Suppose a person is standing on a non-rotating$^1$ oblate spheroid of uniform density. He first stands on one of the poles, then on the equator. In which case is the gravitational force greater?

enter image description here

In Case 1, the distance between the centre of the person and the spheroid is lesser than in Case 2, but in Case 2, there is lesser 'distribution' of force, i.e., more force is concentrated in one direction.


By applying parallelogram of forces, Newton's inverse square law of gravity, and any other rule, in which case will the Gravitational force be greater? Or will they both be equal?


$^1$ The attentive reader may ponder how come the spheroid is oblate in the first place if it is non-rotating and there is hence no centrifugal force to stretch it? Well, it is a hypothetical question. Please accept the slightly unrealistic premise.

  • $\begingroup$ Comment to the question (v1): Is the oblate spheroid spinning? Note that for a planet-sized object an oblate form is typically due to spinning. Related: physics.stackexchange.com/q/141856/2451. $\endgroup$
    – Qmechanic
    Commented Nov 12, 2014 at 16:37
  • $\begingroup$ No. Earlier, the object in question was intended to be a planet (Earth). But I wanted to know what will the result be if it's a perfect oblate spheroid, and not spinning, unlike the planets. $\endgroup$
    – user49111
    Commented Nov 12, 2014 at 17:48
  • $\begingroup$ @imakesmalltalk - Gravity pulls large objects into a spheroidal shape. A non-spinning object will form a sphere rather than an oblate spheroid. The oblate spheroid shape of the Earth results precisely because it is spinning. The Sun and Venus both spin, but very very slowly. They are near-perfect spheres as a result. $\endgroup$ Commented Nov 15, 2014 at 18:24
  • $\begingroup$ @DavidHammen Could be an asteroid. Doesn't have to be large enough for gravity to pull it into a sphere $\endgroup$
    – Jim
    Commented Nov 17, 2014 at 15:13
  • $\begingroup$ An originally rotating planetoid could be made of molten metal that cools to an (almost) rigid oblate spheroid shape while rotating and then artificially the rotation is stopped. Hypothetically possible if the tensile strength of the metals strong enough ;-) $\endgroup$
    – KDP
    Commented Feb 23 at 23:47

8 Answers 8


Using the same notation, assumptions and approximations as my Phys.SE answer here, the gravitational potential (monopole + quadrupole) is

$$U(r) ~\approx~ U_1(r)+U_4(r) ~\approx~ -\frac{GM}{r}\left(1 + \frac{(3s^2-2)fR^2}{5r^2}\right).$$

Here $a$ and $b$ are the equatorial and polar radius of the oblate spheroid; $$0~<~f:=~1-\frac{b}{a} ~\ll~ 1$$ is a flatness parameter, which is assumed to be small for simplicity; $s\equiv \sin(\theta)$; $\theta\in[0,\pi]$ is the polar angle; $R\approx b(1+fs^2) $ is the surface profile.

By symmetry the gravitational acceleration $\vec{g}=\vec{\nabla} U(R)$ has only a radial component at the poles and equator, so it is enough to consider just the radial derivative

$$ \left. \frac{\partial U(r)}{\partial r}\right|_{r=R} ~=~\frac{GM}{R^2}\left(1 + \frac{3}{5}(3s^2-2)f\right) ~\approx~\frac{GM}{b^2}\left(1 - \frac{1}{5}(s^2+6)f\right).$$

We conclude that the polar gravitational acceleration

$$ g(\theta=0)~\approx~\frac{GM}{b^2}\left(1 - \frac{6}{5}f\right)$$

is slightly bigger than the equatorial gravitational acceleration

$$ g(\theta=\frac{\pi}{2})~\approx~\frac{GM}{b^2}\left(1 - \frac{7}{5}f\right) ~\approx~\frac{GM}{a^2}\left(1 + \frac{3}{5}f\right).$$

The ratio between the equatorial and polar gravitational acceleration is therefore less than one:

$$\frac{g(\theta=\frac{\pi}{2})}{g(\theta=0)}~\approx~1 - \frac{1}{5} f.$$

If we wrongly had only considered the monopole contribution, we would wrongly had gotten a gravitational acceleration ratio equal to the (inverse) square of the quotient of radii $$\frac{b^2}{a^2}~=~(1-f)^2~\approx~ 1 - 2 f. $$

tl;dr: The monopole contribution wins; however the quadrupole contribution cancels 90% of the effect coming from the change in radius, so it is a close call.

