Harvesting energy from the earth rotation

[Simon Bridge]

You only need to put the rod across a rift. Attach to one end and rest the other end on rollers - use it to drive a wheel.

You saw the analysis of the torque needed to turn the Earth's sloww rotation into useable energy? Do the same for the much much slower continental drift.

 

[LURCH]

The lunar tide is definitely expelling mass. The Moon. :p The solar tide also results in Sun being "pushed", but calling that one "expelling mass" would be a bit of a stretch, I agree.

This is a good point, and shows one of the changes of view that makes the tidal generator a worthwhile idea. The Moon's mass is being expelled anyway. The tides come in and go out and generate (when taken as a whole) an almost unimaginable amount of energy. Even though the tides come and go relatively slowly, these motions represent an small movement of the entire ocean, which is a tremendous amount of mass. So the energy is being expended whether we put it to use or not.

Likewise with continentle drift. The continents are moving slowly, but with tremendous overall force. The amount of energy it would take to stop the continents from drifting is the amount of energy available to tap; the engeneering challenge would be in gearing down the generator at the end of the rod. But the energy is already moving from one place to another. We can briefly divert that flow to our own use if we are clever about it.

 

[Stanley514]

In particular, you cannot use a gyro to extract energy from Earth's rotation, because the torque would cause the gyro to tip, until its axis of rotation is aligned with Earth's.

If you could speed up a powerball by rotating it with your hand then why you cannot speed up hyro with the Earth rotation?Of course, the Earth makes fewer number of rotations per 24 hours, but theoretically?If you could speed up something then you could harness energy by taking part of energy out and keeping the same speed.Still absolute speed of Eath rotation is not too low.It`s close to 100 meters/sec.

 

[Simon Bridge]

When you speed up the powerball by twisting with your hand, you also have an equal and opposite torque on your hand, and yourself. You muscles provide a counter-torque in reaction - keeping you from rotating. You can do this because your body is braced against the Earth ... so the counter-torque is transferred to the Earth (imparting a small angular momentum to the Earth). Thus - the energy to the ball comes, at least partly, from the rotation of the Earth.

Note: you cannot take energy out of a rotating object without changing the speed that it rotates at.

To exploit the rapid surface tangential speed of the Earth, you need to find some place to stand that is stationary wrt the rotation - and remains stationary under Newton's 3rd Law.

So I can imagine a big gear at the north pole which turns with the Earth and meshes with another gear that, therefore turns the other way. But what is the other gear attached to?

To make it work, I'd have to hold the second gear by the axle and walk round and round the drive gear on the ground. If I go around once a day, the axle I'm holding is now stationary wrt the rotation of the Earth and, indeed, the gear is turning and generating power.

It can take a while to get your head around rotating reference frames and the misconceptions that arise from the common approximation that the Earth is stationary.

 

[mfb]

Without moon, we would have waves from storms (independent of Earth's rotation) and tides from sun. It is possible to extract power from both.

A direct way to harvest energy from Earth's rotation would be given with a space elevator. Lifting mass to geostationary orbit requires energy, but if you keep lifting the object the centrifugal force (in rotating coordinates) will give you energy.

 

[Simon Bridge]

A direct way to harvest energy from Earth's rotation would be given with a space elevator. Lifting mass to geostationary orbit requires energy, but if you keep lifting the object the centrifugal force (in rotating coordinates) will give you energy.

That won't generate energy you didn't give it getting it there.

 

[StephenofCaly]

Nah, we live here. Screw up someone else's backyard. Let's harvest energy from the orbital velocity of Jupiter. So Jupiter's orbit comes in a kilometer or so - who's to care? Somebody do the math.

 

[Simon Bridge]

Jupiter has a big effect on things like keeping the asteroids in their belt.
How about Saturn or Neptune.

But the question remains: how?

 

[mfb]

That won't generate energy you didn't give it getting it there.

