Europa and the Unified Theory

Factual Questions
21

Thanks, Podkayne, that is what I was trying to come up with.

The friction results in the conversion of the system’s mechanical energy to heat. I believe that this loss of mechanical energy eventually causes the orbiting body to end up being tidally locked.

**

First of all, inertia is not a force. Secondly, in an inertial frame, no force is necessary to “keep the orbiting body from falling.” Newton’s First Law states that objects tend to travel in a straight line at constant velocity unless acted upon by a net external force. Thus, an object in a circular orbit must have a net external force acting on it. In an inertial frame, this force is gravity, AND THAT’S IT! The gravitational force is the centripetal force, that is, the force causing the orbiting object to constantly accelerate such that it travels in a circular path. That is the force that the body is “feeling.”

22

You are quite right, as usual.

::clapping hands:: I believe in centrifugal force. I believe in centrifugal force!

robby, you are correct; inertia is not a force. That was poorly worded on my part. Lets take a body in a stable circular orbit. Gravity is the centripetal force that causes the body to accelerate (where this acceleration is a change in the direction rather than a change in the speed). The acceleration is proportional to the distance from the center and is directed toward the center. Because of inertia the accelerating body feels the outward directed centrifugal force. Gravitational force, on the other hand, increases as you get closer to the center. Tidal forces are caused by the combination of the gravity and the centrifugal force.

23

Sorry, esteemed Doctor, but while we’re basically in agreement, I’m going to be a bit pedantic:

Inversely proportional to the distance from the center squared. (Hey, I warned you that I was getting pedantic.)

Not in an inertial frame. In an inertial frame, the correct answer is that the intertia of the body keeps it from falling in. The centrifugal force exists only in a rotating frame.

Imagine a frame corotating with the moon. The planet is in the center, and the moon is stationary. Now, even in a corotating frame, gravity exists–but what’s keeping the moon from falling inward? In this frame, the moon is not moving, so it’s not inertia. The answer is the centrifugal force, which pops naturally out of F=ma when you write it in corotating coordinates.

Now, if one is not being pedantic, I find that when attempting to explain the reason why there is a tidal bulge on the far side of the body, nothing causes Astro 101 students’ lightbulbs to come on like invoking the centrifugal force, which almost everyone has a good physical intuition for. I usually don’t burden them with thinking about rotating vs. inertial frames; I just explain it both ways, and pray that one or the other sticks–but, then, I suppose the Teeming Millions are held to a higher standard. :slight_smile:

24

Precisely. If tiny changes in Europa’s distance from Jupiter leading to tiny changes in Europa’s shape will melt a layer of ice, imagine a poor little moon rotating through its tidal bulge, causing every point on the moon to go up-and-down, from high-tide to low-tide, twice a day! Tidal locking is a (geologically) fast process, which is why almost every moon in the solar system is locked to its planet.

25

What are you talking about? I’m not steamed; I ain’t even slightly miffed. . . Huh? Oh. . . esteemed. Never mind.

I meant the acceleration due to the rotation. This acceleration cannot be inversely proportional to the inverse square of the distance; that would make the acceleration of the center of a rotating disk infinite.

Consider a point rotating with a constant angular velocity. The speed is constant, but the direction changes. This is the acceleration that I meant, and it is, I am pretty sure, proportional to the distance to the center and directed toward the center. The force of gravity is inversely proportional to the distance from the center squared, but the acceleration is proportional to the distance from the center.

26

F=ma, so if the centripetal force goes as 1/r^2, then the centripetal acceleration goes as 1/r^2 as well, right? On the other hand, we’re also taught sometime in high school that centripetal acceleration is v^2/r. And v = rw (w being the angular velocitiy), so then centripetal acceleration can be written as w^2*r. This, I assume, is the source of the disagreement.

The resolution of the problem is that you have to keep in mind that the angular velocity itself depends on how far the object is from the center (remember Kepler’s third law?). For a situation in which you have gravitational attraction, the centripetal acceleration goes as 1/r^2. Only if the central force is linear in r will the centripetal acceleration be linear in r so that you can choose any angular velocity you wish and still get a circular orbit at any radius you want.