  • $\begingroup$ Another way to get that factor of $\frac 1 5 f$ is to use spherical harmonics which results in $g = \frac{GM}{r^2}\left(1-\frac 3 2 J_2\frac{a^2}{r^2}(3\cos^2\theta-1)\right) + O(r^{-6}$ , where $J_2$ is the second dynamic form. The factor of $\frac 1 5$ arises from the value of $J_2$ for a uniform density oblate spheroid, $J_2 = \frac 1 5 (1-b^2/a^2)$. $\endgroup$ Commented May 7, 2017 at 14:20
  • $\begingroup$ @David Hammen: Thx. $\endgroup$
    – Qmechanic
    Commented May 7, 2017 at 16:40

I was revisiting triple integrals, so I decided to give this a go. This is the result:


As expected, the pole wins.


Starting from Newton's law of universal gravitation: $$\mathrm d\mathbf g = G\frac{\mathbf r\ \mathrm dm}{\left|\mathbf r\right|^3}$$

And the parametrization of our oblate asteroid: $$(x,y,z) = \mathbf r(r,\theta,\phi)=\left[\alpha\cdot(r\sin\phi\cos\theta+1),\beta\cdot r\sin\phi\sin\theta,\gamma\cdot r\cos\phi\right]$$

This leaves the origin exactly on the equator or pole, depending on the choices of the semi-major axis lengths $\alpha$, $\beta$ and $\gamma$:

  • $\alpha=\beta>\gamma$ leaves us at the equator.
  • $\alpha<\beta=\gamma$ leaves us at the pole.

Note that the asteroid's centroid is in the x-axis:


(this with radii 0.7, 1.0, 1.0):

Change of variables gives: $$g_x=G\alpha\beta\gamma\rho\int_0^1\int_0^{2\pi}\int_0^\pi\frac{xr^2\sin\phi}{\left(x^2+y^2+z^2\right)^\frac{3}{2}}\mathrm d\phi\ \mathrm d\theta\ \mathrm dr$$

Note I left cartesian variables there, because I was going to do this numerically. Also, naturally $g_y=g_z=0$. Also, $\rho$ is density (I used it to test Earth).

I fired up Scilab and learned the int3d function. Results with Earth were fine, so I decided it was good enough for the plot above. For completeness, Earth data used was:

Equator radius: 6,378,136.6 m
Pole radius: 6,356,751.9 m (source)
Density: 5,520 kg/m³ // (source)
Gravitational constant: 6.67384e-11

Gravity on equator: 9.8289 m/s²
Gravity on pole: 9.8354 m/s²

I think this is good, considering on equator there's ~0.3% less gravity due to Earth's rotation, which yields closer to our usual 9.8 m/s². Also, Earth's almost a sphere, and the results were realistically close to each other I guess.

  • $\begingroup$ +1. The numerical plot seems to agree with the theoretical slope value $\to\frac{1}{5}$ when (minor axis)/(major axis) $\to 1$, cf. my answer. $\endgroup$
    – Qmechanic
    Commented Nov 14, 2014 at 22:22
  • $\begingroup$ @Qmechanic for info: a linear trendline would be of slope .2165 $\left(y=.2165x+.7864\right)$, giving $R^2=.9996$. I wasn't certain if I should convey that or quadratic. Do you think one or the other would be more appropriate? $\endgroup$ Commented Nov 15, 2014 at 0:16
  • 1
    $\begingroup$ @Floris there you go: pastebin.com/DuLEhune. I tried to comment it a little bit, but feel free to ask any questions. $\endgroup$ Commented Nov 15, 2014 at 0:47
  • 2
    $\begingroup$ The numerical slope @ 1 is roughly a fifth: $0.2323-2*0.0176=0.1971\approx\frac{1}{5}$. $\endgroup$
    – Qmechanic
    Commented Nov 15, 2014 at 18:13

This turns out to be a surprisingly hard thing to calculate precisely. The gravitational potential (you need to differentiate this to get the force) at the surface of an oblate spheroid is described on the Wolfram website, but a quick glance at this is likely to put you off this approach.

Anyhow, the answer is that the gravity is stronger at the pole than at the equator. You can sort of rationalise this by considering the following diagram:

Oblate spheroid

I've drawn the sphere that just fits within the oblate spheroid. If you ignore the parts of the planet outside the sphere then it's obvious that gravity is stronger at the pole because the gravitational acceleration is given by:

$$ a = \frac{GM}{r^2} \tag{1} $$

At the pole $r = b$, which is smaller than at the equator where $r = a$, so equation (1) tells us that just considering the sphere the acceleration is higher at the pole.

The next question is how much difference the bits of the planet outside the sphere make when you add their gravity in. This is where it all gets complicated, but the short answer is that while it does reduce the difference gravity remains stronger at the pole.

