Do Lagrange points still exist if there is significant radiation pressure on the third body from the first?

Question

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 $1-\delta$.

Questions:

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 $\delta$?

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 $L_2 \equiv L_3$