Perpetuum Mobile and Gravitation

[shlosmem]

I have fundamental question about what is called the “law of conservation of energy”.
We all hear about the tidal power stations which using the tidal power. The source of the tidal power came from the changes in the gravity field between the moon and the earth. Allegedly, because of the law of conservation of energy this influence must cause to some energy lose in the moon or the earth. And indeed we know that the moon orbit get longer and slower over time. My question is, are we really must say that the energy of the tide and the loss of the kinetic energy of the moon are equal?
According to the general relativity theory the gravitation is the time space curve effect of a big object. This curve is not “energy consuming”, which means basically - two objects can spin around each other in space forever even that such a spin is a change in momentum that should consume energy according to the classic theory. The question is did the tidal effects caused by that “miracle” eternity momentum changes are indeed “energy consuming”?
Lest imagine that instead of the moon there is black hole and the Earth is spinning around it. This can cause to tidal power effects exactly like happen by the moon. This energy coming from the black hole which means that the black hole mass must be reduce according to the equation of e=mc^2. This is against what we know about black holes which are never losing any mass.
But if the answer is “no” that mean in other words that the tidal power stations are kind of “Perpetuum Mobile” - creating energy from nothing. This is of course a weirder conclusion.

 

[Nugatory]

I have fundamental question about what is called the “law of conservation of energy”.
We all hear about the tidal power stations which using the tidal power. The source of the tidal power came from the changes in the gravity field between the moon and the earth. Allegedly, because of the law of conservation of energy this influence must cause to some energy lose in the moon or the earth. And indeed we know that the moon orbit get longer and slower over time. My question is, are we really must say that the energy of the tide and the loss of the kinetic energy of the moon are equal?

We do if we want to understand what's going on here - it is a necessary part of the physical explanation. You also may have a basic misunderstanding about how to apply conservation of energy here:

two objects can spin around each other in space forever even that such a spin is a change in momentum that should consume energy according to the classic theory. The question is did the tidal effects caused by that “miracle” eternity momentum changes are indeed “energy consuming”?

Classical theory does not say that; the classical theory says that in the absence of friction the circular orbit will last forever. The direction of the moon's momentum changes, but its kinetic energy (which is a function of the magnitude of the momentum, not its direction) does not.

Lest imagine that instead of the moon there is black hole and the Earth is spinning around it. This can cause to tidal power effects exactly like happen by the moon. This energy coming from the black hole which means that the black hole mass must be reduce according to the equation of e=mc^2. This is against what we know about black holes which are never losing any mass.

You are also misunderstanding black holes. Whether an object is a black hole or not, the gravitational effects are the same everywhere above its surface (physical surface of a planet, event horizon of a black hole). So the physics of the rotating Earth orbiting a rotating black hole are the same as the physics of the Earth orbiting an ordinary star of the same mass. We can extract energy from the system using tidal power stations, and the power comes not from inside the black hole but from the kinetic energy of the rotating moving objects.

By the way... questions of this sort sometimes degenerate into argments about whether teh current physical understanding of these systems is correct, If that's where this thread is going, expect it to be locked pretty quickly - we're here to explain what the science says and how it works, not to argue its correctness with people who are unwilling to understand either.

 

[jbriggs444]

The source of the tidal power came from the changes in the gravity field between the moon and the earth.

Can you firm that statement up?

The tides cause the Earth to slow its rotation and cause the moon to increase its distance from the Earth. The change increases the gravitational potential energy of the moon with respect to the Earth, making it less and less negative [assuming a zero point at infinity].

If "changes in the gravity field" is the same thing as "change in gravitational potential energy", the above statement gets the sign wrong. The change in gravitational potential energy is a drain on net tidal power, not a contribution.

 

[CWatters]

My question is, are we really must say that the energy of the tide and the loss of the kinetic energy of the moon are equal?

We don't have any reason to believe conservation of energy wouldn't apply.

Proving it by measurement would be hard. Pretty sure there are other things that effect the moons orbit.

 

[shlosmem]

Classical theory does not say that; the classical theory says that in the absence of friction the circular orbit will last forever. The direction of the moon's momentum changes, but its kinetic energy (which is a function of the magnitude of the momentum, not its direction) does not.

Well, you probably must invest some energy in order to change the direction of the moon's momentum . Where do you think this energy is coming from?

 

[A.T.]

Well, you probably must invest some energy in order to change the direction of the moon's momentum.

Changing the direction doesn't require energy.

 

[shlosmem]

Let's say I want to change the direction of some moving object, how can I do it?