  • 4
    $\begingroup$ +1 Thanks for your answer John. I've heard this many times, that $F_G$ is stronger at the poles. Its no brainer that as $r$ is reduced, $F$ increases (as $F_G \propto \frac{1}{r^2}$). But what I wanted to know is that how exactly does gravity remains stronger at the pole? My intuition too says that its obvious. But how can it be shown? $\endgroup$
    – user49111
    Commented Nov 5, 2014 at 8:52
  • $\begingroup$ @imakesmalltalk: you could probably find a derivation somewhere in Googlespace. In principle it's simple as you just split the spheroid up into suitable elements and integrate. In practice you'd be surprised how often basically simple functions turn out to have horribly complicated integrals. That's what happens here. $\endgroup$ Commented Nov 5, 2014 at 9:32
  • $\begingroup$ The reason why celestial bodies are approximately oblate spheroids originates from their rotational velocity, which in their rotating reference frame induces a centripetal force. When you look at the effective surface acceleration you could also include this centripetal force, which would make it even more clear that you will have lower gravity at the equator. $\endgroup$
    – fibonatic
    Commented Nov 14, 2014 at 15:09

Gravitational potential plus the centrifugal potential due to the Earth's rotation at the Earth's surface is approximately given by $$\begin{aligned} U(r,\theta) = -\frac {\mu_e}{r} + \frac {\mu_eJ_2 a^2}{r^3}\frac{3\sin^2\theta-1}{2} - \frac 1 2 \omega^2r^2\cos^2\theta &\qquad&\qquad(1) \end{aligned}$$ where $\mu_e = GM_e$ is the Earth's gravitational parameter, $J_2$ is the Earth's second dynamic form factor, $a$ is the Earth's equatorial radius, $\theta$ is the geocentric latitude, and $r$ is the radial distance to the surface of the Earth at that latitude.

The first two terms in the above are the first two terms in the spherical harmonic expansion of the Earth's gravitational field. I've truncated the higher order terms in the spherical harmonic expansion. The last term is the centrifugal potential. This value would be constant over the Earth's surface if the Earth was in hydrostatic equilibrium. The Earth is not quite in hydrostatic equilibrium, but it is very close. The response to deviations from hydrostatic equilibrium? We call them earthquakes.

Hydrostatic equilibrium means the gravitational potential at the pole and equator should be equal to one another. These are given by $$\begin{aligned} U_{\text{pole}} &= -\frac {\mu_e}{c}\left(1-J_2 \left(\frac a c\right)^2\right) \\ U_{\text{equator}} &= -\frac {\mu_e}{a} \left( 1+\frac 1 2 J_2\right) - \frac 1 2 \omega^2 a^2 \end{aligned}$$ where $c$ is the Earth's polar radius. Using $c=a(1-f)$ where $f$ is the Earth's flattening and equating the above yields a relationship between rotation, size, shape, and distribution of mass: $$f\approx \frac {3 J_2} 2 - \frac {a^3\omega^2}{2\mu_e}$$ The Earth's rotation dictates the shape of the Earth.

Taking the gradient of equation (1) yields gravitational acceleration at the surface of the Earth. As the $\hat \theta$ component of the gradient vanishes at the equator and at the poles, I'll just look at the radial component of the gradient. $$\frac{\partial U}{\partial r} = \frac {\mu_e}{r^2}\left(1-3 J_2 \left(\frac a r\right)^2 \frac{3\sin^2\theta - 1} 2\right) - r\omega^2\cos^2\theta$$ At the pole and equator these become $$\begin{aligned} g_{\text{pole}} &= \frac {\mu_e}{c^2}\left(1-3J_2 \left(\frac a c\right)^2 \right) \\ g_{\text{equator}} &= \frac {\mu_e}{a^2}\left(1+\frac 3 2 J_2\right) - a\omega^2 \end{aligned}$$ Note that the quadrupole term (the term involving $J_2$) is negative at the pole, positive at the equator. One way of looking at that quadrupole term is that it represents the tidal bulge. Equatorial regions are close to the bulge. (They're right on top of it!) Polar regions are quite removed from it. The monopole term overstates gravitation at the poles while it understates it at the equator. The gravitational component of $g$ is still greater at the pole, by about 1.79 cm/s2. The centrifugal component of $g$ is non-existent at the poles and reduces $g$ at the equator even further, by about 3.39 cm/s2. The total makes gravitation about 5.18 cm/s2 greater at the pole than the equator.

Note that a satellite orbiting the Earth doesn't feel that centrifugal term. It only feels the gravitational component of the potential. If the satellite is in a circular orbit, the quadrupole ($J_2$) term will makes the satellite subject to reduced gravitational acceleration over polar regions than over the equator.