It will. In rotating coordinates, the potential is given by V(r)=-1/3mωr^3 - GmM/r. As you can see, this does not have a lower bound, it goes to minus infinity for r→infinity (and has a maximum at geostationary orbit). In practice, the strength of the cable would limit the maximal radius of the space elevator.

 

[Simon Bridge]

Do you mean to say that all you have to do is spin around to get free energy?

 

[mfb]

All you have to do is to build a space elevator and lift material high enough to extract rotational energy from Earth (and find some way to transfer this energy to places where you need it). It is not really free, it slows the rotation of earth. The total energy is 2*10^29 J, so 10GW over 100 years (3*10^19 J) would not influence it in a significant way. But you can see that other energy sources are probably better.

The extracted rotational energy $1/3~m\omega r^3_{final}$ will go to:
- potential energy of the lifted mass (close to GmM/r_earth)
- kinetic energy of the lifted mass $1/3~m\omega^2 r^2_{final}$
- work done to your payload, which you can extract with a dynamo or similar devices

 

[Buckleymanor]

http://mb-soft.com/public2/earthrot.html
The idea has been around for some time and some research done I have other links somewhere if posters are interested.

 

[mfb]

The proposed idea there violates basic laws of physics. As the author does not plan to emit anything to space or at least raise it by significant amounts, the moment of inertia of Earth would not increase. However, the rotational energy is linked to energy by $2E=I\omega^2 = \frac{L^2}{I}$. L is fixed in closed systems, so if you want to extract energy, you have to increase I or link the Earth to systems with a larger I.
The author repeats over and over that energy conservation would not be violated, but apparently he never heard about angular momentum.

C Johnson, Theoretical Physicist, Physics Degree from Univ of Chicago

:yuck:

Apart from that, he is a bit optimistic about the energy content. Lengthening the day by 1 billionth would give a maximal amount of 4*10^21 J, that is about 10 times the current world energy consumption.

Well, the whole site seems to be very close to the crackpot-region with a lot of CAPS and repeated complaints that nobody cares about things which are impossible or too small to use them.

 

[StephenofCaly]

The proposed idea there violates basic laws of physics...Well, the whole site seems to be very close to the crackpot-region with a lot of CAPS and repeated complaints that nobody cares about things which are impossible or too small to use them.

I enjoy that remark. The problem here seems to be in imagining the movement of objects on a fixed radial vector from the Earth's center, passing through a continuous spot on the Earth's surface at the equator, and sweeping into space. An imaginary clock hand, as it were, and one that can lock things somehow to its line, so that stuff must slide up and down the vector.

At about 24,000 miles, this vector touches an orbit called the geosynchronous orbit. Objects orbiting at this height require the proper angular momentum to remain stable on the vast imaginary vector.

Objects locked onto the vector at a closer distance lack the angular momentum to stay at that distance, and will tend to fall back down the vector. Objects locked onto the vector at a further distance would tend to RISE, as the vector - if it can lock onto the object - sweeps it higher and higher like a xistera - the jai-alai thingy - does to a jai-alai ball.

It takes a certain amount of work to go from the Earth's surface to geosynchronous orbit - which then could somehow harvest the work of going from geosynchronous to infinity.

My guess is that it's a lose - it takes more work to get up to the zero-point than could be harvested, even by flinging something off into west of eternity.

PS: I shall leave it off there, as it seems that personal theories and the commenting upon them is suspect.

 

[mfb]

It is possible, and I already mentioned and evaluated it, just read page 3 of this thread.

My guess is that it's a lose - it takes more work to get up to the zero-point than could be harvested, even by flinging something off into west of eternity.

It is not, see my previous posts.

PS: I shall leave it off there, as it seems that personal theories and the commenting upon them is suspect.

This is not a personal theory, it is just an application of Newton's Laws of motion and gravity.

 

[Simon Bridge]

Those aren't references supporting what you say - they are just references to you saying it...

The PE calc looks like the difference between gravitational PE and Centrifugal effect.

