@OrganicMarble's answer to 1959 Peanuts cartoon about the Fischer ellipsoid (Earth is “pear-shaped”)? finally let me see the cartoon that Mathematician, Engineer and Geoscientist Irene Fischer wrote about in Geodesy? What's That?: My Personal Involvement in the Age-Old Quest for the Size and Shape of the Earth. In it there is a passage that mentions it in relation to the shape of the Earth as deduced from radio signals received from Sputnik-1 and other early spacecraft.

I remember reading about Earth's "pear-shapedness" in school books and never understanding it because the Earth is pretty much described by an oblate spheroid, a shape which is symmetric about the equator.

The disconnect is that it's been known since Newton (or before?) that the Earth is mostly an oblate spheroid (it couldn't not be) but after you subtract that off, what's left looks a little bit like a pear, appearently (sic).


In the essay The Relativity of Wrong The Skeptical Inquirer, Fall 1989, Vol. 14, No. 1, Pp. 35-44 Isaac Asimov eloquently as always explains:

Even the oblate-spheroidal notion of the earth is wrong, strictly speaking. In 1958, when the satellite Vanguard I was put into orbit about the earth, it was able to measure the local gravitational pull of the earth--and therefore its shape--with unprecedented precision. It turned out that the equatorial bulge south of the equator was slightly bulgier than the bulge north of the equator, and that the South Pole sea level was slightly nearer the center of the earth than the North Pole sea level was.

There seemed no other way of describing this than by saying the earth was pear-shaped, and at once many people decided that the earth was nothing like a sphere but was shaped like a Bartlett pear dangling in space. Actually, the pear-like deviation from oblate-spheroid perfect was a matter of yards rather than miles, and the adjustment of curvature was in the millionths of an inch per mile.

Question: But what is the nature of this pear-shapeeness? Does it come from the shape of the Earth's surface, or is it really just a way of saying that one of the components of Earth's octapole moment known as J₃ is non-zero?

The activities of the famous International Geophysical Year involved both analyzing radio signals from Sputnik-1 and Explorer-1 and painstaking geographic surveys of Earth's surface that spanned large chucks of Earth. So I'm not sure which type of data lead to the fruitful epiphany of Earth's pear shape.

The International Geophysical Year (IGY; French: Année géophysique internationale) was an international scientific project that lasted from 1 July 1957 to 31 December 1958. It marked the end of a long period during the Cold War when scientific interchange between East and West had been seriously interrupted.

International Geophysical Year Source

Related elsewhere in SE:

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    $\begingroup$ I've anecdotally heard that part of it is due to the mass of Antarctica and its ice sheet $\endgroup$ Commented Jul 14, 2020 at 22:02
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    $\begingroup$ I recommend A Tapestry of Orbits by Desmond King-Hele if you're interested in reading more about the early days of satellite geodesy. $\endgroup$
    – Chris
    Commented Jul 16, 2020 at 18:45
  • $\begingroup$ @Chris Yes that is exactly the book I want to read now, thank you very much for the reference! From Wikipedia: Desmond King-Hele: "... is considered 'one of the pioneers of space geodesy'.[4] Based on satellite geodesy, King-Hele refined the estimate for Earth's pear shape, finding a 45 m difference between north and south polar radii.[5][6]" and reference 5 is Refining the Earth's Pear Shape! $\endgroup$
    – uhoh
    Commented Jul 17, 2020 at 0:41

1 Answer 1


In this answer dimensionless gravitational multipolar moments for the Earth are given. It is noted there that after the quadrupolar contributions $J_2, J_{2,2}$ (but not $J_{2,1}$) the next largest multipoles are $J_3, J_4, J_{3,1}$; all of the latter are similar in magnitude to each other and three orders of magnitude smaller than the quadrupolar terms but substantially greater than other higher-order contributions. The large-scale geometric features of Earth, such as mountain ranges, drive these higher-order components and we need multiple contributions to capture these features.

Of these identified principal components, only $J_3$ is indeed the main pear-shaped contributor. This is based on the features of the spherical harmonic functions that contribute to these gravitational potential terms:

  • $J_n$ where $n$ is even and $J_{n,m}$ where $n-m$ is even involve spherical harmonic functions are symmetric under reflection through the Equator, so have no pear-shaped component. This includes the $J_4$ and $J_{3,1}$.

  • $J_1$ vanishes in the gravitational potential (no gravitational dipole) and $J_{n,n-1}$ ($n\ge 2$) involve spherical harmonics that are symmetric with respect to reflection through the Equator plus a rotation. This symmetry again does not allow a pear shape.

  • That leaves $J_n$ with $n$ odd and $>2$, and $J_{n,m}$ with $n-m$ odd and $>2$. This includes $J_3$ as the lowest order multipole that generates a pear shape, and as one might expect this is also the largest such contribution. But the pear is lumpy because the non-pear-shaped $J_4$ and $J_{3,1}$ terms are similar in size. As mentioned earlier, this lumpiness comes about because the shape of the Earth is not a simple ellipsoid or a pear but a more complex form generated by its large-scale geography.

  • $\begingroup$ This is great news, thank you! I don't know what "difference between total nodes (3) and meridian nodes (0) which is odd and greater than 2" means exactly nor how it is confirmation of pear-shapedness; is it possible to elaborate a bit, since it seems to be central to your answer? Thanks! $\endgroup$
    – uhoh
    Commented Jul 16, 2020 at 10:13
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    $\begingroup$ It's from the theory of spherical harmonic functions. Gravitational potential fields of Earth and other hydrostatically equilibrated objects are decomposed into these functions, whose coefficients are then the moments we use. A spherical harmonic function has $n$ total nodal surfaces, out of those $m$ are planes that pass through the North and South Poles and thus lie along meridians. The other nodes are planes or cones with other orientations, and it is these nodes that mold a pear-shape or not. $\endgroup$ Commented Jul 16, 2020 at 10:20
  • $\begingroup$ "...and it is these nodes that mold a pear-shape or not" but How? Why? Where does this come from? What does it mean? I don't understand. $\endgroup$
    – uhoh
    Commented Jul 16, 2020 at 10:21
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    $\begingroup$ Take a look at this sketch. To make a pear shape a spherical harmonic must have positive values (red, yellow) differently shaped in the southern versus northern hemisphere, and out of the functions shown in the sketch only (3,0) does that. If we put more functions we would find the right property only for (n+3,m), (m+5,m), etc. Functions with an even difference are symmetric between the hemispheres and those with a difference of 1 have the southern hemisphere just a rotation of the northern one. $\endgroup$ Commented Jul 16, 2020 at 10:37

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