Lagrange points are a mathematical consequence of the The Circular Restricted Three Body Problem (CR3BP or CRTBP); two massive bodies orbiting around their center of mass and a third massless "test particle" that responds to their gravity.

In addition to these there are halo orbits and Lissajous orbits associated with these points and a heck of a lot of other three-body periodic, closed orbits including rectilinear halo orbits of Artemis fame.

Mass concentrations or "mascons" were first dealt with orbital-mechanically after an object in lunar orbit crashed into the surface much sooner than expected. The lumpy gravity field of the Moon is caused by concentrations of enhanced mass below the lunar surfac.

Begin question

Instead of the CR3BP's $m_1$ and $m_2$ being mathematical points, they could be replaced by uniform density spheres (or even spheres of only radially-varying density) as Newton's shell theorem tells us.

As long as the third body's orbit doesn't intersect one of the surfaces, it wouldn't know the difference.

Suppose the large radius of say $m_1$ became so large that it enveloped the small radius of $m_2$. At this point it's a uniform density sphere with a small, higher density sphere just under the surface.

Question: Could a body with a very large mascon have Lagrange-like points and halo-like orbits in principe? If so, what are the contrainst on the problem? Could this happen at the Moon?

Q: Why do you say "Lagrange-like" and "halo-like"? A: The rotation rate of the planet is not necessarily equal to the orbital period of free $m_1$ and $m_2$ so this needs thought. Proper, classical Lagrange points won't apply to a rigid rotor with a different rotation rate than free-orbiting bodies.

Further reading:


2 Answers 2


Provided that "enveloped" means the radially-varying density spheres have their mass distribution additively combined, and the rate of rotation is equal to that of a two body system, the setup is identical to the CR3BP, for regions outside the bodies.

...which doesn't help much since $L_1$, $L_3$, $L_4$ and $L_5$ are now inside $m_1$.

But even $L_2$ doesn't look very promising, as the surface of $m_1$ is rotating above orbital velocity, quickly deforming the system into a Jacobi ellipsoid, violating the shell theorem.

So the equal rotation rate case does not seem useful.

For the rigid rotor case, objects with only a single mascon in the equator plane would at least have $L_2$ and $L_3$ "equivalents", being the points where the stationary orbit intersects the centre-mascon line.

The $L_4$ and $L_5$ equivalents exist too, but that requires some math!

L4 and L5 equivalents

$L_4$ and $L_5$ forces still work out with a slower rotation rate, the triangle merely gets stretched vertically and balance is preserved due to symmetry.

To prove this:

Let $h$ be the height of the isosceles triangle and $\omega$ be the angular velocity. Without loss of generality, we can also set the distance between $m_1$ and $m_2$ to 1, and also let the sum of the reduced masses equal 1.

Acceleration along both the x-axis and y-axis of the co-rotating frame then needs to equal 0 for some value of $h$.

$$\sum F_x = \frac{m_1}{h^2 + \frac{1}{4}} \cdot \frac{1}{2\sqrt{h^2 + \frac{1}{4}}} - \frac{m_2}{h^2 + \frac{1}{4}} \cdot \frac{1}{2\sqrt{h^2 + \frac{1}{4}}} - \omega^2 \sqrt{h^2 \ \frac{1}{4} + m_2^2 - m_2} \cdot \frac{\frac{1}{2} - m_2}{\sqrt{h^2 \ \frac{1}{4} + m_2^2 - m_2}} = 0$$

$$\frac{\frac{m_1 - m_2}{h^2 + \frac{1}{4}}}{2\sqrt{h^2 + \frac{1}{4}}} -\omega^2\left(\frac{1}{2} - m_2\right)= 0$$

$$\frac{1}{\omega^2} = \sqrt{\left(h^2 + \frac{1}{4}\right)^3}$$

And vertically:

$$\sum F_y = \frac{m_1}{h^2 + \frac{1}{4}} \cdot \frac{h}{\sqrt{h^2 + \frac{1}{4}}} + \frac{m_2}{h^2 + \frac{1}{4}} \cdot \frac{h}{\sqrt{h^2 + \frac{1}{4}}} - \omega^2 \sqrt{h^2 \ \frac{1}{4} + m_2^2 - m_2} \cdot \frac{h}{\sqrt{h^2 \ \frac{1}{4} + m_2^2 - m_2}} = 0$$

$$\frac{\frac{h}{h^2 + \frac{1}{4}}}{\sqrt{h^2 + \frac{1}{4}}} -\omega^2h= 0$$

$$\frac{1}{\omega^2} = \sqrt{\left(h^2 + \frac{1}{4}\right)^3}$$

The math checks out!