If you just fired something straight up, in a rotating reference frame, it does not go straight. Put it on a radial track, though, and push it straight up, then the rotation gives you some help ... though you lose energy from the thing rotating. Now your rotating reference frame no longer has a constant angular speed wrt an inertial frame - did you take this into account?

The further you run the mass up the track, the more energy you draw off from the rotation. This is the same effect as an ice-skater pushing their arms out to slow their spin. At some stage, you'll slow to a stop.

At the end of the track, above geostationary, the mass seems to accelerate madly away for a bit - but isn't that just a classical relativity effect? The mass just heads off at a constant velocity in the inertial frame.

There is a common mistake people make when they consider rotational gravity - they may have someone start out in the "low g" section of a cylindrical habitat moving slowly radially and then figure that since gravity depends on the distance from the center, that the person experiences an increasing gravitational pull as they move outwards.

 

[mfb]

Those aren't references supporting what you say - they are just references to you saying it...

Those are calculations supporting what I say.

Now your rotating reference frame no longer has a constant angular speed wrt an inertial frame - did you take this into account?

The slowdown of Earth is negligible for any reasonable mass lifted with a space elevator. I did not take it into account, for anything below ~10^18 kg the effect is smaller than 0.1% (the rotational energy of Earth is the same as the kinetic energy of 7*10^21kg, moving around Earth with 100.000km radius once a day like a giant slingshot)

The further you run the mass up the track, the more energy you draw off from the rotation. This is the same effect as an ice-skater pushing their arms out to slow their spin. At some stage, you'll slow to a stop.

Right. You certainly do not want that to happen, and stop the device if the days become too long. The geostationary orbit will move upwards a bit.

At the end of the track, above geostationary, the mass seems to accelerate madly away for a bit - but isn't that just a classical relativity effect? The mass just heads off at a constant velocity in the inertial frame.

This is just Newtonian mechanics in rotating systems.

There is a common mistake people make when they consider rotational gravity - they may have someone start out in the "low g" section of a cylindrical habitat moving slowly radially and then figure that since gravity depends on the distance from the center, that the person experiences an increasing gravitational pull as they move outwards.

If they keep stationary relative to the rotating cylinder (and their mass is small compared to the cylinder), this is true. If they are in free fall, it is wrong. Conclusion: You need a space elevator as rotating "floor", otherwise it will not work.

 

[Simon Bridge]

The slowdown of Earth is negligible for any reasonable mass lifted with a space elevator.

Not if you want to to go to arbitrarily high kinetic energy as per your previous posts.

Right. You certainly do not want that to happen, and stop the device if the days become too long. The geostationary orbit will move upwards a bit.

If, however, you return mass by the same route, you can recover the lost energy.

If they keep stationary relative to the rotating cylinder (and their mass is small compared to the cylinder), this is true. If they are in free fall, it is wrong. Conclusion: You need a space elevator as rotating "floor", otherwise it will not work.

You have to modify the equation for an elevator frame that is not classically rigid. It will flex as the mass travels up.

In general you don't expect the beanstalk-type elevator to be rigid ... or even to go up strictly radially. From Earth, carbon-nanofibers are not strong enough ... though I have seen calculations that make such a thing more feasible for, say, Mars.

We often neglect small reaction in our calculations - eg. when we jump, we neglect the recoil of the earth. The trouble comes when doing the same for large distances and energies. I'm not so much concerned about your ideas but that someone less versed in physics will read what you have written and think "Wow: free energy!" I seem to attract people so this is a kind of self defense ;)

The last set of calculations I saw had a space elevator go twice the distance to geostationary ... 71,400km. I couldn't help thinking of tidal drag... and what would happen if it broke! I have Endless Frontier VolII, Jerry Pournelle (ed). as a lay-accessible optimistic take on the possibilities.

 

[mfb]

Not if you want to to go to arbitrarily high kinetic energy as per your previous posts.

