Fast Efficient Orbital Transportation?

[SciNoodle]

Image Earth, our big ball. One flywheel is on the left (in a geostationary orbit), another flywheel on the right, perfectly symmetrical. The only thing that is asymmetrical is that the left flywheel is spinning fast and has two loads/parcels on 2 arms ready to let go of. At the right timing, they are let go of. I imagine they can be let go in a way that has them travel in orbit to the other awaiting flywheel in a orbit with the same height from Earth. Then they attach to the awaiting flywheel. This would lose minimal energy, while obtaining fast transportation possibly for advanced space construction for artificial intelligence.

 

Is this possible, and if not, why?

 

If it lets go of its parcels, it loses mass, but it also loses some of the forward momentum, which could be evened out (if not already) so the orbit is the same. While the parcels that leave have less mass but also less momentum too.

[Strange]

This would lose minimal energy

 

 

Why? Surely it would have to lose exactly the same amount of energy as any other method of getting the loads up to the same speed.

[Bender]

Catching the load with the second flywheel is a nonelastic collision, conserving momentum but not necessarily kinetic energy. The difference in energy has to be absorbed somehow by the collision (possibly in a spring so you can recuperate it?).

 

Why do you want to do this?

Edited January 19, 2017 by Bender

[SciNoodle]

I'm adding it to a list of all the possible advanced capabilities artificial intelligence will have.

 

Yes you must spin it up, but then once it's spun up, it stays spinning, and then it can be let go, travel in an orbit fast, and re-attach to this 2cd flywheel while another re-attaches to the 2cd flywheel on the other arm and like no energy was lost.

[Janus]

Image Earth, our big ball. One flywheel is on the left (in a geostationary orbit), another flywheel on the right, perfectly symmetrical. The only thing that is asymmetrical is that the left flywheel is spinning fast and has two loads/parcels on 2 arms ready to let go of. At the right timing, they are let go of. I imagine they can be let go in a way that has them travel in orbit to the other awaiting flywheel in a orbit with the same height from Earth. Then they attach to the awaiting flywheel. This would lose minimal energy, while obtaining fast transportation possibly for advanced space construction for artificial intelligence.

 

Is this possible, and if not, why?

 

If it lets go of its parcels, it loses mass, but it also loses some of the forward momentum, which could be evened out (if not already) so the orbit is the same. While the parcels that leave have less mass but also less momentum too.

 

If the Idea is two transport a package from one geosynchronous object to another opposite side of the Earth, you are limited into how to do this. The easiest method is to fire the parcel backwards in the Opposite direction of your orbital motion at a relative velocity of ~765.23 m/s. This will put it into an orbit that will take 12 hrs to complete and return to the same point of its orbit, In that 12 hrs, the other object will have completed 1/2 orbit and will now be where your original launch platform is. The package will make one full trip around the Earth, in the time it takes the Earth and the Geosynchronous stations complete 1/2 a rotation.)

 

Firing a package backwards will add orbital velocity to the launch platform and will change its orbit, The new orbit will be slightly larger and elliptical with a longer period. How much of a difference depends on the mass difference between package and launch platform. You could cancel this out by simultaneously firing another parcel of equal mass in the other direction, but this will package will enter a larger orbit with a period of 81.3 hrs. This is not fit evenly into the period of the geosynchronous orbits, So this package wouldn't meet up with the other object and would just keep orbiting independently. continual launches of this type would result in a number of these packages in independent orbits and cause a hazard. You could instead use rockets mounted on the fly wheel to counter the momentum change, but since they would take the same amount of fuel as it would to just accelerate the package in the first place, this doesn't make much sense.

 

Then there is the matter of what exactly it would take for a flywheel to be able to toss a package at a relative speed of 765.23 m/s. If you wanted to keep the g-force experienced while spinning at the end of the arm to 1g the arm would have to be ~60 km long. If the package could withstand 10g, it still would have to be 6km long. This is a lot of stress for just the structure itself to withstand.

 

In the end, I just don't see it as being practical and serves little purpose. Just what advantage is there in just traveling from one geosynchronous orbital object to another?

[SciNoodle]

The advantage. - Couldn't the flywheels be really close to Earth enough so that the orbit to stay up in space is a fast one and can go from one side to the other pretty fast? With less energy wasted than if by rocket pulses.

