View Full Version : Stupid question...

ksodbartman

2003-May-16, 09:19 AM

So I've got a friend who wondered if everyone would fly off the earth if it stopped rotating. The obvious answer is NO! But then I got to thinking about how gravity MIGHT be affected by a non-rotating earth, and I wondered about the loss of energy associated with the spin.

A: Would that not affect earth's gravity?

B: If it would, how much?

Thanks!

Argos

2003-May-16, 01:18 PM

I think the answer is not too obvious. If the Earth stopped all of a sudden everything on the surface would continue moving in a tangent, and moving faster than the speed of sound. The people at NASA use the Earth motion to add speed to rockets. It helps saving fuel. But, remember, there are severe restrictions on stopping Earth. It is not an easy thing to do. So, forget about Joshua´s Long Day.

A rotating body drags the space-time. The effect is more noticeable for very massive bodies, like neutron stars and Black Holes. Thus, a non- rotating Earth would affect gravity to some extent.

russ_watters

2003-May-16, 01:52 PM

Yes, if the earth stopped suddenly, we'd all slam into our monitors at 1000 miles an hour. But if the earth stopped slowly (decelerating over a period of say a day) we'd notice nothing.

Except: gee, its pretty dark this morning...

Glom

2003-May-16, 02:55 PM

Well Argos got all relativistic on us, but if you want a classical take, in the polar regions you would feel no change between Earth spinning and Earth not spinning once you get over the deceleration, but on the equator, you would get heavier.

Just consider. On the equator, you have are travelling in a circle radius 6×10^6 m. Your period is 24 hours (sidereal) so your angular speed is 6.3×10^-5 rad/s. In order to be held in that circle gravity needs to apply an acceleration of w²r which is 3.2×10^-2 m/s². Acceleration due to gravity is 9.8m/s², so on the equator, your net acceleration due to gravity relative to the rotating frame is... well... pretty much the same, but you get the idea. You feel just that little bit lighter in theory although you'd probably not really notice it much.

For the benefit of the viewers at home, here's those equations again.

mean radius of Earth is r.

latitude of the gimp is L

angular speed of Earth is w

gravitational field strength is g

mass of the gimp is m

so, radius at any latitude = r cosL°

acceleration to maintain circle = w²r cosL°

net acceleration in the rotating frame = g - w²r cosL°

apparent 'weight' of the gimp = m(g - w²r cosL°)

Spaceman Spiff

2003-May-16, 03:28 PM

Back in the early '80's the group "Modern English" had a hit called "I'll stop the world and melt with you..." These guys must have taken some physics, because Earth's rotational kinetic energy (about 6 x 10^29 Joules if I crunched the numbers right) has to go somewhere.

Glom

2003-May-16, 03:46 PM

Let's see. Moment of inertia of a sphere is ¼mr², isn't it? Radius is 6E6m. Mass is 6E24kg. w = 2pi/(24×60×60)

kE = ½Iw² = 1.4×10^29J

Okay, I got a quarter what you got. :-?

mutant

2003-May-16, 03:51 PM

My head just imploded from trying to read and understand Glom's post. Man, some of you people are way too smart for me. I need to go back to school.

That seems to be my main problem, I developed an interest in astronomy way too late in my life. Now I have this old brain that refuses to soak up any new info.

kilopi

2003-May-16, 04:22 PM

Let's see. Moment of inertia of a sphere is ¼mr², isn't it?

For a sphere of constant density that should be 2/5mr², but the Earth doesn't have a constant density, so it's closer to .3mr² I think.

Glom

2003-May-16, 04:41 PM

Sorry about your minor cranial implosivities.

Maybe I should explain a little for the benefit of others.

In mechanics, we can use work done, with a bit of integration and fancy substitutions to show that the kinetic energy possesses by an object rotating about a fixed axis is.

kE=½Iw²

where w is the angular speed of the object and I is the moment of inertia.

