When an imperfect sphere can be obtained by rotating an ellipse around an axis, it is called a spheroid. There are two types of spheroids, oblate ones and prolate ones.

Most solar bodies can be assimilated to an oblate spheroid, for the purpose of first approximation computations, such as the prediction of satellite movements orbiting that body. The Earth is one of them. On the other hand, some moons in the solar system have a prolate approximation.

When a planet is approximated by an oblate spheroid, its oblateness is characterized by a single coefficient often referred to as $J_2$. When $J_2$=0, the planet's shape is close to a sphere. High positive $J_2$ means that the eccentricity of the ellipse of revolution is high. For example the J2 of Mars is almost twice that of Earth, according to this NASA site.

Can $J_2$, and it alone, be used to characterize the shape of an prolate body? If yes, is there any concrete example of such a characterization?

This question was triggered by Any exact analytical solution for non-Keplerian orbits ..., where it can be read

These refer to Kelperian-like orbital parameterization of orbits around an oblate/prolate spheroid characterized by $J_2$.

At first reading, my interpretation is that the sentence implies that oblate and prolate spheroids differ only by the value of a single characteristic coefficient. In addition, in further comment threads, @uhoh seems to claim that the sign of $J_2$ is what makes the difference.

  • 1
    $\begingroup$ The $J_{n}$ terms are Legendre polynomial expansions. Yes, the term will have a negative sign if the body is prolate rather than oblate. So? $\endgroup$
    – Jon Custer
    Nov 5, 2021 at 19:07
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    $\begingroup$ @John Custer, I can imagine that it is trivial for those familiar with Legendre polynomials. However, I would think that a concrete example would help. If it is too much to ask, never mind. – $\endgroup$
    – Ng Ph
    Nov 5, 2021 at 20:45
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    $\begingroup$ +1 to help look for examples I've just asked in Astronomy SE List of J₂ for solar system bodies beyond the planets; are there any prolates? $\endgroup$
    – uhoh
    Nov 5, 2021 at 22:45
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    $\begingroup$ @uhoh, thanks, but concrete examples are just a way to be convinced quickly of the correct conclusion. The really nice thing is that someone could walk me (and perhaps some others) through the math, in a way that my head doesn't spin. $\endgroup$
    – Ng Ph
    Nov 5, 2021 at 23:04
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    $\begingroup$ Very roughly speaking, $J_2$ is a measure of the following: if one point is over the body's pole, and another point is over the equator at the same distance from the body's center (not surface!), how much is the first point "uphill" compared to the second point w.r.t. the body's gravity? For an oblate body, the first point is "uphill" (roughly speaking, because it's further from the body), so $J_2$ is positive; for a prolate body, the first point is "downhill", so $J_2$ is negative. $\endgroup$
    – Litho
    Nov 6, 2021 at 16:06

2 Answers 2


This is an expansion of my comment.

Very roughly speaking, $J_2$ is a measure of the following: if one point is over the body's pole, and another point is over the equator at the same distance $R$ from the body's center (not surface!), how much is the first point "uphill" compared to the second point w.r.t. the body's gravity?

Let us first consider the case of two equal point masses. At any point which is not on the same line with them and not on their plane of symmetry, the total gravity due to those masses does not point at their midpoint, but deviates somewhat towards the near mass (see picture 1 in the attached image: $A$ and $B$ are the masses; $O$ is their midpoint; $X$ is the point where we measure the gravity; sorry for the poor quality). For the sum of the gravity accelerations $\mathbf{g}_A$ and $\mathbf{g}_B$ to point to $O$, the ratio of their magnitudes $\frac{g_A}{g_B}$ would have to be the same as $\frac{|XA|}{|XB|}$ (pic. 2), i.e., $g_A$ would have to be smaller than $g_B$. But $g_A$ is in fact greater than $g_B$, so the sum $\mathbf{g}_A+\mathbf{g}_B$ points toward some point on the segment $OA$.

It means that if you move along a quarter-circle with the center at $O$ from a point above the "equator" to a point above the "pole" (pic. 3), the angle between the direction of your movement and the direction of local gravity is always (except at the endpoints) smaller than $90^\circ$, i.e., you're moving "downhill". It means that $J_2$ is negative in this case.