 

**But what about a flywheel that is spinning two loads and releases them BOTH at the same time AND neither in the direction of velocity nor from the back end but rather from the sides? To then orbit to a *second flywheel (the loads travel 50% around the Earth circumference) even as polar so there "is" one waiting there.

 

I see this as being fast transportation (for construction by a advanced artificial intelligence), and with minimal energy loss be "letting go, making sue they orbit to the next flywheel arms, spin, let go again, repeat. You have a constant velocity either transporting matter or attaching into these "battery" flywheels.

[Janus]

The advantage. - Couldn't the flywheels be really close to Earth enough so that the orbit to stay up in space is a fast one and can go from one side to the other pretty fast? With less energy wasted than if by rocket pulses.

orbits don't work that way. In order to go around the Earth in less time, you have to drop into a lower orbit which means you have to lose speed.

**But what about a flywheel that is spinning two loads and releases them BOTH at the same time AND neither in the direction of velocity nor from the back end but rather from the sides? To then orbit to a *second flywheel (the loads travel 50% around the Earth circumference) even as polar so there "is" one waiting there.

Again, orbits don't work that way. If you launch payloads at right angles to your orbital velocity, you are are doing a vector addition to your own orbital velocity. At geosynchronous orbit you are moving at better than 3 km sec If you launch payloads at 1 km per sec at right angles to your orbit they will have an orbital velocity of sqrt(3^2+1^2) = 3.16 km/sec putting it into an orbit with a higher average altitude and a longer period. If you fire them at 3 km/ sec , they will be moving at 4.24 km/sec This is actually escape velocity for this altitude and they will leave Earth orbit entirely. The only way to fire them so that they travel half way around the Earth at a right angle to your orbit (relative to its center, not its surface) and end up at the same altitude when they get there is to fire them backwards at at an angle of 45 degrees, at a 4.24 km/sec. Then they will take 12 hrs to get there, but then so will you. You will spend a lot of Delta v just to have it come back to you. You can't think of orbits as being relative to the Surface the Earth, they are relative to a non-rotating frame at its center and the Earth turns underneath independently

I see this as being fast transportation (for construction by a advanced artificial intelligence), and with minimal energy loss be "letting go, making sue they orbit to the next flywheel arms, spin, let go again, repeat. You have a constant velocity either transporting matter or attaching into these "battery" flywheels.

Orbital mechanics is fairly complex and not always intuitive. Trust me when I tell you that your system will not work as you think it will.

[SciNoodle]

If you put it closer to Earth, theirs more gravitational pull, meaning it must be moving faster to orbit.........if you lose speed your orbit will frop but your'll head to the ground since it's no-long 50-50 it'll be ex. 47 for satelite since slowed and 53 for gravity since closer TOO...Your'll need MORE speed than before's orbit once "placed" at the orbit height...

 

You think my idea won't work (not possible)? Off the idea and looking at normal space factories simply constructing in space is all - is it safe to say then (for my AI ideas/project) "AIs will be able to efficiently construct things in space ex. megastructures"?

[Bender]

No, Janus is right: for a lower orbit, you have to slow down. Your speed will be lower, but the distance you need to travel is also smaller, since it is a smaller orbit. Net result is that you go around the Earth faster. Simply speeding up or slowing down will result in an elliptical orbit without course corrections.

 

Just curious:

Why would there be a difference when AI's want to build a megastructure vs humans wanting to do the same?

 

Why would they put something in orbit at the wrong side of the earth and then move it to the other side?

[Janus]

If you put it closer to Earth, theirs more gravitational pull, meaning it must be moving faster to orbit.........if you lose speed your orbit will frop but your'll head to the ground since it's no-long 50-50 it'll be ex. 47 for satelite since slowed and 53 for gravity since closer TOO...Your'll need MORE speed than before's orbit once "placed" at the orbit height...

Like I said, Orbital mechanics is not always intuitive. If I am in a circular orbit and wish to move to a lower circular orbit, I doing the following:

First I fire my engines in the direction I am orbiting to slow down. This puts my speed to being less than needed to maintain a circular orbit at my altitude, so I drop into an elliptical orbit with a perigee closer to the Earth than my present altitude.

The amount I slow down will determine what my this perigee will be, so I make sure to do so by the amount that puts the perigee at the new altitude I wish to orbit.(the only way to fall all the way to the Earth is if I slowed down enough to reduce the perigee so that it dips far enough into Earth's atmosphere.)

