Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: hamdani yusuf on 19/05/2017 10:32:29
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Harvesting energy from tides is known to make the Earth to rotate slower, but what is the effect on the moon's orbit? Measurements shows that the moon is moving away from Earth, but that means its potential energy is increasing?
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And its kinetic is decreasing
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...and the Earth's kinetic energy is also decreasing.
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For 1 Joule tidal energy harvested, how much earth's rotational energy lost, how much loss in moon's kinetic energy, how much gain in moon's potential energy?
How about earth's potential energy?
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For 1 Joule tidal energy harvested
There is a huge amount of tidal energy involved in daily flexing the solid Earth, and dragging the world's oceans across their ocean floors.
If we harvested 1 Joule (or 1 TJ) of this energy, it would make an insignificant change in the amount of energy lost to tides every day.
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For 1 Joule tidal energy harvested
There is a huge amount of tidal energy involved in daily flexing the solid Earth, and dragging the world's oceans across their ocean floors.
If we harvested 1 Joule (or 1 TJ) of this energy, it would make an insignificant change in the amount of energy lost to tides every day.
ok, let's make it 1 terajoule.
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In order for an object to reach a higher and therefore longer orbit its velocity must increase by some amount to get out of its current orbit. The new orbit will require a lower velocity than the original one. So it has to slow down again. This describes an oscillation in speed which involves jerk.
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In order for an object to reach a higher and therefore longer orbit its velocity must increase by some amount to get out of its current orbit. The new orbit will require a lower velocity than the original one. So it has to slow down again.
This imagines that the effect of tides is the same as firing a rocket motor for a short time.
If you are in a circular orbit around Earth, and wish to reach a higher circular orbit around Earth using a chemical rocket, you:
- fire the rocket for a short time, putting you into an elliptical orbit whose highest point above Earth is the altitude of the new orbit you want.
- On this elliptical orbit, you start moving farther from the Earth; your velocity away from the Earth declines towards zero due to gravitational attraction to the Earth.
- when you reach the maximum altitude of this elliptical orbit, and are about to fall back to Earth, you fire your rocket again, for a short time, to give you the (lower) velocity you need to stay in orbit at this higher altitude.
- As the comment says, this involves (two) bursts of acceleration, plus a period of coasting in zero-g, when you slow down again.
- You feel "jerk" at the start and end of the rocket burns (and because rockets are chaotic, you feel some high-frequency jerk during the burn, which does not affect your altitude).
However, tidal effects on the Moon are continuous, rather than in short bursts, and so you would not feel any measurable jerk.
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According to my calculations, the Moon gains about 3.826 x 10^18 joules from the Earth due to tidal action every year. It gains 7.564 x 10^18 joules of gravitational potential energy due to the increase in distance from the Earth (3.82 centimeters per year) and loses 3.738 x 10^18 joules of kinetic energy due to the slowing of its orbital speed. That assumes I didn't make any kind of calculation error. The Moon has been doing this for billions of years and will continue to do it for hundreds of millions of years yet. Of course, the rate has varied in the past (it would have taken energy from the Earth at a much faster rate in the distant past when it was much closer to the Earth). In the future, that rate will slow down. The rotational kinetic energy of the Earth is 2.138 x 10^29 joules.
It's pretty safe to conclude that nothing humans could do in the immediate future would come remotely close to exhausting such a resource.
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According to my calculations, the Moon gains about 3.826 x 10^18 joules from the Earth due to tidal action every year. It gains 7.564 x 10^18 joules of gravitational potential energy due to the increase in distance from the Earth (3.82 centimeters per year) and loses 3.738 x 10^18 joules of kinetic energy due to the slowing of its orbital speed. That assumes I didn't make any kind of calculation error. The Moon has been doing this for billions of years and will continue to do it for hundreds of millions of years yet. Of course, the rate has varied in the past (it would have taken energy from the Earth at a much faster rate in the distant past when it was much closer to the Earth). In the future, that rate will slow down. The rotational kinetic energy of the Earth is 2.138 x 10^29 joules.
