14
$\begingroup$

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

  1. The centrifugal "force" cancels out the gravity minimally, more so at the equator than at the poles.
  2. 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 gravity at the equator be greater as there is more mass pulling a body perpendicular to the tangent as more mass is aligned along this axis?

$\endgroup$
13
$\begingroup$

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 the gravitational constant on the surface of Earth.

$\endgroup$
  • 1
    $\begingroup$ Does the argument "you are closer to the center of mass" work? $\endgroup$ – Floris Oct 17 '14 at 18:18
  • 2
    $\begingroup$ 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. $\endgroup$ – Ben Crowell Oct 17 '14 at 18:25
  • 1
    $\begingroup$ @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). $\endgroup$ – David Hammen Oct 17 '14 at 19:13
  • $\begingroup$ @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. $\endgroup$ – Floris Oct 17 '14 at 19:30
  • $\begingroup$ 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. $\endgroup$ – David Hammen Oct 17 '14 at 19:46
10
$\begingroup$

Lots of places state that the Earth's gravity is stronger at the poles than the equator for two reasons:

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

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

  1. 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.

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

  3. 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.

$\endgroup$
  • $\begingroup$ 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 would not be true if the Earth had a uniform mass distribution. $\endgroup$ – Peter Mortensen May 7 '17 at 7:12
  • 1
    $\begingroup$ @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-rotating oblate spheroid. $\endgroup$ – David Hammen May 7 '17 at 14:26
  • 2
    $\begingroup$ 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. $\endgroup$ – uhoh Sep 22 '17 at 11:42
3
$\begingroup$

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.

$\endgroup$
-1
$\begingroup$

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

$\endgroup$

protected by Qmechanic May 13 '15 at 14:01

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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