Would a space elevator theoretically be possible on a tidally-locked planet?

Question

It's my understanding that many of the theorized methods of creating a space elevator rely on the upward centrifugal force of earth's rotation on a counterweight high above to keep the structure taut and counter the planet's gravity. So, is there any possibility that a space elevator could be at all plausible on a planet with similar mass and size of earth that's tidally-locked to a dwarf star? If so, where would it need to located on the planet (disregarding any potential planet-side environmental factors the material would be exposed to, such as extreme heat or cold)?

Answer 1

Let's start with the naive estimate where we neglect the perturbations of the space elevator by the gravity of the star. The radius of the geostationary orbit is given by, \begin{equation} r=\sqrt[3]{\frac{Gm}{\omega^2}} \end{equation} where $m$ is the planet mass and $\omega$ its rotation angular velocity. For the tidally locked planet $\omega$ coincides with the angular velocity of its orbital motion around the star. Assuming that the orbit is circular with radius $R$ and denoting the star mass as $M$ this angular velocity equals, \begin{equation} \omega^2=\frac{GM}{R^3} \end{equation} This yields, \begin{equation} r=R\sqrt[3]{\frac{m}{M}} \end{equation} Assuming that $m\ll M$ we get that $r\ll R$ i.e. the orbital elevator stays close to the planet. One may then naively expect that we indeed approximately describe it omitting the star gravity from our consideration.

However this is not true because we have to describe the motion on the timescales similar to the planet orbital period. Then even small differences in gravitational force acting on the planet and space elevator will lead to the significant differences in their trajectories around the star. Therefore we need to describe the space elevator motion taking into account both gravity of the planet and gravity of the star.

Then we remember that there are 5 points - the Lagrangian points, where objects remain stationary with respect to planet and the star. Taking the image from wikipedia

If the planet is tidally locked to the sun the Lagrangian points are also stationary from the point of view from the surface! Then we may build a space elevator connecting the tether to an asteroid located in one of the Lagrangian points. The most obvious choice are L1 or L2. However these Lagrangian points are unstable and our asteroid would tend to leave them thus requiring constant corrections of its orbit. From this point of view the stable points L4 and L5 seem more favourable. Of course they are much farther from the planet and this may compensate this advantage completely.

Answer 2

Forget the star.

Imagine a space elevator from Pluto to its moon, Charon, which is phase locked with it. Charon always shows the same face to Pluto, and a Charonian day is exactly equal to a Charonian month, is equal to a Plutonian day. This means that if you built a space elevator from the point on pluto closest to Charon's surface, you could go to Charon's surface, and both ends of the elevator would stay stationary with respect to each other.