As I fall in toward the Earth, I gain speed. By the time I reach perigee, I will be moving too fast to maintain maintain a circular orbit. If I do nothing I will complete the other half of the ellipse and will eventually return to where I started.

Instead, I fire my engines again in the same direction to slow down again. this reduces my speed to that for a circular orbit at my new altitude. My new orbital velocity will be more than it was in my old orbit.

In effect, I apply the brakes twice and end up moving faster.

I could, if I wanted, back off my engines so that instead of slowing down all at once, I did it slowly over a long time. Done right, I could be constantly "applying the brakes" as I moved in towards the Earth and actually be gaining speed the whole time.( The opposite is happening to our Moon, It is being accelerated forward by tidal interaction with the Earth, climbing higher and losingspeed as it does so.)

The trick is that you are trading kinetic energy(and velocity) for a change in Gravitational potential energy.

[MigL]

Falling straight through ( radially or even elliptically ) through the Earth is a valid 'orbit'.

Has a lot more challenges but would be vastly quicker.

Have you seen the new version of Total Recall ( no Arnold, but Kate Beckinsdale !!!) ?

[SciNoodle]

I suspected the answer was elliptical.

 

Anyhow......forget my original orbit concept........................I still want to write something (true) in my book I'm making though.. Is it safe to say that the AIs could efficiently construct megastuctures in space? Cus things stay moving when you push them (low gravity).

[Bender]

Sure. Humans could too

 

(btw, I retract my previous statement about slowing down in lower orbit and instead point to Janus' clear and more correct explanation)

[J.C.MacSwell]

Like I said, Orbital mechanics is not always intuitive. If I am in a circular orbit and wish to move to a lower circular orbit, I doing the following:

First I fire my engines in the direction I am orbiting to slow down. This puts my speed to being less than needed to maintain a circular orbit at my altitude, so I drop into an elliptical orbit with a perigee closer to the Earth than my present altitude.

The amount I slow down will determine what my this perigee will be, so I make sure to do so by the amount that puts the perigee at the new altitude I wish to orbit.(the only way to fall all the way to the Earth is if I slowed down enough to reduce the perigee so that it dips far enough into Earth's atmosphere.)

As I fall in toward the Earth, I gain speed. By the time I reach perigee, I will be moving too fast to maintain maintain a circular orbit. If I do nothing I will complete the other half of the ellipse and will eventually return to where I started.

Instead, I fire my engines again in the same direction to slow down again. this reduces my speed to that for a circular orbit at my new altitude. My new orbital velocity will be more than it was in my old orbit.

In effect, I apply the brakes twice and end up moving faster.

I could, if I wanted, back off my engines so that instead of slowing down all at once, I did it slowly over a long time. Done right, I could be constantly "applying the brakes" as I moved in towards the Earth and actually be gaining speed the whole time.( The opposite is happening to our Moon, It is being accelerated forward by tidal interaction with the Earth, climbing higher and losingspeed as it does so.)

The trick is that you are trading kinetic energy(and velocity) for a change in Gravitational potential energy.

IIRC (I know you will know this), you also waste (brake) half your trade of Gravitational energy to get to a lower circular orbit.

 

I'm less sure (don't clearly see it, but am aware that is what is expected) how the moon gains the extra energy required for a higher/slower orbit...I assume it is robbing some of the Earth's angular momentum with total angular momentum conserved but some loss in kinetic energy (heat etc) beyond the gain in gravitational potential?

Edited February 10, 2017 by J.C.MacSwell

[Bender]

I'm less sure (don't clearly see it, but am aware that is what is expected) how the moon gains the extra energy required for a higher/slower orbit...I assume it is robbing some of the Earth's angular momentum with total angular momentum conserved but some loss in kinetic energy (heat etc) beyond the gain in gravitational potential?

Correct, the moon exerts tidal force on the Earth, slowing down its rotation (you can see this on a gps clock, which is, without correction, 18 seconds ahead of UTC at the moment). In return (Newton's third law), the Earth pulls the moon to a higher orbit, which indeed conserves total angular momentum.

[Janus]

Correct, the moon exerts tidal force on the Earth, slowing down its rotation (you can see this on a gps clock, which is, without correction, 18 seconds ahead of UTC at the moment). In return (Newton's third law), the Earth pulls the moon to a higher orbit, which indeed conserves total angular momentum.