$$h = \sqrt{\sqrt[3]{\omega^4} - \frac{1}{4}}$$

For $\omega = 1$ this yields the normal $L_4$ and $L_5$ points at $h = \frac{\sqrt{3}}{2}$, no solution exist if $\omega \gt 2\sqrt{2}$, and for rigid rotators $\omega \lt 1$ this results in an acute isosceles triangle with $h$ tending towards infinity as $\omega$ approaches zero.

But here we see why this (and the general case too!) wouldn't work for the Moon, as all such points by definition would have to be static relative to the surface: The Moon is tidally locked to the Earth, so these locations must coincide with the $EML_{1-5}$ points.

  • 1
    $\begingroup$ @uhoh I did the math, L4 and L5 equivalents confirmed. $\endgroup$ Commented Aug 18, 2021 at 11:59
  • $\begingroup$ just fyi I didn't get your notification because I haven't yet commented here. For this situation you have to leave a comment under the question. $\endgroup$
    – uhoh
    Commented Aug 18, 2021 at 14:16
  • $\begingroup$ So... It's late and I will look at the math tomorrow, did you show that Lagrange-like points can still exist for a range of $\omega$ other than 1? Cool! Will you then go on to show that some kind of Halo-like orbits can exist as well? $\endgroup$
    – uhoh
    Commented Aug 18, 2021 at 14:18

Your thought problem is interesting.

Refresher on Halo orbits

First, it's important to recall that the CR3BP allows one to rewrite the equations of motion in a non-inertial rotating frame. This is important because it also means that the Lagrange points themselves exist solely in this rotating frame. This superb visualization by Dr. Diane Davis shows at the 1min 08s mark what an NRHO looks like in a non-rotating frame, specifically in the Earth centered J2000 inertial frame:


Second, the CR3BP is an approximation of real world dynamics, it's a useful tool for preliminary work. Detailed analysis and operational work is done in an inertial frame when accounting for the mascons of the Earth and the Moon, the gravity of the Sun (and maybe even Jupiter), solar radiation pressure, thruster performance issues (ramp up/down of the burn), etc.

Your thought experiment

You would need the mascon to be large enough for the celestial body to have a valid approximation of two distinct masses. At this point, the problem becomes an astrophysics problem: can such a celestial object exist and, crucially, can it exist far enough away from other celestial bodies such that the gravity of those external bodies does not perturb the orbital motion too much (thereby breaking the CR3BP approximation).

I believe the answer here is most likely not. For weirdly shaped objects (asteroids, comets, small moons like Phobos), it typically is not possible to simply orbit just one such object. When solar systems are in their infancy, they have a bunch of angular momentum causing the accretion disks of planetoids to concentrate in a spheroid. Weirdly shaped objects will form later and won't be a primary celestial body, typically a smaller object in a cloud of others (again moons, asteroids, comets). Therefore, the CR3BP approximation won't be valid when accounting for the many other objects in similar orbits.

To dig further, I recommend reading the research by Dr. McMahon who leads the ORCCA lab at CU Boulder: https://www.colorado.edu/faculty/mcmahon/research . Lots of the lab's research is focused on mapping the mascons of asteroids while orbiting them (they worked on O-Rex a bunch).

  • $\begingroup$ I'm torn between the two current answers. Yours is certainly the "voice of reason" and best in terms of reality, but the other one addresses the "In principle?" part with so much rigor, including that the rotation period would not have to be the two-body orbital period that I couldn't help going for the more abstract answer. $\endgroup$
    – uhoh
    Commented Aug 23, 2021 at 1:02

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