It's pretty safe to conclude that nothing humans could do in the immediate future would come remotely close to exhausting such a resource.
Thanks for your effort. Can you share your calculation?
Did you calculate based on earth-moon barycenter? IMO, when the moon is moving away from earth, it also means that the earth is moving away from the barycenter too, which makes total potential energy of the system increase.
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Thanks for your effort. Can you share your calculation?
I found an equation that predicted orbital velocity based on the distance a satellite is from a planet and the planet's mass. The equation is: orbital velocity (m/s) = square root ((gravitational constant x mass of the planet (kg)) / orbital radius (m)).The orbital velocity I calculated for the Moon's semi-major axis of 384,399,000 meters was 1,018.20824493 meters per second. Then I used Newton's kinetic energy equation to calculate how much kinetic energy the Moon (with a mass of 7.342 x 10^22 kilograms) would have at that velocity. The equation is: kinetic energy (J) = 0.5 x mass (kg) x (velocity (m/s))^2. The result was 38,059,020,182,894,347,263,576,879,000 joules.
Then I repeated the calculations with the Moon being 3.82 centimeters further out (384,399,000.0382 meters), resulting in an orbital velocity of 1,018.20824488 meters per second and a kinetic energy of 38,059,020,179,156,504,796,530,624,000 joules. I subtracted the two to get a kinetic energy decrease of 3,737,842,467,046,255,000 joules per year.
For the gravitational potential energy, I likewise used the required equation as found by Google. The equation is: gravitational potential energy (J) = (-(gravitational constant) x (mass of Earth (kg)) x (mass of Moon (kg))) / orbital radius (m). First I calculated the potential energy for the Moon at 384,399,000 meters from Earth and then calculated it for the Moon at 384,399,000.0382 meters from Earth. The first result was 76,118,040,365,661,721,284,394,600,402 joules and the second was 76,118,040,358,097,421,841,161,601,579 joules. I subtracted the two to get an increase of 7,564,299,443,232,998,823 joules of gravitational potential energy. I then subtracted the kinetic energy from the potential energy to get 3,826,456,976,186,743,823 joules.
Did you calculate based on earth-moon barycenter? IMO, when the moon is moving away from earth, it also means that the earth is moving away from the barycenter too, which makes total potential energy of the system increase.
I didn't. I just used a straight Earth-to-Moon distance.
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In order for an object to reach a higher and therefore longer orbit its velocity must increase by some amount to get out of its current orbit. The new orbit will require a lower velocity than the original one. So it has to slow down again.
This imagines that the effect of tides is the same as firing a rocket motor for a short time.
If you are in a circular orbit around Earth, and wish to reach a higher circular orbit around Earth using a chemical rocket, you:
- fire the rocket for a short time, putting you into an elliptical orbit whose highest point above Earth is the altitude of the new orbit you want.
- On this elliptical orbit, you start moving farther from the Earth; your velocity away from the Earth declines towards zero due to gravitational attraction to the Earth.
- when you reach the maximum altitude of this elliptical orbit, and are about to fall back to Earth, you fire your rocket again, for a short time, to give you the (lower) velocity you need to stay in orbit at this higher altitude.
- As the comment says, this involves (two) bursts of acceleration, plus a period of coasting in zero-g, when you slow down again.
- You feel "jerk" at the start and end of the rocket burns (and because rockets are chaotic, you feel some high-frequency jerk during the burn, which does not affect your altitude).
However, tidal effects on the Moon are continuous, rather than in short bursts, and so you would not feel any measurable jerk.
It wouldn't be measurable and only apply where the path of the moon wasn't perpendicular to the direction of the earth's gravitational field.
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Thanks for your effort. Can you share your calculation?
I found an equation that predicted orbital velocity based on the distance a satellite is from a planet and the planet's mass. The equation is: orbital velocity (m/s) = square root ((gravitational constant x mass of the planet (kg)) / orbital radius (m)).The orbital velocity I calculated for the Moon's semi-major axis of 384,399,000 meters was 1,018.20824493 meters per second. Then I used Newton's kinetic energy equation to calculate how much kinetic energy the Moon (with a mass of 7.342 x 10^22 kilograms) would have at that velocity. The equation is: kinetic energy (J) = 0.5 x mass (kg) x (velocity (m/s))^2. The result was 38,059,020,182,894,347,263,576,879,000 joules.
