Would a planet's tidal forces really push away a moon out of its Hill Sphere?

Question

If a moon orbits in the same direction as its planet's rotation (i.e. a prograde orbit), and its orbital period is longer than its planet's rotational period, then tidal forces would cause the moon to slowly drift away. This is true for the vast majority of moons in the Solar System, and for every large round moon bar Triton.

The moon's orbit would continue expanding, and the planet's rotation would continue slowing down, until both bodies are tidally locked to each other. This is what's expected to happen to the Earth-Moon system if it is not engulfed by the Sun as a red giant.

However, what if the moon just can't tidally lock its primary? Whether that's because the planet is hundreds of thousands of times more massive than the moon, or because its rotation is too fast to be slowed down in time.

Would the planet's tidal forces seriously cause its moon's orbit to expand so much that it's outside its Hill Sphere? That seems counterintuitive to me. Tidal forces drop by the distance cubed, and as the moon got farther away, the tidal forces that caused it to drift away in the first place would get exponentially weaker.

I've heard the titular claim multiple times, but to me it seems more likely that a moon would asymptotically approach a distance within the Hill Sphere that it can't be pushed farther out of, as the tidal forces would become negligible.

Answer 1

You can think about this in terms of energy conservation. The orbiting moon has a certain amount of kinetic and potential energy. The spinning planet has a certain amount of kinetic energy in its rotation. The kinetic energy of the rotation will be transferred to the kinetic and potential energy of the moon, until such time as the orbit period of the moon equals the rotation time of the planet. This distance depends on the masses and rate of rotation of the bodies, It doesn't depend on the mass and distance to the star. (Here I've adopted a simple model in which the tides don't dissipate energy. That they do dissipate energy changes the values of the numerical calculation, but not the reasoning that the limiting distance isn't equal to the Hill sphere)

The Hill sphere depends on the mass of the planet, star and their distance. It doesn't depend on the rotation rate of the planet.

So you see that if there was no star, the moon would move further away until no more energy was being transferred. This distance might be larger than the radius of the Hill sphere. If that's the case then tidal forces will push the moon out of orbit.

There's nothing special about the Hill sphere for the matter of tides. Yes, tidal forces decrease rapidly with distance, but while the planet is spinning faster than the orbit of the moon, the moon will raise tides on the planet, the rotation of the planet will pull these tides ahead of the moon, and the moon will be accelerated, causing the moon to move further. There's no reason why the maximum distance the moon will be pushed to must be equal to or less than the Hill sphere's radius.

For the Earth-Moon system, Wikipedia gives an estimate that the maximum distance the moon could reach is when it has an orbital period of 47 days, corresponding to an orbital distance of 610000km, which is still well inside the Hill Sphere relative to the sun (1.5 million km). But by the time the moon reaches that distance, the sun will be long gone.

Answer 2

Would the planet's tidal forces seriously cause its moon's orbit to expand so much that it's outside its Hill Sphere? That seems counterintuitive to me. Tidal forces drop by the distance cubed, and as the moon got farther away, the tidal forces that caused it to drift away in the first place would get exponentially weaker.

The two concepts, Hill Sphere versus tidal acceleration, have nothing to do with one another. The Hill Sphere is an approximation of where a very small body can stably orbit a not quite so small body in the presence of a much more massive third body. For example, the Earth orbits the Sun, and Moon orbits the Earth (but with significant perturbations from the Sun). The Hill Sphere is concerned with the radial gravitational acceleration of the smallest body toward the intermediate sized body versus that of the radial gravitational acceleration of the smallest body toward the larger but more remote third body.

Tidal acceleration addresses the fact that the gravitational acceleration toward the central body (e.g., the Moon toward the Earth) is not exactly radial. The Moon raises tides on the Earth, both in the oceans and the in Earth as a whole. These tides more or less rotate with the Earth. Because the Earth is rotating, the tidal gravitational influences on the Moon on average lead the Moon's orbital position. This results in a secular transverse (non-radial) component of the Moon toward the Earth. Note well: This transverse acceleration has nothing to do with the Sun.

Answer 3

While I agree with everything said in the answer by James K, there is an important detail to be added to this story.

Numerical analysis by Astakhov et al. (2003) and Domingos et al. (2006) indicated that the orbits too closely approaching the Hill radius become unstable in the long term, while those confined to a smaller domain, the reduced Hill sphere, remain stable. Therefore, when analysing if a tidally receding moon can reach a stable synchronous configuration, we should check if synchronism is attained within the reduced Hill sphere, not just within the ordinary Hill sphere.

The radius of the reduced Hill sphere (`the reduced Hill radius') for a circular ($e_m=0$) orbit of the moon is $$ r^{\,'}_H=B\,a_p\,(1-b\,e_p)\,\left(\frac{M_p}{3\,M_{\ast}} \right)^{1/3} $$ and depends on the direction of the orbit: $$ B\;=\;0.49\;\,,\;\;b\,=1.03\;\,,\quad\mbox{for prograde-orbiting moons}\,;\; $$ $$ B\;=\;0.93\;,\;\;b\,=1.08\;\,,\quad\mbox{for retrograde-orbiting moons}\,. $$ Here $a_p$ and $e_p$ and are the planet's semimajor axis and eccentricity, while $M_p$ and $M_\ast$ are the planetary and stellar masses.

This calculation demonstrates that, for the Moon, synchronism is established barely below the reduced Hill radius.