In mechanics, we have a thing called inertia. That is a measure of how hard it is to move something. In linear mechanics (ie moving in a straight line), this inertia happens to be equivalent to the mass of the object. It makes sense. You can push your shopping trolley easily when it's empty but by the end, when it's full, it's much harder.

The same thing goes for rotational mechanics, the mechanics of making things spin, in this case, about a fixed axis. Except we call it the moment of inertia, denoted by some, including me, by I. Just like linear inertia, the moment of inertia is a measure of how hard it is make something spin. Now this is dependant on two things. First the mass and second the distance from the axis squared. So for a particle moving in a circular path about a fixed centre, it's moment of inertia is mr².

For rigid shapes, moments of inertia are derived analytically. You can imagine dividing the shape up into an infinite number of infinitessimally small particle and simply ending up the moments of inertia. Calculus enables us to do that for simply shapes.

IIRC, the result for a sphere rotating about a diameter is that the moment of inertia is ¼mr². Given that the mass of Earth is around 6×10^24kg and that its radius is around 6×10^6m, the moment of inertia is about equal to 5.6×10^36kgm².

Next step is to work out w, which we measure in radians per second. A radian is equivalent to about 51°. There are exactly 2pi radians in a complete circle.

We know Earth's period of rotation. 24 sidereal hours (24 solar hours is close enough for our purposes). This translates to 86400 seconds. This means that in 86400s, the sphere of Earth rotates through 2pi radians. So, how many radians does it sweep out in a second? Simply divide 2pi by 86400s and get 7.3×10^-5 radians per second.

So, sub this into the equation for kinetic energy,

kE=½Iw² and therefore kE=1.4×10^29 joules of energy.

Soupdragon

2003-May-16, 06:01 PM

Is that why I put on a few pounds in Singapore?

Or was it the beer? :wink:

ksodbartman

2003-May-16, 06:11 PM

I don't know if I got my question across accurately. It had more to do with relativistic notions, such as energy having an effect on the curvature of spacetime along with mass. Since a spinning earth has more kinetic energy than a non-spinning one, wouldn't that reduce the earth's curvature of spacetime?

Glom

2003-May-16, 06:46 PM

Let's see. Moment of inertia of a sphere is ¼mr², isn't it?

For a sphere of constant density that should be 2/5mr², but the Earth doesn't have a constant density, so it's closer to .3mr² I think.

Yeah that sounds familiar. It doesn't make a great deal of difference though. 17×10^29J now.

BTW for a circular lamina rotating about a diameter, the moment of inertia is ¼mr² isn't it? That being the case, isn't a sphere rotating about a diameter just equivalent to a load of circular laminas rotating about their diamters so the moment of inertia of a sphere is just the sum of the individual moments of inertia of the laminas? Therefore, isn't it just a mass scale-up? Why is it not therefore ¼mr² for a sphere rotating about a diameter?

Spaceman Spiff

2003-May-16, 07:17 PM

Let's see. Moment of inertia of a sphere is ¼mr², isn't it?

For a sphere of constant density that should be 2/5mr², but the Earth doesn't have a constant density, so it's closer to .3mr² I think.

Yeah that sounds familiar. It doesn't make a great deal of difference though. 17×10^29J now.

BTW for a circular lamina rotating about a diameter, the moment of inertia is ¼mr² isn't it? That being the case, isn't a sphere rotating about a diameter just equivalent to a load of circular laminas rotating about their diamters so the moment of inertia of a sphere is just the sum of the individual moments of inertia of the laminas? Therefore, isn't it just a mass scale-up? Why is it not therefore ¼mr² for a sphere rotating about a diameter?

The moment of inertia of a uniform density sphere is 2/5 MR^2. The Earth's density is not uniform; it is compact, with about 1/3 of its mass contained in the core. So the real moment of inertia is somewhat less than this. I was mainly going for a ballpark number, so I just ignored the constant out in front.