To be accurate, only some of the Earth's lost angular momentum is transferred to the Moon, The rest is lost to tidal heating and radiated away.

[J.C.MacSwell]

To be accurate, only some of the Earth's lost angular momentum is transferred to the Moon, The rest is lost to tidal heating and radiated away.

Is the loss of angular momentum a significant amount? I assume the energy lost to tidal heating is comparatively more (say as a percent of the exchanges...)

What is the mechanism of the angular momentum loss due to radiation?

[Janus]

Is the loss of angular momentum a significant amount? I assume the energy lost to tidal heating is comparatively more (say as a percent of the exchanges...)

What is the mechanism of the angular momentum loss due to radiation?

The loss to radiation makes up all of the difference not gained by the Moon. Tidal heating raises the temp of the Earth, the Earth then radiates more. At some point, the equilibrium is met. The extra energy caused by tidal heating is being radiated away as fast as it is generated (otherwise the Earth would just keep getting hotter and hotter.)

 

The mechanism can be explained this way Imagine a photon being emitted from the Equator at a right angle to the surface. You are watching this from a frame that is not rotating with the Earth. Thus from your frame the source (the surface) has a velocity in the east direction and the photon will show an aberration. ( it will be traveling at a slight angle to the line going straight to the center of the Earth. Also due to the motion of the surface with respect to you, the photon will undergo a slight Doppler shift increasing its frequency/momentum. For momentum to be conserved, the Earth must lose some momentum to compensate.

 

You can also consider two photons, one sent East and the other West. The photon sent East is Doppler shifted to a higher frequency and the one to the West is shifted to a Lower frequency. Again since the momentum of a photon is tied to its frequency, the net momentum of the pair of photons will be in the Easterly direction.

 

So how does this translate in to angular momentum considering that the photons are traveling in straight lines? To see this, draw a straight line from the center of the Earth to the photon. As the photon travels away from the Earth, this line will sweep with an angular velocity of decreasing rate. The line will also be increasing in length. Angular momentum is dependent on the angular velocity times the radius. In this case, one decreases as the other increases, the result is that the product of the two remains constant. This means that a mass that has a constant linear momentum relative to a point (such as the center of the Earth) also has a constant angular momentum.

[J.C.MacSwell]

 

 

You can also consider two photons, one sent East and the other West. The photon sent East is Doppler shifted to a higher frequency and the one to the West is shifted to a Lower frequency. Again since the momentum of a photon is tied to its frequency, the net momentum of the pair of photons will be in the Easterly direction.

 

So how does this translate in to angular momentum considering that the photons are traveling in straight lines? To see this, draw a straight line from the center of the Earth to the photon. As the photon travels away from the Earth, this line will sweep with an angular velocity of decreasing rate. The line will also be increasing in length. Angular momentum is dependent on the angular velocity times the radius. In this case, one decreases as the other increases, the result is that the product of the two remains constant. This means that a mass that has a constant linear momentum relative to a point (such as the center of the Earth) also has a constant angular momentum.

I don't have any problem with the geometry of it (second paragraph above makes sense), but although the bolded seems straight forward and what I would expect on the one hand...on the other how is this any different from translational momentum?

 

For example a radiating particle in space viewed from any reference frame but it's own...using the same thinking should it not be being slowed by it's radiation? (obviously not...but what am I missing?)

[J.C.MacSwell]

I don't have any problem with the geometry of it (second paragraph above makes sense), but although the bolded seems straight forward and what I would expect on the one hand...on the other how is this any different from translational momentum?

 

For example a radiating particle in space viewed from any reference frame but it's own...using the same thinking should it not be being slowed by it's radiation? (obviously not...but what am I missing?)

I think I see this after thinking about it...there would be no net force on the remaining mass of the particle or spinning Earth, just an angular momentum loss associated with the loss in mass from radiating. Since this happens at the Earth's surface that would be a slightly higher loss than average associated with the Earth...but a very small effect overall.

 

I hope these last two posts made sense. Even though the photons have different frequencies they provide no net force on leaving...yet their associated net momentum continues East as it did before. Since that alignment is offset from the Earth's axis, they carry away angular momentum as well.

Edited February 11, 2017 by J.C.MacSwell

[Bender]

To be accurate, only some of the Earth's lost angular momentum is transferred to the Moon, The rest is lost to tidal heating and radiated away.

Interesting. Do you have any source on the relative size of this effect with respect to the torque the moon and the earth exert on each other?