 

[A.T.]

Let's say I want to change the direction of some moving object, how can I do it?

By applying a force perpendicular to the object's velocity.

 

[Drakkith]

Well, you probably must invest some energy in order to change the direction of the moon's momentum . Where do you think this energy is coming from?

If energy is expended, then that energy must go somewhere. In the case of a perfectly circular orbit, the velocity, and thus the kinetic energy, is the same at all points in the orbit. So where did the energy go? The answer is nowhere. No energy was expended.

Consider that energy is the potential to do work, and that work is: w=fd, where f is force and d is displacement. This only works if the force is applied along the same axis that the displacement takes place in. In the case of a circular orbit, the force is ALWAYS perpendicular to the displacement, so no work is performed on the orbiting object and no energy is expended.

 

[russ_watters]

Can you firm that statement up?

The tides cause the Earth to slow its rotation and cause the moon to increase its distance from the Earth. The change increases the gravitational potential energy of the moon with respect to the Earth, making it less and less negative [assuming a zero point at infinity].

I'll try:

Power generators are often named after the carrier of the energy, not the original source of the energy. Wind power and hydroelectric power originate from the sun, for example (also: "steam engine" doesn't tell you what is powering it). Similarly, "tidal power" is only carried by the movement of the tides: the energy that is being turned into electricity is the rotational kinetic energy of the earth. Tidal friction and tidal power plants slow the rotation of the earth.

The lunar recession is an additional consequence of that, but my understanding (less certain...) is that energy is conserved in that interaction: the moon is gaining momentum in its orbit, but the total energy of the orbit (kinetic and potential) stays constant.

http://www.talkorigins.org/faqs/moonrec.html

 

[jbriggs444]

The lunar recession is an additional consequence of that, but my understanding (less certain...) is that energy is conserved in that interaction: the moon is gaining momentum in its orbit, but the total energy of the orbit (kinetic and potential) stays constant.

That's clearly incorrect. The kinetic energy of an object in [circular] orbit is half of its escape energy. Worded slightly differently, it is half of its potential energy deficit. The higher the orbit, the lower the potential energy deficit and the higher the total energy.

The moon is losing momentum in its orbit, going slower and slower but orbitting higher and higher. The net impact is that its angular momentum is increasing over time.

Edit: clarified that the above applies for a set of different circular orbits. Russ may be recalling that for a single elliptical orbit the sum of potential and kinetic energy is a constant.

 

[willem2]

I'll try:


The lunar recession is an additional consequence of that, but my understanding (less certain...) is that energy is conserved in that interaction: the moon is gaining momentum in its orbit, but the total energy of the orbit (kinetic and potential) stays constant.

The force of the tidal bulges on the moon is in the direction of the velocity of the moon, so the total energy of the moon must increase.

Combining mv^2 = GMm/r^2 with E = (1/2)mv^2 - GmM/r will get you:

E = - (1/2)GmM/r for the total energy. (with 0 at infinity)

 

[Khashishi]

Gravitational waves do carry energy. We've never measured gravitational waves directly, but there's strong indirect evidence from study of binary stars. The orbit gradually decays to a smaller, faster orbit. http://www.space.com/17346-gravity-waves-binary-stars-speed.html

AFAIK, all orbits decay this way, but the decay is very, very slow for the Earth/moon, that other effects are dominant.

 

[davenn]

interesting article, but it doesn't account for the earth-moon distance increasing, rather than decaying

from wiki ... Due to tidal acceleration, the orbit of the Moon around Earth becomes approximately 2.2 cm more distant each year.

Dave

 

[shlosmem]

If energy is expended, then that energy must go somewhere. In the case of a perfectly circular orbit, the velocity, and thus the kinetic energy, is the same at all points in the orbit. So where did the energy go? The answer is nowhere. No energy was expended.

Consider that energy is the potential to do work, and that work is: w=fd, where f is force and d is displacement. This only works if the force is applied along the same axis that the displacement takes place in. In the case of a circular orbit, the force is ALWAYS perpendicular to the displacement, so no work is performed on the orbiting object and no energy is expended.

So you are basically saying I can open any screw without spending any energy? this is good news.

 

[willem2]

So you are basically saying I can open any screw without spending any energy? this is good news.

If there's no friction, this is indeed possible.

 

[shlosmem]

If there's no friction, this is indeed possible.

Why the friction changs the fact that the force "is ALWAYS perpendicular to the displacement"?