Well, a real space elevator would have a finite length anyway - even if you ignore all engineering issues, at 350,000km it would crash into moon. You can gain energy if the cable extends to >150,000km if my calculation is right.
My old posts have an error - the energy should be 1/2 ω^2 r^2 instead of 1/3 ω r^3

If, however, you return mass by the same route, you can recover the lost energy.

Mass at geostationary orbit is perfect - you can gain energy by lowering or raising it.

You have to modify the equation for an elevator frame that is not classically rigid. It will flex as the mass travels up.

I neglected this, too, with the assumption that most mass of the elevator is required to support the elevator, and only a small fraction of its mass can be transported on it.

We often neglect small reaction in our calculations - eg. when we jump, we neglect the recoil of the earth. The trouble comes when doing the same for large distances and energies. I'm not so much concerned about your ideas but that someone less versed in physics will read what you have written and think "Wow: free energy!" I seem to attract people so this is a kind of self defense ;)

It is not free energy, but energy which is available in large amounts. This is similar to fusion power, for example - I think that fusion is way better than slowing earth, but that is not the topic here.

 

[K^2]

The fact that your equation gives arbitrary amount of energy should immediately make you suspicious. Earth has finite amount of energy available. Seems like a thing to take into account. Let's throw angular momentum conservation in.

$$\left(I+mR^2\right)\omega = \left(I + mr^2\right)\omega '$$

Using ω' in equation for effective potential.

$$U = -\frac{MG}{r} -\frac{\omega '^2 r^2}{2} = -\frac{MG}{r} - \left(\frac{I+mR^2}{I+mr^2}\right)^2 \frac{\omega^2 r^2}{2}$$

Note that this potential goes to 0 at r->inf, which is the sort of thing you'd expect. Rotation just doesn't matter if you take it to infinity.

For m<<M, this potential has 6 extreme points, 3 of which are on positive real axis, tow of which correspond to equilibria. First is unstable equilibrium at ~42,000 km, corresponding to geostat. Assuming Earth can be modeled as homogeneous sphere for purposes of moment of inertia estimate, the second equilibrium point is stable and is located about 10^16 km above Earth. At that point, the potential is at about -6x10^28/m J, which means, you can extract up to about 1/4 of Earth's total rotational energy.

 

[Simon Bridge]

Mass at geostationary orbit is perfect - you can gain energy by lowering or raising it.

Nooooo ... by that argument you can set a mass going up and down in a cycle and eventually stop the Earth rotating. Remember where the energy comes from, and small numbers add up to big ones.

I neglected this, too, with the assumption that most mass of the elevator is required to support the elevator, and only a small fraction of its mass can be transported on it.

You are hoping that the energy transferred to this very small mass, from the Earth, via the elevator is also small compared with the transverse rigidity (on this scale the tower is more like a rope - stretchy and flexy) ... but then, right now, such a thing would need magic materials just to exist.

 

[mfb]

The fact that your equation gives arbitrary amount of energy should immediately make you suspicious.

It should simply show you the limits of the equation.

the second equilibrium point is stable and is located about 10^16 km above Earth.

=1057ly. Sounds a bit... impractical :D.
If you plan to lift significant fractions of Earth's mass, the gravitational energy needs a modification, too.

Mass at geostationary orbit is perfect - you can gain energy by lowering or raising it.

Nooooo ... by that argument you can set a mass going up and down in a cycle and eventually stop the Earth rotating.

No. Geostationary orbit is just a local maximum of the potential (for m<<m_earth). If you balance a pen on its tip you can gain energy by letting it fall in every direction. Does that imply any possible cycle?

 

[K^2]

The entire proposal isn't practical, so that's a moot point. The question was, can energy be gained using this method in principle. If you don't factor in conservation of angular momentum, you end up with a result that diverges to infinity, which is wrong and obviously doesn't tell you about whether the actual result is a net gain or net loss. For a sufficiently massive object with sufficiently low angular velocity, the second equilibrium will actually be at higher potential than surface potential, despite your formula still saying you can gather energy from it.