27

Podkayne, excellent description. Better than mine on the specifics for Europa. I especially like the rubber ball analogy.

DrMatrix, I do not have the description wrong, however we are defining “sides” differently. I was using the poles as top and bottom, and sides being the equator - near side and far side. You’re using the gravity well to define top and bottom, so sides are the poles. That make sense now?

It is a fictitious force because there is no acceleration. What is the centrifugal acceleration? The centripetal acceleration is gravity. Write the force balance on the free body. There is the linear path (tangent), and then there is the force curving the path. However, Podkayne drags in non-inertial frames, which is illegal. :wink:

Not really. They can be considered the effect of a gravity gradient - greater pull and less pull.

Okay, sometimes fictitious forces are useful.

28

For the following, I’m thinking of Europa as a point of mass, which is obviously wrong as far as tides go (and probably not what DrMatrix is thinking of), but this will make my thoughts more clear (I hope.) I’m repeating a lot of what g8rguy is saying, but maybe in a slightly different way.

For a point mass in the inertial frame you can’t have an acceleration due to rotation. The rotation/revolution of the moon is due to acceleration–the acceleration of gravity. The rotation cannot cause another acceleration. (I realize this isn’t what you were saying, Doc, but this is why I was balking.)

Indeed–for a rigid rotating disk, where the force accelerating the points on the disk is the mechanical strength of the disk itself, you’re right. However, in an orbital system, the acceleration is indeed 1/r[sup]2[/sup]. You can’t get an infinite acceleration because eventually you’d be inside the planet and the mass above you wouldn’t be contributing to the gravitational force. (This all breaks down for a black hole, which are infinitely dense–the acceleration does go to infinity at r=0.)

Maybe the two points of view could be reconciled by remembering that Europa is, of course, not a point mass, and all points on Europa are not orbiting the planet independently; the center of mass of Europa is orbiting the planet, and all the mass of Europa is forced to follow the same path as the center of mass; thus all points have the same angular velocity, w, across Europa the centripetal acceleration is proportional to the distance from Jupiter.

This may have been what DrMatix was trying to say all along.

In review, the acceleration of the center of mass of Europa is proportional to 1/r[sup]2[/sup]. The whole body is accelerated in this way, producing the angular velocity, w. Now, ignoring Jupiter’s gravity, considering only Europa’s motion, the force a point in Europa feels is depends linearly on its distance from Jupiter.

For the center of mass, a=GM/R[sup]2[/sup]=w[sup]2[/sup]R where R is the distance from Jupiter to the center of mass, leading to w[sup]2[/sup]=GM/R[sup]3[/sup]. Then for any other point on Europa, at a distance r from Jupiter, so the accleration DrMatix was talking about is a=GMr/R[sup]3[/sup]. Right?

How does this help us understand tides? There’s an acceleration, pointing in the same direction as gravity, but a bit stronger.

But luckily, this is irrelevant; Europa isn’t a rigid body; it’s a fluid body–otherwise there’d be no tides. To some degree, each point on Europa is orbiting on its own, but is constrained by Europa’s self-gravity (not predominatly by mechanical strength). The points are not accelerated as a=GMr/R[sup]3[/sup]; they accelerate as a=GM/r[sup]2[/sup](in Jupiter’s direction)-Gm/p[sup]2[/sup] (toward Europa’s center), where m is Europa’s mass and p is the distance from the point to Europa’s center of mass.

Think of Europa as (almost) a sphere. Then the pull toward Europa’s center is the same for all points at the surface, while the acceleration of Jupiter’s gravity is much stronger at the front than the back. Thus the front falls toward Jupiter faster, and the back falls more slowly, and Europa stretches.

My head hurts and I am hungry. There may be errors above, and I hope you will point them out.

29

I think g8rguy is right about the source of the disagreement.