Consider that,

  1. The force of gravity falls off with distance (the inverse square law).
  2. Gravity is a property of the components, not the whole.

So it is the atoms of your oblate spheroid that are exerting gravity on the person (and these forces sum over the whole object) and one way to work this out is to calculate the force of gravity exerted by each atom on the person and sum them.

This is not very easy to calculate but we can approximate by making a few assumptions:

  1. Let's work in 2 dimensions as its much simpler to calculate than 3!
  2. That the mass of the spheroid is evenly distributed (e.g. no iron core, etc).
  3. That the object is not spinning.

So we don't know where the atoms are but we can simply subdivide the spheroid into equal regions and work with those as an approximation. Let's start with the simplest case...

Simply draw a box around the oblate spheroid and divide into 4 equal areas. If an area is more than half matter (black) then place a center point and work out the distance to the person. Calculate the force of gravity for this distance and sum all forces for each area (we're assuming each area has an equal mass) to find which case has the higher gravity.

Gravity is inversely proportional to the square of the distance so it falls off quickly the further away you get. So distance is key.

Since gravity is dependent on the distance we can approximate again by simply averaging the distances. The shortest average distance (to each of the areas) should have the higher gravity (as the person is closer to more mass). This is much easier to calculate and to picture intuitively.

This is a very rough approximation but you can divide the surrounding box into more and more areas to get a better approximation.

Finally this only works if all the matter is on one side of the person. E.g. the smallest average of all distances to the 'atoms' is actually if the person is in the center of the mass where gravity is zero (mass on either side starts to cancel out the force as you move into the body).

Even with just dividing into 4 areas you can see that the person is further away from the areas (matter) in case 2 than in case 1. I think this is intuitively why we know the right answer without knowing the reason why.


If the density of that oblate spheroid is constant then gravitational force is bigger in Case 1 because in that case the person stands closer to the center of the mass.

  • 4
    $\begingroup$ That argument ("closer to the center of mass") does not always yield the right answer. Take for example a double star. If you are close to the "center of gravity" (barycenter) of the pair, you feel very little gravitational attraction to either. You need to dig a little deeper for the right answer. $\endgroup$
    – Floris
    Commented Nov 14, 2014 at 23:10

For when the guy stands on the pole, slice the planet into circular disks, with their centers on the axis, each with radius = r = f(h) and thickness = dh, where h is distance of the center of the disk from one pole, and f(h) is function that represents shape of oblate spheroid. Then slice each disk into multiple rings, nested one within the other, each ring is of radius r which increases from 0 to the radius of the disk.

Now double Integrate (add) the gravitational contribution from each ring within each disk as h goes from 0 to the height of the planet (H), and r goes from 0 to the radius of the specific disk. The integral will be adding The gravitational contribution of one ring, which can be calculated simply since every part of the ring is the same distance from the lucky person standing on the pole.

For when the guy is on the equator, I don't see any obvious way to do this with only a double integration, so you will need a triple integral. so you might as well use the same scheme as above, but then divide each ring into many segments (like pie slices) of radial thickness dt, and integrate the Gravitational contribution of each tiny cubical piece which is dh by dr by dt

  • 1
    $\begingroup$ This is simply $Gm\int\int\int_\mathrm{planet}\frac{\mathbf r}{r^3}\ \mathrm dM$, right? I don't think this helps much: if the OP knew calculus, he'd have done this by himself. To avoid being downvoted, you could attempt the integration yourself and see what pitfalls are there. This would be more useful. $\endgroup$ Commented Nov 14, 2014 at 14:16
  • $\begingroup$ How can you do that without knowing function that describes shape of oblate spheroid ? ... or does the phrase "oblate spheroid" uniquely specify some shape ? $\endgroup$ Commented Feb 9, 2016 at 0:22
  • $\begingroup$ Sure it does: mathworld.wolfram.com/OblateSpheroid.html $\endgroup$ Commented Nov 25, 2017 at 6:02
  • $\begingroup$ @Andres, I assumed that any spheroid, generated by rotating any defined curve about an axis, could be described by the phrase "oblate spheroid". I was not aware that the phrase only applies to bodies generated by rotating an ellipse. Thanks. $\endgroup$ Commented Nov 25, 2017 at 16:08

Neglecting rotation, it seems to me that if one takes the case where the diameter at the equator is very much larger than that at the pole, we can consider it to be a disk.

Standing on the edge, all the mass is directly below you and all of it contributes to attracting you to the centre. Standing at the centre of the disk, the overwhelming bulk of the disks mass is at the periphery, and the gravitational effect you experience from it is only the component which points to the centre which will be tiny. Most of the effect will be at 90 degrees to this and balance out.

I would expect one to to be heavier at the equator than at the pole.


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