Then I repeated the calculations with the Moon being 3.82 centimeters further out (384,399,000.0382 meters), resulting in an orbital velocity of 1,018.20824488 meters per second and a kinetic energy of 38,059,020,179,156,504,796,530,624,000 joules. I subtracted the two to get a kinetic energy decrease of 3,737,842,467,046,255,000 joules per year.
For the gravitational potential energy, I likewise used the required equation as found by Google. The equation is: gravitational potential energy (J) = (-(gravitational constant) x (mass of Earth (kg)) x (mass of Moon (kg))) / orbital radius (m). First I calculated the potential energy for the Moon at 384,399,000 meters from Earth and then calculated it for the Moon at 384,399,000.0382 meters from Earth. The first result was 76,118,040,365,661,721,284,394,600,402 joules and the second was 76,118,040,358,097,421,841,161,601,579 joules. I subtracted the two to get an increase of 7,564,299,443,232,998,823 joules of gravitational potential energy. I then subtracted the kinetic energy from the potential energy to get 3,826,456,976,186,743,823 joules.
There's an easier way. The total orbital energy can be found by:
E= -GMm/2a
where a is the semimajor axis. By using 3.987e14 for GM (we actually know this product more accurately than we know either G or the mass of the Earth) and calling it U, you can find the energy difference by
-Um/2 (1/a1-1/a2)
For your given Moon mass, this works out to 3,783,812,701,003,630,000 Joules.
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The variable U is known as the standard gravitational parameter and depends upon mass.
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Surely the 38.2mm per year of extra separation would not be constant and would increase year on year.... therefore the amount of energy that holds the moon in place would need to increase but as it is decreasing because of all the wave farms, the rate of change of yearly position would also increase, of course this ignores any external forces that may reduce that distance impacts from meteorites for instance ...
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There's an easier way. The total orbital energy can be found by:
E= -GMm/2a
where a is the semimajor axis. By using 3.987e14 for GM (we actually know this product more accurately than we know either G or the mass of the Earth) and calling it U, you can find the energy difference by
-Um/2 (1/a1-1/a2)
For your given Moon mass, this works out to 3,783,812,701,003,630,000 Joules.
Interesting. Looks like our two calculations aren't too far off from each other.
Surely the 38.2mm per year of extra separation would not be constant and would increase year on year.... therefore the amount of energy that holds the moon in place would need to increase but as it is decreasing because of all the wave farms, the rate of change of yearly position would also increase, of course this ignores any external forces that may reduce that distance impacts from meteorites for instance ...
No, it wouldn't be constant. As it gets further away, the rate would actually slow down since (1) the amount of energy available to be extracted from the Earth decreases as its amount of rotational kinetic energy decreases over time and (2) the increasing distance means that the strength of the tidal interactions decreases over time as well.
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Teach us more in this field. After reading on here, we can learn more in this topic.
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Teach us more in this field. After reading on here, we can learn more in this topic.
I´m glad you enjoy learning ... But watch out: I´m afraid many things said here may be erroneous. But discussion bring light! (literally translated from Spanish).
I can´t actually see how tidal harvesting on Earth could increase our Moon´s speed… Locally harvested energy is a small part of the energy acquired by water (certainly thanks to Moon´s pull … and Sun´s), which if not harvested would be wasted mainly as friction heat. How could that “tele-affect" the Moon?
Our planet tides do affect it, making days a little longer after eons. Similarly, when the Moon was first forming (much closer to Earth than now) , on Moon there were tides of molten lava, and that made Moon´s days longer and longer with time.
That said, as far as I can understand the reason of Moon´s distance slow increase is related to Earth tides, but independently of any possible harvesting.