Also, Earth's rotation period is not 86,400 s (that's a mean solar day), but 86,164 s. And Earth is not a sphere, nor is it completely solid. But none of this really matters all that much. I was just introducing another consideration as to the consequences of a rotating Earth that stopped in some short amount of time. But maybe that's not what the original question had intended.

Grey

2003-May-16, 07:49 PM

I don't know if I got my question across accurately. It had more to do with relativistic notions, such as energy having an effect on the curvature of spacetime along with mass. Since a spinning earth has more kinetic energy than a non-spinning one, wouldn't that reduce the earth's curvature of spacetime?

The answer is yes. The rotational energy would contribute to the gravitational field. However, that figure of 1.7 x 10^29 J of energy corresponds to about 1.9 x 10^12 kg of matter. That sounds like a pretty large amount, but the Earth's mass is about 6 x 10^24 kg, so the contribution from the rotational energy is 12 orders of magnitude less than the physical mass. I doubt that anyone would notice. :)

Donnie B.

2003-May-16, 08:46 PM

Let's see. Moment of inertia of a sphere is ¼mr², isn't it? Radius is 6E6m. Mass is 6E24kg. w = 2pi/(24×60×60)

kE = ½Iw² = 1.4×10^29J

Hmmm...

A Hiroshima-sized nuclear weapon yields about 1.2x10^13 Joules of explosive energy. It's interesting that this is 16 orders of magnitude too small to stop the Earth's rotation (assuming all the energy could be appropriately and efficiently coupled to have that effect). Even using 1-megaton equivalents, we'd need 140,000,000,000,000 bombs (that's 140 thousand billion) to stop the world.

This amount of energy seems a bit excessive to be produced by a passing celestial body that doesn't even come into contact with the Earth. It also seems unlikely that so much energy could be expended in a short time without some pretty significant side effects. Even if spread out evenly over a week's time, it is equivalent to 230,000,000 1-megaton bombs going off per second during that week. Without looking it up, I'd guess this is a not-insignificant fraction of the Sun's entire output.

A PXer might argue (if they had enough brains to do so) that only the crust has to be stopped; the mantle and core could keep on spinning. This would provide the way to restart the rotation, too... crust-mantle friction would do the job. Hope everybody has their asbestos shoes handy... :lol:

DrTypo

2003-May-17, 12:04 PM

A PXer might argue (if they had enough brains to do so) that only the crust has to be stopped; the mantle and core could keep on spinning. This would provide the way to restart the rotation, too... crust-mantle friction would do the job. Hope everybody has their asbestos shoes handy... :lol:

Aren't PXer claiming that the Earth will stop spinning for just one day and then restart?

You've just provided a scientific explanation of PXer claims! I hope you're proud of yourself.

--

DrTypo

Donnie B.

2003-May-17, 03:31 PM

A PXer might argue (if they had enough brains to do so) that only the crust has to be stopped; the mantle and core could keep on spinning. This would provide the way to restart the rotation, too... crust-mantle friction would do the job. Hope everybody has their asbestos shoes handy... :lol:

Aren't PXer claiming that the Earth will stop spinning for just one day and then restart?

You've just provided a scientific explanation of PXer claims! I hope you're proud of yourself.

--

DrTypo

I don't think you understood what I was saying. First, it's ludicrous to imagine that the entire Earth could stop rotating in response to a passing celestial body. Even if the enormous energy were available, there's no mechanism that could couple that energy into stopping the rotation. And with no other effects except a few earthquakes and tornadoes? It's fantasy.

So I proposed that a true believer might say that only the crust would stop rotating. What would this involve? For the moment, let's imagine that the crust is a thin shell with a mass (and therefore, angular momentum) of 0.01% of the total mass of the Earth, and that there's no friction between the crust and the mantle to resist the stoppage. Even with these very generous assumptions, you'd still need the energy equivalent of 23,000 1-megaton bombs per second for a full week to stop the crust's rotation.