A small body in a stable circular orbit is moving inertially. It is in freefall with no net acceleration. Or to put it another way, since gravity and acceleration cannot be distinguished, the force of gravity is equal to the force generated by its acceleration. There are two equal opposing forces acting on the body. Let’s consider them separately. Ignoring, for the moment, the body’s motion, the body is subject to the force of gravity, which is inversely proportional to the distance from the center squared. To see the force opposing gravity, we ignore gravity and consider just the motion; a body moving in a circle with angular velocity w experiences an acceleration of w [sup]2[/sup] r towards the center. Where these two forces are equal, we have a stable orbit. For a large body, the center of mass will move in a circle. For parts of the body inside this circle, the force due to gravity will be stronger than the centrifugal force due to motion; for parts of the body outside the path, the centrifugal force will be stronger. This is the source of the tidal forces.

Well, that’s the way I see it. I hope the above makes sense.

A force inversely proportional to the cube of the distance? :confused: You’ve totally lost me here.

30

Sanity check: The Sun is 2.72E7 times the mass of the moon, and about 390 times as far away. If 1/R^3 is correct, the tides due to the Moon should be about 390^3 / 2.72E7 = 2.18 times as large as those due to the Sun. I recall the factor being about 2, so this checks.

31

The phrase “moving inertially” doesn’t mean anything to me. Either an object is moving in a straight line at constant speed (i.e. moving solely because of its inertia), or it’s moving under the influence of a force. The moon is moving in a circle, therefore gravity is acting on it.

It’s not standing still or moving in a straight line. That mean’s it’s accelarating–unless, again, you’re in a corotating frame. You don’t get to average over the orbit before you decide whether or not it’s accelerating!

Whoa, cowboy. Either we’re doing Newtonian mechanics or relativistic mechanics. You can’t mix and match.

In “force generated by its acceleration,” if “it” refers to the “frame corotating with Europa” then you are correct.

Yes, if we’re in a corotating frame. In an inertial frame there’s one force: gravity. There is no force pushing outward.

The two forces (again, assuming a corotating frame) must only be equal for a circular orbit. Every orbit in a two-body system is stable–no need to invoke exotic balancing forces. Gravity+inertia=stable orbit, always.

Everything you say is exactly true, in a corotating frame. In the intertial frame, there is no centrifugal force; there is only inertia. I am being pedantic and nitpicky; you obviously do understand what you’re talking about at a non-technical level, and your explanation would fly in Astro 101.

But there is no outward-pushing force in an inertial frame. The moon is falling. It is accelerating. The acceleration is not balanced by another acceleration. Nothing is pushing the moon upwards. It just happens to be traveling perpedicular to its fall fast enough that it gets over the horizon before it hits the planet.

Replace the centrifugal force from the corotating plane with the back end of the planet’s desire to keep going in a straight line, and you’re 100% correct in an inertial frame.

Careful: two r’s–big-R is the distance from Jupiter to the center of mass of Europa; little-r is the distance from Jupiter to the specific point on Europa. Since Europa is small, little-r almost equals big-R. In other words, there’s an extra big-R (or pretty close to it) hidden in the little r in the numerator, so roughly still a goes like R[sup]2[/sup].

Zenbeam’s sanity check applies to the tidal force, the difference in the gravitational force on the for any two points at two different radii, which is proportional to R[sup]-3[/sup]. That’s not quite what we’re talking about here, though.

32

For what it’s worth, DrMatrix is right: A body in a gravitational orbit is moving along a spacetime geodesic, with no forces acting on it: This is what he means when he says it’s moving inertially. Or, the frame where the body is at rest is, in fact, an inertial frame.

But Doc? Let’s just let the classicalists have their way on this one… It makes things simpler all around. I mean, black holes? Relativity, all the way. Europa? Don’t give yourself a headache.

If you’re going to use the Newtonian framework, here (as even I, a relativist, would recommend), then Podkayne is exactly right, and came up with a better explanation than I could hope to give, to boot.