Where does it's gravitational force come from to maintain its constant somewhat spherical shape?

Question

How dose a Tidally locked gas giant have a balanced gravitational force to keep it's spherical shape when one side is constantly facing the sun? Where does it's gravitational force come from to maintain its constant somewhat spherical shape?

Answer 1

First off, a planet has a spherical shape because of its own gravity. This answer perhaps has too much detail on the shapes of planets - but I'm trying to give a correct understanding of what effects planets' shapes. The answer, in the end, is that because of the tidal effect of the sun, the planet will be very slightly wider in the direction facing toward the sun.

Let's start with the right framework - the surface of a planet is an approximate "equipotential" - meaning that the the gravitational potential energy of something sitting on the equator is the same as the gravitational potential of something sitting on the poles. This ensures that "downhill" isn't in any particular direction - if the poles were "downhill" of the equator, mass would move from the equator to the poles to fix that (especially in a gas giant, where mass is free to move in any direction)

With nothing else around, a body will be spherical, and the potential at the surface will be $-GM/R$. If it spins, we add our first most significant correction, the centrifugal potential $-\omega^2r^2\cos^2(\theta)/2$, where $\theta$ is the longitude and $\omega$ is the angular velocity of the rotation. This (small) perturbation makes the planet slightly wider at the equator than at the poles. To see how, let $R$ be the radius of the planet at the poles, then the potential of the surface is $-GM/R$. The potential of the gravity of the planet is still $-GM/r$, where $r(\theta)$ will be the radius of the planet as a function of longitude. Then we need to find $r(\theta)$ such that $$-GM/r-\omega^2r^2\cos^2(\theta)/2=GM/R$$ Assuming $r=R+\delta(\theta)$ (the correction is small) we get $$GM\delta/R^2-\omega^2r^2\cos^2(\theta)/2=0$$ $$GM\delta/R^2-\omega^2(R^2+2R\delta)\cos^2(\theta)/2=0$$ $$\delta=\frac{\omega^2R^2\cos^2(\theta)}{2(GM/R^2-2\omega^2R\cos^2(\theta))}$$ NOTE: this calculation is off by a factor of two due to the effect described here Why is the Earth so fat?

Finally we get to tidal effects, which are even smaller than rotational effects - a few meters on earth as opposed to 20km. It's worth keeping in mind that your gas giant is orbiting the sun - the main effect of the gravity between the sun and the gas giant is just to make it orbit. Tidal effects come from the gravity from the sun being slightly stronger on the near side and slightly weaker on the far side. The effective potential from this effect is the difference between the gravity of the sun in the center of the planet and the gravity of the sun in other places - so (with $d$ the distance between the sun and some spot on the planet; $D$ being the distance between the sun and the center of the planet; $\theta$ being the angle formed by a point on the surface, the center of the planet, and the sun) $$ \frac{GM_s}{d}-\frac{GM_s}{D}=\frac{GM_s}{\sqrt{(D-R\cos(\theta))^2+(R\sin\theta)^2}}-\frac{GM_s}{D} $$ I won't work out all the details again, but suffice to say this perturbation is smaller than the one before because $GM_s/D^3$ is smaller than $\omega^2$ (on earth the ratio of the two is 8e-5). Essentially the tidal bulge of this hypothetical planet will work the same way as the tidal bulge on earth - the earth is slightly wider in the direction toward/away from the sun and slightly thinner in the perpindicular directions (the poles, and the sunset side). Small caveat, on earth the moon does a bigger tidal effect than the sun by a factor of 2ish. The only difference will be that the tidal bulge of your tidally locked gas giant will be constant - not moving, as it does on earth (giving you the experience of water coming toward and moving away from your coast). So in the direction facing the sun, the planet will be permanently a tiny bit wider.

PS are there tidally locked gas giants? Not in our solar system.