Now consider: what sort of effect would such an event have? Let's imagine some magical machine that could put all that energy into stopping the crust without any significant damage, and (while we're fantasizing) we'll say that the oceans and atmosphere are stopped too, to get around the problem of mega-hurricanes and tsunamis. So what do we have?

We have a thin layer of rock stopped wrt the fixed stars, and a vast sphere of mantle and core spinning merrily away underneath, at a rate of 1100 mi/hr (at the equator). Have you ever touched a spinning automobile tire? Remember what it did to your hand?

The crust wouldn't blithely start to spin up again. It would melt, spectacularly. Hence the need for fireproof footwear.

I hope you now understand what I've tried to say. The notion that the Earth's rotation could be stopped and restarted is totally ridiculous, whether we're talking about the whole planet or just the crust.

DrTypo

2003-May-17, 09:07 PM

Even if the enormous energy were available, there's no mechanism that could couple that energy into stopping the rotation.

Are you sure? What about tidal-lock?

Have you ever touched a spinning automobile tire? Remember what it did to your hand?

I've never done that. It probably hurts a lot.

The crust wouldn't blithely start to spin up again. It would melt, spectacularly. Hence the need for fireproof footwear.

Yes, but you considered that there was no friction between the crust and the mantle. So nothing would happen.

I hope you now understand what I've tried to say. The notion that the Earth's rotation could be stopped and restarted is totally ridiculous, whether we're talking about the whole planet or just the crust.

I was just being silly.

However I'm really curious about tidal effects. Maybe an ultra-massive object (The Mother Of All Black Holes) flying by could catch the Earth, which would become its tidally-locked satellite.

--

DrTypo

Donnie B.

2003-May-18, 12:20 PM

Even if the enormous energy were available, there's no mechanism that could couple that energy into stopping the rotation.

Are you sure? What about tidal-lock?

DrTypo, I'm not sure from your posts whether you're joking or serious. Since you don't use emoticons to indicate irony or comedy, I'm going to assume that you're serious.

Tidal effects are tiny when it comes to the exchange of angular momentum. Consider: the Sun has had five billion years to work on the Earth, and its tides are about half as big as lunar tides; yet the Earth is not tidally locked to the Sun. A fleeting close encounter with PX would have no significant effect.

Or, just for argument's sake, let's say that PX caused tides a billion times larger than solar tides. Even if this were enough to stop the Earth's rotation quickly (which it's not), I don't think we'd care much, as we'd all be exterminated by the tidal bulges (oceanic and crustal).

Have you ever touched a spinning automobile tire? Remember what it did to your hand?

I've never done that. It probably hurts a lot.

Only if roasting human flesh bothers you.

The crust wouldn't blithely start to spin up again. It would melt, spectacularly. Hence the need for fireproof footwear.

Yes, but you considered that there was no friction between the crust and the mantle. So nothing would happen.

I assumed that only to simplify the calculation of the minimum energy required to stop the crust. In fact, there would be enormous friction between the crust and mantle. The melting would occur as the rotation stopped rather than when it restarted. Really, the whole scenario is so preposterous that your objection seems absurd to me.

I hope you now understand what I've tried to say. The notion that the Earth's rotation could be stopped and restarted is totally ridiculous, whether we're talking about the whole planet or just the crust.

I was just being silly.

However I'm really curious about tidal effects. Maybe an ultra-massive object (The Mother Of All Black Holes) flying by could catch the Earth, which would become its tidally-locked satellite.

Sure, and monkeys could fly out of my <oops, it's a family board>.

The thing I find so bizarre about the rotation-stoppage claim is that it's just about the least likely outcome of a close encounter with a massive object. Disruption of the Moon's orbit would be much more plausible, as would ejection of the Earth from the solar system.

Has anyone noticed the similarity between this claim of Nancy's and the virtually identical claim of Velikovsky? The only difference is that the former claim is a prediction, whereas Velikovsky claims it already happened to produce the Biblical story of Joshua and the battle of Jericho.