As far as accounting for Earth's loss of mass, my formula does account for it. Starting mass of Earth is M+m.

And if we are talking about practical limitations, there is no material out of which you can build the elevator like this. Nothing will provide sufficient resistance to bending over such distance. The way a space elevator works around that is by having an anchor point slightly above geostat to provide tension. Placing an anchor for your method would eat up more energy than you can harness.

 

[Simon Bridge]

No. Geostationary orbit is just a local maximum of the potential (for m<<m_earth). If you balance a pen on its tip you can gain energy by letting it fall in every direction. Does that imply any possible cycle?

Ah now I get you ... you mean: starting from geostationary you gain kinetic energy in either direction. Excuse me.

 

[Simon Bridge]

Placing an anchor for your method would eat up more energy than you can harness.

Isn't that, if you got the anchor mass from the Earth?... though I understood that the elevator idea is not really for generating energy but to get easier subsequent access to space.

Also - wouldn't such a cable try to lie along an equipotential - so it would have to be tethered at both ends? (Otherwise we are describing a long thin satellite which is not only in geostationary orbit but must rotate to always present one end to the Earth.)

I'm reading mfb's proposal as "having paid the enormous cost of getting one in the first place - we can use the space elevator to draw off energy from the Earth rotation ..." which would satisfy OPs question - sort of.

 

[mfb]

I'm reading mfb's proposal as "having paid the enormous cost of getting one in the first place - we can use the space elevator to draw off energy from the Earth rotation ..." which would satisfy OPs question - sort of.

This.

Building a space elevator purely as power plant is highly impractical. But if that structure is ever built for cheap space launches, it would be possible to use it as power plant, too (engineering issues are a different thing). The first cable would extend to some point a bit above GEO to keep the whole thing in place, afterwards you can extend it outwards with material lifted on the space elevator itself.

For a sufficiently massive object with sufficiently low angular velocity, the second equilibrium will actually be at higher potential than surface potential, despite your formula still saying you can gather energy from it.

I think you have to extend the equation a bit in the general case, as the center of mass and rotation shifts, too. Assuming I of Earth already contains the space elevator (or neglecting its mass):
The initial center of mass relative to the geometric center of Earth is Rm/(M+m), assuming that the mass to lift just lies around on the surface. It becomes important once you lift your mass over large distances as done in your calculation:

The center of rotation will be at x=r m/(M+m), this gives a total moment of inertia of I + Mx^2 + m(r-x)^2 (neglecting momentum of inertia of m).

Using the reduced mass $\mu=\frac{Mm}{M+m}$, the r-dependent angular momentum becomes
$$\left(I+M\frac{r^2m^2}{(M+m)^2}+m\frac{r^2M^2}{(M+m)^2}\right)\omega(r) = (I+\mu r^2)\omega(r)$$Short:
$$(I+\mu R^2)\omega_0 = (I+\mu r^2)\omega(r)$$
Solving this for ω(r) gives
$$\omega(r)=\omega_0\frac{I+\mu R^2}{I+\mu r^2}$$

The total potential+kinetic energy at distance r now is given by
$$U =\frac{-GMm}{r} + \frac{1}{2}(I+\mu r^2)\omega_0^2 \left(\frac{I+\mu R^2}{I+\mu r^2}\right)^2 =\frac{-GMm}{r} + \frac{\omega_0^2(I+\mu R^2)^2}{2}\frac{1}{I+\mu r^2}$$

Fixing m to some value and introducing positive constants a,b and c, the general shape is
$$U=\frac{-a}{r} + \frac{b}{c+r^2}$$
Derivative:
$$U'=\frac{a}{r^2} + \frac{-2br}{(c+r^2)^2}$$
Solving for U'=0:
$$a(c+r^2)^2 = 2br^3$$
Small angular momentum or high mass of Earth gives small b or large a, while c is constant (given by the geometry of earth), therefore the equation will not have any solution.
Ok, I can confirm your result.

