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As we know, an object that is beyond the Roche limit doesn't disintegrate (obviously) because the tidal forces upon the object are weaker than the gravitational pull of the object towards its centre, so the planet remains intact.

My question is: if that's so, why does the body deform even before crossing the Roche limit?

If the gravitational pull of the particles in the object outweighs the tidal forces, wouldn't the object have to stay totally round? There couldn't have been any deformation because the tidal forces are totally cancelled out.

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  • $\begingroup$ Total force matters. And you can elongate a spring without need to break it. $\endgroup$ – Alchimista May 11 '19 at 5:42
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When we do mathematical calculations, sometimes terms cancel and we get to draw a line through them.

But in the real world cancellations and "total cancellations" don't really happen.

Forces just add. Often there's a big force and one or more little forces, and as long as the big force substantially dominates the little forces, we tend to notice only the big force. But it doesn't cancel, it just covers up the effects of the others to the casual observer.

Let's think about gravity. We are so used to moving around in 1 g of gravity that we don't notice it. You could say that our brain "cancels" out 1 g from our conscious awareness and we compensate for it automatically. We don't feel and notice the 1 g when we stand or sit or lay down, even though we make use of it to stay put.

However, get in an elevator or an amusement ride and add a little acceleration, and we notice it right away! That sinking feeling when a fast elevator starts accelerating downward for example watch (but don't listen to) the video The Equivalence Principle

The Equivalence Principle screenshots youtube.com/watch?v=m8iP5WsiSbc

The purpose of the above discussion is to discount the idea of "total cancellations". A bit of rock or soil on the surface of a body feels the gravitation of the rest of the body pulling down, the "centrifugal force pulling" it up if the body is rotating, and the gravitation of nearby bodies (e.g. Sun or Moon or planet) pulling up or down or sideways depending on orientation. It feels all of these all the time.

If the net force, the sum of all the forces, pulls up, then the body starts to disintegrate. If the net force is still down, it doesn't.

Think of it as a threshold, rather than a cancellation.

...if that's so, ¿why does the body deforms even before crossing the Roche limit?

As long as the body is made of material capable of deforming or flowing (which rock certainly is, especially when hot) then it will continue to move or flow until the sum of all the forces reaches zero. Gravity usually dominates, but it's the force of the compressed layer below pushing up that counters the sum of all the forces.

If the body is rotating, then the net force is weaker on the equator than at the poles, and so less pressure is needed to compensate. That means the higher pressure at the poles will cause some of the material to move to the equator.


To read more about several of the competing forces pulling on us all the time as we stand on the Earth, see:

Once you are comfortable with the idea of equilibrium, multiple forces adding and pressure, then have a look at all the interesting answers to these particular Roche limit-related questions:

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Slightly long for a comment, so I'll put it here.

A comparison could be drawn to the equatorial bulge. A planet that rotates is fatter at the equator than the poles. Nothing "lifts" off the surface because the outward force from the rotation is generally much weaker than the gravity, but the planet still bulges because the forces get added together and the planet effectively has less gravitation on it's axis of rotation.

Just as Earth's rotation, which on a person's weight might only be a fraction of a lb between equator and pole, that variation is still enough to give the Earth a 42 km tidal bulge.

Equatorial bulge of the planets

0.5% weight loss at the Equator vs North Pole. (note 80% of your equatorial weight loss is due to rotation, 20% is due to further distance from the center).

The tidal force on Earth is quite a bit weaker than that and the bulge is measured in feet, not 21 km radius, but Earth being as large as it is, even a small force over the entire surface creates a bulge. Here's a fun answer on the tidal force calculated on the surface of Phobos.

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It may also help your mental picture to consider a phenomenon many people see every day.

The Earth orbits the Moon quite a distance outside the Moon’s Roche limit. Nevertheless, it is distorted by this. The distortion is a bit less than a metre: that is, a perfectly fluid frictionless inertialess material would settle into an ellipsoidal shape pointing directly towards (and away from) the Moon. Water is not frictionless and inertialess, and in addition the oceans are shallow containers rather like a soup plate, and the water in them sloshes like soup which is why the tides are often much higher (and sometimes non-existent).

The aim of this picture is to show that distortion outside the Roche limit is not hard to believe but, on the contrary, an everyday experience.

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