Lagrange point diagram

From this answer:

To obtain the distance to L1, find the smallest value of $r$ such that

$$\frac{M_2}{r_1^2} + \frac{M_1}{R^2} - \frac{r_1(M_1+M_2)}{R^3} - \frac{M_1}{(R-r_1)^2} = 0.$$

To obtain the distance to L2, find the smallest value of $r$ such that

$$\frac{M_1}{R^2} + \frac{r_2(M_1+M_2)}{R^3} - \frac{M_1}{(R+r_2)^2} - \frac{M_2}{r_2^2} = 0.$$

Above is how to calculate distances from $M_2$ to the $L_1$ and $L_2$ points. These solutions represent the balance between gravitational and and centripetal forces in the co-rotating frame.

Now suppose that the third body experiences a reduced inverse square force from $M_1$ by a factor $\delta$ which might be the case if it felt radiation pressure from the Sun. The force from $M_2$ would be unchanged but from $M_1$ it would be scaled by a factor $1-\delta$.


  1. Can it be shown (rather than just stated) that the Lagrange points still exist and behave the same way, but be in a different location?
  2. If they would, what is the equation that would need to be solved for the new $r_1$ and $r_2$ for a given $\delta$?

"bonus points:" can it be shown (rather than just stated) that halo orbits would still exist and behave in a similar way for non-zero $\delta$?

  • $\begingroup$ Lagrange points are a mathematical abstraction that happens to correspond to points in space that have certain useful gravitational properties. But these locations in space may be not practically usable due to a slew of other reasons. Some of these reasons will cause another location in space to behave like the Lagrange point was supposed to. But that location isn't actually the Lagrange point. The Lagrange points exist always, even if they happen to be under the planet surface. $\endgroup$
    – SF.
    Commented Jan 22, 2020 at 9:29
  • $\begingroup$ @SF. so we'll call them Uhoh points if necessary, but I really don't think Lagrange would mind a small change in the forces used in an otherwise identical derivation. $\endgroup$
    – uhoh
    Commented Jan 22, 2020 at 9:56
  • $\begingroup$ Call them force equilibrium points and I don't think anybody will mind. $\endgroup$
    – SF.
    Commented Jan 22, 2020 at 10:11
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    $\begingroup$ @SF. perhaps we can still call them libration points since I've only asked about 1 and 2, that way we can keep the "L" $\endgroup$
    – uhoh
    Commented Jan 22, 2020 at 10:15
  • 1
    $\begingroup$ Eh, this is stackexchange, so I edited both. Take a second look, and do a rollback if I'm completely out of my mind. $\endgroup$ Commented Jan 22, 2020 at 12:28

1 Answer 1


1) and 2) are easy to show, the bonus is very hard and I will not attempt it.

A $L$iberation point can be seen as a balance between three accelerations in a rotating frame of reference.

  1. Gravity from $M_1$
  2. Gravity from $M_2$
  3. Centrifugal acceleration.

For $L_2$, the first two are $-\frac{(1 - \delta)M_1}{(R + r_2)^2}$ and $-\frac{M_2}{r_"^2}$ respectively. Your $\delta$ included.

The third acceleration would be $\omega^2r_{centre}$, where $\omega^2 = \frac{\mu}{R^3}$ and $r_{centre} = \frac{RM_1}{\mu} + r_2$

We then have:

$$-\frac{M_2}{r_2^2} - \frac{(1 - \delta)M_1}{(R + r_2)^2} + \frac{M_1 + M_2}{R^3} \left(r_2 + \frac{RM_1}{M_1 + M_2}\right) = 0$$

Which simplifies to:

$$\frac{M_1}{R^2} + \frac{r_2(M_1 + M_2)}{R^3} - \frac{(1 - \delta)M_1}{(R + r_2)^2} -\frac{M_2}{r_2^2}= 0$$

Which unmistakeably resembles your second formula.

For completeness sake, here are $L_1$:

$$\frac{M_1}{R^2} + \frac{r_1(M_1 + M_2)}{R^3} - \frac{(1 - \delta)M_1}{(R + r_1)^2} +\frac{M_2}{r_1^2}= 0$$

And $L_3$:

$$-\frac{M_1}{R^2} - \frac{r_3(M_1 + M_2)}{R^3} + \frac{(1 - \delta)M_1}{(R + r_3)^2} +\frac{M_2}{r_3^2}= 0$$

This derivation should answer 2). But does it exist?

A considerably simpler argument can be used for that.

Say we are moving $L_2$ inwards towards the second body:

  1. The gravity from $M_1$ grows, but only towards the fixed value at the distance of the second body.
  2. The gravity from $M_2$ grows, and it quickly tends towards infinity as the $L_2$ approaches the point mass
  3. The centrifugal acceleration decreases.

It follows that any increase in the acceleration from $M_1$ can be countered with the arbitrary high value for the combination of the two other accelerations.

The same argument can be made for moving away from the second body. The centrifugal acceleration grows linearly arbitrary high, while the countering gravity shrinks with the square of the distance until the equation reaches equilibrium.

$L_2$ does always exist

The same, however, is not true for $L_1$. While an increase in acceleration from $M_1$ can be countered by moving $L_1$ arbitrarily close the the second body, a decrease in acceleration beyond $1 - \delta = 0$ will result in all acceleration being in the same direction. In fact, one would have to be on the opposite side of the central body, in which case $L_2 \equiv L_3$

  • 1
    $\begingroup$ @uhoh I think you will like this plot of where in a two-body system a solar sail can keep hovering: i.sstatic.net/ECF0V.png $\endgroup$ Commented Jan 22, 2020 at 21:17
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    $\begingroup$ Stable solar sail regions with delta 0 to 0.1: i.sstatic.net/HbPX3.png $\endgroup$ Commented Jan 22, 2020 at 21:22
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    $\begingroup$ An animation: i.fiery.me/sWMiL.webp $\endgroup$ Commented Jan 22, 2020 at 22:38
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    $\begingroup$ I'll be needing some coffee to process these; I had plans for today, now it looks like I'll be doing this, thanks a lot! ;-) $\endgroup$
    – uhoh
    Commented Jan 23, 2020 at 0:24
  • 1
    $\begingroup$ @uhoh All of them are for remaining fixed in the rotating frame. The regions with very high variations of delta are caused by the very bad working angle required to stay in place. The required acceleration vector is almost perpendicular to the Sun, due to the planet's gravity. Hence, the solar sail is nearly pointing it's edge towards the sun, getting very little acceleration. $\endgroup$ Commented Feb 6, 2020 at 23:10

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