Moon related high tide doesn´t happen when moon is exactly over considered meridian, but with some delay. That´s because the top of the tidal bulge can´t catch up with below moon meridian, due to the high linear velocity of sea surface because of earth rotation (40,000 km/24h at the equator), and it always is some angular distance East to the meridian below Moon.
Although main component of attraction between moon and C.G. of the tidal bulge is vertical, a relatively tiny tangential component does exist. Naturally, in both senses (Newton´s 3rd Law).
That is why Moon´s tangential velocity, slowly but surely, increases.
It´s a kind of “chainless" hammer throwing effect, without releasing non existing chain ...
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Teach us more in this field. After reading on here, we can learn more in this topic.
I´m glad you enjoy learning ... But watch out: I´m afraid many things said here may be erroneous. But discussion bring light! (literally translated from Spanish).
I can´t actually see how tidal harvesting on Earth could increase our Moon´s speed… Locally harvested energy is a small part of the energy acquired by water (certainly thanks to Moon´s pull … and Sun´s), which if not harvested would be wasted mainly as friction heat. How could that “tele-affect" the Moon?
Our planet tides do affect it, making days a little longer after eons. Similarly, when the Moon was first forming (much closer to Earth than now) , on Moon there were tides of molten lava, and that made Moon´s days longer and longer with time.
That said, as far as I can understand the reason of Moon´s distance slow increase is related to Earth tides, but independently of any possible harvesting.
Moon related high tide doesn´t happen when moon is exactly over considered meridian, but with some delay. That´s because the top of the tidal bulge can´t catch up with below moon meridian, due to the high linear velocity of sea surface because of earth rotation (40,000 km/24h at the equator), and it always is some angular distance East to the meridian below Moon.
Although main component of attraction between moon and C.G. of the tidal bulge is vertical, a relatively tiny tangential component does exist. Naturally, in both senses (Newton´s 3rd Law).
That is why Moon´s tangential velocity, slowly but surely, increases.
It´s a kind of “chainless" hammer throwing effect, without releasing non existing chain ...
Harvesting tidal energy involves using something like a water turbine to to drive a generator. These turbines will add a bit of extra tidal drag. The present tidal bulge position is determined by a balance between the Earth trying to drag the bulge with it and the Moon trying to align it with itself. The extra drag will cause the tidal bulge to be dragged by the rotating Earth just a bit more ahead of the Moon. This in turn, causes it pull forward on the Moon a little more, increasing the rate at which the Moon gains orbital energy. (One of the reasons you can't just extrapolate from the Moon's present distance and recession rate to find its distance from the Earth long ago, say a billion years, is that during that time there has been continental drift, and the positional shifts of the continents has an effect on the magnitude of the tidal drag.
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...These turbines will add a bit of extra tidal drag. The present tidal bulge position is determined by a balance between the Earth trying to drag the bulge with it and the Moon trying to align it with itself. The extra drag will cause the tidal bulge to be dragged by the rotating Earth just a bit more ahead of the Moon. This in turn, causes it pull forward on the Moon a little more, increasing the rate at which the Moon gains orbital energy.
Thank you.
I would agree with you if:
1) Tidal bulge water were really moving towards the Moon meridian, kind of counter-rotating around Earth C.G., not rotating at 2π radians a day as the rest of ocean waters ...
2) ... and turbines were always installed with the direction of local parallel, as water supposedly moved.
But those hypothesis have basic errors:
On the one hand, almost 100% of tidal bulge water moves ONLY vertically. It´s the "shape" of bulge surface which changes, appearing as if water itself were moving horizontally.
On the other hand, those turbines are mounted in many different directions, depending on local geographical features ...
E.g., the case I know best it´s an estuary south of Spain, were stronger currents have app. north to south direction. There a turbine would dragg water in a direction pararell to Moon´s meridian. No extra pull on the Moon !!
Surely there will be some cases with suitable direction, and some extra pull will occur ... but only in one of the senses. Tidal currents are normally used in both senses, but in those last cases in one of the senses the current would be due only to Earth´s gravity (high tide filled the "bath", and when Moon´s effect disappears water simply goes down), and a turbine would decrease pull on the Moon instead ...