Now, if we consider a flat disc, we can divide it into pairs of small sections located symmetrially w.r.t. to the center (pic. 4). For any two such sections, the sum of the gravity accelerations due to them does not point to the center of the disc, but deviates toward the near section. As a result, the total gravity of the disc points to a point inside the near half-disc. So if you again move along a quarter-circle with the center at the center of the disc from a point above the "equator" to a point above the "pole" (pic. 5), the angle between the direction of your movement and the direction of local gravity is greater than $90^\circ$, i.e., you're moving "uphill". So $J_2$ is positive in this case.

enter image description here

  • $\begingroup$ I had some difficulty understanding at the beginning. But, once I have been able to digest the key fact, it is eye-openning! If we move perpendicular to the vector $g$, we remain on the same potential. If we move at <90° wrt to $g$, it's "downhill". And if the angle is >90°, it's "uphill". With that reasoning, I think we can convince anybody without resorting to sophisticated math (just take him climbing a hill). The rest is a little bit of geometry. $\endgroup$
    – Ng Ph
    Nov 8, 2021 at 14:22

For simplicity of calculation let's simulate a prolate body as a central potential plus two smaller masses north and south of center

And let's simulate an oblate body as a central potential plus an equatorial ring of point masses.

If we set $G=1$ for convenience and calculate the gravitation potential $GM/r$ we can see that it peaks at the equator for the oblate case and has a minimum at the equator for the prolate case.

These wiggles are not pure sinusoids, but we can see that their behaviors are mostly like $\sin^2(\theta)$ and $-\sin^2(\theta)$ and since $J_2$ is the coefficient in front of such a term (from Geopotential model between Equations 9 and 10):

$$u_{J2} = J_2 \frac{1}{r^3}\frac{1}{2} \left( 3 \sin^2(\theta)-1 \right)$$

we can see that for an oblate body $J_2$ will be positive and for a prolate body it will be negative.

enter image description here

enter image description here

import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D 

def phi_central(x, m):
    """central potential"""
    x0 = np.zeros(3)
    r = np.sqrt(((x - x0)**2).sum(axis=-1))
    return m / r, x0

def phi_polar(x, m, h):
    """quadrupole potential (axial)"""
    zhat = np.array([0, 0, 1])
    x1, x2 = h * zhat, -h * zhat
    r1 = np.sqrt(((x - x1)**2).sum(axis=-1)) # top
    r2 = np.sqrt(((x - x2)**2).sum(axis=-1)) # bottom
    return m * (1/r1 + 1/r2), x1, x2
def phi_equatorial(x, m, r, N=100):
    """quadrupole potential (equatorial (xy))"""
    theta = np.linspace(0, 2*np.pi, N+1)[:-1]
    # ring in the xy plane
    ring = np.vstack([r * f(theta) for f in (np.cos, np.sin, np.zeros_like)])
    r = np.sqrt(((x[..., None] - ring)**2).sum(axis=-2)) # equator
    return m / r.mean(axis=-1), r, ring

mcen, mpol, meq = 1, 0.1, 0.3
hpol, req = 0.3, 0.9

theta = np.linspace(0, 2*np.pi, 361)
R = 1.0
N = 1000

xz_plane = np.stack([R * f(theta) for f in (np.sin, np.zeros_like, np.cos)], axis=-1)

phi_c, xcen = phi_central(xz_plane, mcen)

phi_p, xp1, xp2 = phi_polar(xz_plane, mpol, hpol)

phi_e, r, ring = phi_equatorial(xz_plane, meq, req, N=N)
fig, ax = plt.subplots(1, 1)
ax.plot(np.degrees(theta), phi_c)
ax.plot(np.degrees(theta), phi_p)
ax.plot(np.degrees(theta), phi_e)
ax.set_ylim(0, 1.1)
ax.set_xlabel('theta (deg)')
ax.set_ylabel('gravitational potential')

fig = plt.figure()
ax  = fig.add_subplot(1, 1, 1, projection='3d', proj_type = 'ortho')
ax.plot([0], [0], [0], 'ok')
x, y, z = zip(xp1, xp2)
ax.plot(x, y, z, 'or')
x, y, z = ring
ax.plot(x, y, z, '-b')
x, y, z = xz_plane.T
ax.plot(x, y, z, '-g')
  • $\begingroup$ This looks really like a good "experimental" exercise. Unfortunately (for me), I have to install Python (and learn how to use it). One think that still bugs me though, if we take an oblate body with a certain $J_2$, then flip the sign of the coefficient while retaining the magnitude, which prolate body results from that operation? $\endgroup$
    – Ng Ph
    Nov 7, 2021 at 19:24
  • $\begingroup$ @NgPh there's an infinite number of shapes that would have that $J2$, even if you specify that the mass doesn't change. There could be a mass concentration near one (or both) poles of a spherical planet, it could be prolate, it could be a binary asteroid... some details of the shape would begin to emerge by looking at the next few dozen coefficients. $J_2$ has unique and special status only when the body is reasonably slowly rotating and in hydrostatic equilibrium in which means that $J_2$ dominates. This is how people encounter it so they start associating $J_2$ with oblateness. $\endgroup$
    – uhoh
    Nov 7, 2021 at 21:27
  • $\begingroup$ @NgPh for more about that see Why is Nodal precession affected by the rotational period of the planet? $\endgroup$
    – uhoh
    Nov 7, 2021 at 21:33

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