And if we are talking about practical limitations, there is no material out of which you can build the elevator like this. Nothing will provide sufficient resistance to bending over such distance.

Bending is significant only if you want to lift a lot of mass at the same time.

 

[Simon Bridge]

Bending is significant only if you want to lift a lot of mass at the same time.

That is just the definition of "a lot of mass" in this context.
(Of course, still need to define "significant".)

How much mass were you thinking of lifting in one go?
Apollo CSM was 30Tonnes or so. How much lateral rigidity would be needed? Ballpark figure?

Of course you need not go that way - you could imagine a structure that flexes like a kite-string with an anchor "out there" maintaining tension. You can still run mass up and down that sort of thing. But it changes the equations.

 

[mfb]

Ballpark figure: Lifting 100 tons with 100m/s requires an initial power of 10MW and gives a Coriolis force of $F_c = 2m \vec{v}\times\vec{\omega} = 1300N$. This would probably bend the space elevator a little bit within some meters of its length.
I found this image for illustration - a 1°-bend in the cable would require an internal tension of ~60*1300N = 80kN, or the gravitational force of 8 tons. I would expect that a space elevator maintains more minimal tension as safety margin (especially when weights of 100 tons have to be lifted).

 

[Simon Bridge]

Pournell's beanstalk was 2x geostationary long ... 1deg is a deflection of over 625km at 1x geo. (picturing the deflection as a kink in the line). Unless the cable stretched, this would pull the midpoint in by 5.4km.

Pournell's (and others) solution was to transport much smaller masses ... 100s of kg to 1T at a time ... in a chain of buckets. Ship components and do your assembly in orbit.

However - I think this topic is just about done to death.

 

[johnmn3]

Supposing we had a gyroscope with 100% mechanical efficiency -- say it is in a vacuum chamber, suspended in place by some frame of superconductors.

Also suppose this gyroscope has magnets on it, facing outward from the walls of the spinning surface (where the white paint on a car's white-wall tires would be), such that we can externally apply force against the axis of the gyroscope's spin, without causing friction.

Now, it is not a stretch to imagine how, as the Earth rotates, the gyroscope's magnets will apply force against a set of magnets connected to a motor that is planted on the earth.

I guess the question is, does applying force against the gyroscope's axis of rotation cause the rotation to slow down? Even if the spin is 100% mechanically efficient?

 

[Simon Bridge]

Welcome to PF;
What is keeping the frame from moving?

 

[johnmn3]

Thanks. Pleasure to be here. The frame would be secured to an object on the earth. It would have a gear system that puts pressure on the gyro as it tries to turn. Would the rotation slow down if the gyro were pushed against its spinning axis?

 

[mfb]

I don't understand the details of your setup (sketch?), but you cannot gain energy with it:
Reducing the angular velocity of Earth requires an increased angular velocity of the gyro (angular momentum conservation). But that violates energy conservation unless you use some external energy source to do so.

 

[johnmn3]

Sure, let me pull out my handy dandy Galaxy Note.

Attached is a PNG of the sketch. A shows the gyro on the Earth's equator. In B you see the gyro inside its frame, attached to the gears (I'm no artist). The frame is a vacuum sealed enclosure, using magnets (and/or superconductors) to 1) let this internal disc spin freely, while 2) allow the application of frictionless force against the walls of the spinning disc, in order to turn the gears.

C shows how the gyro attempts to maintain a constant orientation as the Earth spins.

D shows how the gears apply resistance to the relative spin of the gyro.

The question is, does applying force against spinning axis of the gyro add to or take away from the energy in the spinning axis.

I know there's an input axis and an output axis that I don't fully understand yet. But if the spinning axis can gain and lose energy via the input and output axis, I would argue one could position the generator such that the Earth's turn pushes into the input axis. Then the generator could simply bleed energy from the spin, using coils on the frame and the magnets on the disc.

 

[johnmn3]

Looks like these guys have already put some thought into it: http://www.google.com/patents/US5313850