Keep in mind that, even in those very few useful cases, some sixty minutes before and after both high and low tide water is practically still.
So we have only a few cases, relatively small amount of water involved, and being efficient only some eight hours a day.
But in the natural "hammer- throwing" like mechanism explained by me, the huge amount of tidal bulge water counts, and works twenty-four-seven.
By the way, instead of "chain" I should have said "wire", what actually links hammer and athlete. It was a lapsus due to the fact that I´d recently used the case of a rotating chain when discussing centrifugal forces (to analyze internal stresses in different links)
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It gains ... and loses 3.738 x 10^18 joules of kinetic energy due to the slowing of its orbital speed. That assumes I didn't make any kind of calculation error.
I found an equation that predicted orbital velocity based on the distance a satellite is from a planet and the planet's mass. The equation is: orbital velocity (m/s) = square root ((gravitational constant x mass of the planet (kg)) / orbital radius (m)).The orbital velocity I calculated for the Moon's semi-major axis of 384,399,000 meters was 1,018.20824493 meters per second. Then I used Newton's kinetic energy equation to calculate how much kinetic energy the Moon (with a mass of 7.342 x 10^22 kilograms) would have at that velocity. The equation is: kinetic energy (J) = 0.5 x mass (kg) x (velocity (m/s))^2. The result was 38,059,020,182,894,347,263,576,879,000 joules.
Then I repeated the calculations with the Moon being 3.82 centimeters further out (384,399,000.0382 meters), resulting in an orbital velocity of 1,018.20824488 meters per second and a kinetic energy of 38,059,020,179,156,504,796,530,624,000 joules. I subtracted the two to get a kinetic energy decrease of 3,737,842,467,046,255,000 joules per year.
There's an easier way. The total orbital energy can be found by:
E= -GMm/2a
where a is the semimajor axis. By using 3.987e14 for GM (we actually know this product more accurately than we know either G or the mass of the Earth) and calling it U, you can find the energy difference by
-Um/2 (1/a1-1/a2)
For your given Moon mass, this works out to 3,783,812,701,003,630,000 Joules.
Sorry, but in those calculations you have considered the further the Moon, the smaller its linear speed ... But, as far as I can understand, that´s not actually the case.
You have made calculations which seem to be correct, but ONLY if Moon were rotating "freely" as a satellite, without experiencing neither other force than gravity towards Earth´s C.G. (or the barycenter of Earth/Moon), nor an external torque ...
But, as I explained in my two last posts, constantly a torque is being exerted on the Moon due to the position of the C.G. of the massive bulge of the Earth´s high tide caused by Moon´s attraction.
Although the other bulge at the antipodes also causes a torque, but opposite, this is smaller due to the facts of being further from the Moon, and a smaller angle out of line between centers of G. (smaller tangential component of a smaller attracting force).
So, Moon´s kinetic energy actually increases with its speed and distance to Earth, at the expense of Earth´s energy.
As the hammer going to be thrown, at the expense of the athlete energy !!
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I think this video answers the question pretty nicely.
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I think this video answers the question pretty nicely.
Yes, pretty nicely ... but not completely right !!
Just now I´ve finished drawing and sending following comment to relevant youtube site:
" Your video has many interesting, correct things, some of them rarely mentioned by even physicists. One of them: the interaction between the Moon and Earth´s tidal bulges ...
But you've got a detail wrong. You say:
"The Earth loses a lot of energy due to friction as its insides and the oceans slosh around
during this whole process, which also slows its spin".
Quite right so far. But then you add:
"Since the Earth’s rotation is slowing down, some mass in the Earth-Moon system moves farther from the center to compensate—and that mass, in this case, is the Moon".
And here you are forgetting that that the rotational momentum conservation principle is valid ONLY when in an isolated system, without any external energy input or output ... And tidal friction energy is lost, as heat ...
When you say:
"You know how a spinning figure skater slows down when they extend their arms?"
you are also quite right ... But eventually they would slow down even not extending their arms, due to friction with ice and air !!"