's implicit in the argument, because the equipotential is an equipotential in the rotating frame.

## @David Hammen 2014-10-17 19:13:00

@Floris - The argument that "you are closer to the center of mass" kinda-sort works, where kinda-sorta means about 3/2 (as opposed to one) in this case. About 2/3 of the reduction at the equator is attributable to the equator being 21 km further from the center of the Earth. The other 1/3 is directly due to centrifugal force (and of course that first 2/3 is indirectly due to centrifugal force).

## @Floris 2014-10-17 19:30:40

@DavidHammen - I guess that in my books "gravity" is just the attraction between two massive objects; the force experienced by a mass on the surface of the earth is modulated both distance and rotation, but only the former is "gravity" in my books. Further since OP stated he understood the rotation part, I was really suggesting to focus on the simplest way to state the second part.

## @David Hammen 2014-10-17 19:46:23

I think Lubos long ago wrote an answer that somewhat explains why gravitational due to the equatorial bulge is different than one would naively think. I'll see if I can dig up that answer.

## @David Hammen 2014-10-17 19:48:43

And here it is: physics.stackexchange.com/a/8088/52112 .

## @David Hammen 2014-10-17 19:54:59

@Floris -- Your book disagrees with the books used by geologists and geophysicists. As far as they are concerned, "gravity" includes both the inward gravitational force and the outward centrifugal force due to the Earth's rotation. The canonical value, 9.80665 m/s$^2$ is roughly how much a ball in vacuum appears to accelerate by an Earth-fixed observer at the latitude of Paris.

## @Floris 2014-10-17 20:09:43

As you know I am not a geophysicist; but it doesn't surprise me at all that they should use a "practical" definition. As for the canonical value, I assume it is "averaged over the tidal variations of moon and sun"? Thanks for the link to Lubos's answer; the link from there to Newton led to some fun reading. I loved the line (Newton discussing a dubious observation by Couplet): "But this gentleman's observations are so gross that we cannot confide in them". Just great.

## @Qmechanic 2014-10-17 20:17:20

@David Hammen: For the record: The digging was unnecessary: I already linked to that Phys.SE post in version 1.

## @user4552 2014-10-17 19:17:57

Here's a simple argument that doesn't require any knowledge of fancy stuff like equipotentials or rotating frames of reference. Imagine that we could gradually spin the earth faster and faster. Eventually it would fly apart. At the moment when it started to fly apart, what would be happening would be that the portions of the earth at the equator would at orbital velocity. When you're in orbit, you experience apparent weightlessness, just like the astronauts on the space station.

So at a point on the equator, the apparent acceleration of gravity $g$ (i.e., what you measure in a laboratory fixed to the earth's surface) goes down to zero when the earth spins fast enough. By interpolation, we expect that the effect of the actual spin should be to decrease $g$ at the equator, relative to the value it would have if the earth didn't spin.

Note that this argument automatically takes into account the distortion of the earth away from sphericity. The oblate shape is just part of the interpolation between sphericity and break-up.

It's different at the poles. No matter how fast you spin the earth, a portion of the earth at the north pole will never be in orbit. The value of $g$ will change because of the change in the earth's shape, but that effect must be relatively weak, because it can never lead to break-up.