O_Desfibrador

2003-May-19, 03:04 PM

BTW for a circular lamina rotating about a diameter, the moment of inertia is ¼mr² isn't it? That being the case, isn't a sphere rotating about a diameter just equivalent to a load of circular laminas rotating about their diamters so the moment of inertia of a sphere is just the sum of the individual moments of inertia of the laminas? Therefore, isn't it just a mass scale-up? Why is it not therefore ¼mr² for a sphere rotating about a diameter?

That would be true if all the laminas had the same radius, but in a sphere as we get away from the axis the laminas grow smaller and smaller and end up being just a point. And that´s not all: only one lamina would rotate about its diameter, the others would rotate about an axis parallel to its diameter but that doesn´t "touch" them.

A load of laminas rotating about the diameter of the central one is a cilinder rotating about a diameter that passes through its center of mass. (I=M/4(R^2+h^2/3))

A load of laminas rotating about a perpendicular diameter that crosses their centers is a cilinder rotating about its CM. (I=1/2MR^2)

The moment of inertia of a sphere is I=2/5MR^2

If the sphere should be hollow I=2/3MR^2

P.D.:I´m studying for my Physics finals and I should be able to proof all this.

kilopi

2003-May-19, 03:53 PM

BTW for a circular lamina rotating about a diameter, the moment of inertia is ¼mr² isn't it? That being the case, isn't a sphere rotating about a diameter just equivalent to a load of circular laminas rotating about their diamters so the moment of inertia of a sphere is just the sum of the individual moments of inertia of the laminas? Therefore, isn't it just a mass scale-up? Why is it not therefore ¼mr² for a sphere rotating about a diameter?

That would be true if all the laminas had the same radius, but in a sphere as we get away from the axis the laminas grow smaller and smaller and end up being just a point.

I think what glom (http://www.badastronomy.com/phpBB/viewtopic.php?p=86506#86506) is envisioning is a bunch of discs rotating about a common axis--but that is not equivalent to a sphere, as it would have to be constructed of semicircular wedges, of width zero at the axis, rather than constant width.

Welcome to the BABB, O_Desfibrador, and good luck with the finals.

BlueAnodizeAl

2003-May-22, 01:39 AM

Negligible, the force of gravity of the earth is largely dependant on the mass of the Earth. See some of my other posts. Newton's Law of Universal Gravitation says F = (G*Me*m)/(r^2), measure r from an objects C.G.

Sorry I did not post this before. I was replying to the initial question: How much would gravity be effected if the Earth ceased rotating. If anything there might be a marginal INCREASE in local gravity.

kilopi

2003-May-22, 03:20 AM

Negligible, the force of gravity of the earth is largely dependant on the mass of the Earth. See some of my other posts. Newton's Law of Universal Gravitation says F = (G*Me*m)/(r^2), measure r from an objects C.G.

Negligible for what?

xriso

2003-May-22, 04:33 AM

Okay, it is true that a disc lamina has a moment of inertia of mr^2/4. As for the sphere, we can indeed decompose it into multiple laminas, but there are two things to watch for:

1) Not all of the laminas are being rotated around a diameter! (use parallel axis theorem to add correcting factor).

2) Not all of the laminas have the same radius, or the same mass.

It would be difficult to integrate the sphere's moment of inertia using these lamina. Instead, I would suggest using the moment of inertia through the axis of the discs (mr^2/2), and go from the top of the sphere to the bottom. No parallel axis theorem needed, then. Still a difficult integration, mind you! You will get 2mr^2/5.

Oh, and back on topic... to clarify, how much of the Earth is being stopped? If we stopped the whole thing, then the people, being part of the Earth, would stop too.

kilopi

2003-May-22, 10:28 AM

1) Not all of the laminas are being rotated around a diameter! (use parallel axis theorem to add correcting factor).

When Glom first mentioned this, I'm pretty sure that he was imagining a sum of discs each rotating about the same axis (one of the diameters)--and thought that their effect would sum to a sphere.

The MI of a disk around the central axis perpendicular to the disk is 1/2MR^2

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