[Janus]

I don't have any problem with the geometry of it (second paragraph above makes sense), but although the bolded seems straight forward and what I would expect on the one hand...on the other how is this any different from translational momentum?

 

For example a radiating particle in space viewed from any reference frame but it's own...using the same thinking should it not be being slowed by it's radiation? (obviously not...but what am I missing?)

Sorry, but I Probably wasn't clear enough with my explanation. As I said, the Earth gives up momentum in emitting the photons, but I neglected to state how it loses that momentum. I never meant to imply that the loss was in the form of a decrease in angular velocity, or that the emission of the photons caused a slowing of the Earth's rotation. It was in the form of the Relativistic momentum expressed by the energy given up by the emission of the photons.

Basically, the process goes like this: Tidal friction slows the Earth's rotation, which in turns heats the Earth, this slower rotating but warmer Earth has the same angular momentum as it had before, some of it is now in the form of relativistic momentum due to the heat energy. The warmer Earth radiates away that heat as photons which carries away angular momentum with them.( the heat lost is constantly being replaced by the above mentioned tidal friction.

Interesting. Do you have any source on the relative size of this effect with respect to the torque the moon and the earth exert on each other?

You can pretty much figure it out for yourself. Just work out how much angular momentum the Moon gains as it climbs away from the Earth and compare that to how much angular momentum the Earth loses in the same time by its slowing. Both these rates are well known.

[J.C.MacSwell]

Sorry, but I Probably wasn't clear enough with my explanation. As I said, the Earth gives up momentum in emitting the photons, but I neglected to state how it loses that momentum. I never meant to imply that the loss was in the form of a decrease in angular velocity, or that the emission of the photons caused a slowing of the Earth's rotation. It was in the form of the Relativistic momentum expressed by the energy given up by the emission of the photons.

Basically, the process goes like this: Tidal friction slows the Earth's rotation, which in turns heats the Earth, this slower rotating but warmer Earth has the same angular momentum as it had before, some of it is now in the form of relativistic momentum due to the heat energy. The warmer Earth radiates away that heat as photons which carries away angular momentum with them.( the heat lost is constantly being replaced by the above mentioned tidal friction.

Right. I think I eventually figured it out as I posted the next day. But that is (on the scale of things and compared to the energy loss), a pretty small amount angular momentum wise. Is it not a very small effect? (not the energy loss, but the associated angular momentum loss, since it would only take away what it had...no "net recoil" on the remaining Earth from radiating)

 

I guess what I am saying is the vast majority of the angular momentum exchange would be between the Earth Moon orbit and the Earth's spin, with the very small difference radiated away.

Edited February 12, 2017 by J.C.MacSwell

[Janus]

Right. I think I eventually figured it out as I posted the next day. But that is (on the scale of things and compared to the energy loss), a pretty small amount angular momentum wise. Is it not a very small effect? (not the energy loss, but the associated angular momentum loss, since it would only take away what it had...no "net recoil" on the remaining Earth from radiating)

 

I guess what I am saying is the vast majority of the angular momentum exchange would be between the Earth Moon orbit and the Earth's spin, with the very small difference radiated away.

As I said in the last post, you can work it out for yourself. If you do, I think you'll be surprised.

[J.C.MacSwell]

As I said in the last post, you can work it out for yourself. If you do, I think you'll be surprised.

 

 

You can pretty much figure it out for yourself. Just work out how much angular momentum the Moon gains as it climbs away from the Earth and compare that to how much angular momentum the Earth loses in the same time by its slowing. Both these rates are well known.

I don't think you can simply take the difference in these two numbers and say the difference is from radiating the waste heat of the tidal process.

 

Here is my reasoning:

 

1. The dissipation/radiating takes place near the Earths surface. Even at the equator the spin speed is a very small fraction of the speed of light, so the redshift/blueshift effect has to be very small. (If all this radiation was directed East, the effect would be some 650,000 times greater)

 

2. Any difference would be much more due to absorption/dissipation/re-radiation of solar energy hitting the Earth...still related to v/c for momentum, but the energy exchange of solar is (50,000 times if my sources are contextually correct) many times greater than tidal.

 

So I am thinking the Earth would lose some angular momentum anyway, the angular momentum robbed by the moon is a much greater effect, but the angular momentum loss from dissipating tidal energy is almost insignificant in comparison.