 

[Drakkith]

So you are basically saying I can open any screw without spending any energy? this is good news.

No, you have a force in the direction of displacement (up). When you turn the screw, the grooves exert a force on the material the screw is in that forces the screw up and out of the material. Friction drastically increases the amount of energy required to get the screw out.

 

[A.T.]

Why the friction changs the fact that the force "is ALWAYS perpendicular to the displacement"?

Because friction is parallel to displacement.

 

[shlosmem]

No, you have a force in the direction of displacement (up). When you turn the screw, the grooves exert a force on the material the screw is in that forces the screw up and out of the material. Friction drastically increases the amount of energy required to get the screw out.

If we ignore the "up" direction (let's say our screw does not have grooves or we just try to rotate a wheel or change the course of some asteroid) what you saying is that we not have to invest energy to do so?
Anyway the moon’s orbit is not a perfect round.

 

[A.T.]

we just try to rotate a wheel or change the course of some asteroid

These are two different things. Spinning the wheel requires tangential forces. Changing linear direction requires normal (or centripetal) forces.

 

[shlosmem]

These are two different things. Spinning the wheel requires tangential forces. Changing linear direction requires normal (or centripetal) forces.

In both cases if the force is perpendicular to the displacement no energy will be spend.

 

[A.T.]

In both cases if the force is perpendicular to the displacement no energy will be spend.

If the force is perpendicular to the velocity there is no kinetic energy gain. Whether you spend energy to create that force depends on how efficient you are.

To spin up a wheel you have to apply tangential forces, which are not perpendicular to the tangential velocity.

 

[Drakkith]

If we ignore the "up" direction (let's say our screw does not have grooves or we just try to rotate a wheel or change the course of some asteroid) what you saying is that we not have to invest energy to do so?
Anyway the moon’s orbit is not a perfect round.

No, the force is directly in line with the displacement of any point on the screw. In other words, the force applied by the screwdriver on the screw points in the same direction the screw is turning.

In the case of a non-circular orbit, potential energy is changed to kinetic energy and vice versa as the object recedes or approaches the body it orbits. No energy is expended here because the sum of the potential and kinetic energy is always the same.

 

[shlosmem]

No, the force is directly in line with the displacement of any point on the screw. In other words, the force applied by the screwdriver on the screw points in the same direction the screw is turning.

In other words, gravity can change the momentum of an object without additional energy, but we can’t?

 

[A.T.]

In other words, gravity can change the momentum of an object without additional energy, but we can’t?

Gravity is a conservative field, while humans are very dissipative machines. But if a ball hits you on the head, you change its momentum, without investing much energy.

 

[CWatters]

Good explanation of where tidal energy comes from here...

http://curious.astro.cornell.edu/question.php?number=124

 

[shlosmem]

But if a ball hits you on the head, you change its momentum, without investing much energy.

If a ball hits my hand, some of the kinetic energy of the ball will be passed to my hand and move it. I’m asking about the process when we apply a force over an object but no energy is spent.

 

[A.T.]

I’m asking about the process when we apply a force over an object but no energy is spent.

By "we" you mean humans with their muscles? Muscles spent energy even when not producing force at all. They also spent energy when producing static forces that do no work. All the spent energy is dissipated as heat, and none goes to the KE of the object.

 

[shlosmem]

humans with their muscles or with their machines

 

[A.T.]

humans with their muscles or with their machines

Human made things can certainly change the momentum of objects, without spending energy. A stick connecting two masses rotating in space changes their momentum.

 

[Nugatory]

I’m asking about the process when we apply a force over an object but no energy is spent.

Consider a heavy object resting on a table. The table is applying an upwards force to the object (otherwise the object would move downwards because of gravity) but spends no energy doing so.

The table can support the weight forever without getting tired.

 

[shlosmem]

Ok, Thank you all.
I really appreciate it.

 

[Nugatory]

So you are basically saying I can open any screw without spending any energy? this is good news.

No, we are not saying that.

To turn a screw, you will apply tangential forces in opposite directions on opposite sides. As the screw turns, the point of application of these forces moves, so the distance in $W=Fd$ is non-zero. The force is obviously also non-zero, so there's work being done.

It's a different story if the screw doesn't turn. If it's so stuck that the force you're applying doesn't move it, then it's just like the table supporting a weight that I mentioned above - you can lean on the end of your wrench all day without doing any useful work. You will get tired because human muscles burn energy just sitting there, but if you were to hang a weight off the end of the wrench it could sit there forever, twisting the bolt but not moving it.