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Fast Efficient Orbital Transportation?


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#1 SciNoodle

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Posted 19 January 2017 - 10:05 PM

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.


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#2 Strange

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Posted 19 January 2017 - 10:21 PM

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.


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#3 Bender

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Posted 19 January 2017 - 10:46 PM

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 by Bender, 19 January 2017 - 10:46 PM.

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#4 SciNoodle

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Posted 20 January 2017 - 12:27 AM

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.


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#5 Janus

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Posted 20 January 2017 - 09:21 PM

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?


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#6 SciNoodle

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Posted 21 January 2017 - 04:09 AM

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.


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#7 Janus

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Posted 21 January 2017 - 06:09 AM

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.
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#8 SciNoodle

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Posted 21 January 2017 - 10:16 AM

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"?


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#9 Bender

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Posted 21 January 2017 - 09:30 PM

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?


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#10 Janus

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Posted 22 January 2017 - 12:16 AM

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.
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#11 MigL

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Posted 22 January 2017 - 01:18 AM

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 !!!) ?


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#12 SciNoodle

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Posted 22 January 2017 - 04:00 AM

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).


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#13 Bender

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Posted 22 January 2017 - 06:17 PM

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)


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#14 J.C.MacSwell

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Posted 10 February 2017 - 12:55 AM

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 by J.C.MacSwell, 10 February 2017 - 01:30 AM.

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#15 Bender

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Posted 10 February 2017 - 10:01 AM

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.


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#16 Janus

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Posted 10 February 2017 - 06:45 PM

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.


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#17 J.C.MacSwell

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Posted 10 February 2017 - 08:23 PM

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?


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#18 Janus

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Posted 10 February 2017 - 08:58 PM

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.


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#19 J.C.MacSwell

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Posted 10 February 2017 - 09:34 PM

 

 

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?)


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#20 J.C.MacSwell

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Posted 11 February 2017 - 04:58 PM

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 by J.C.MacSwell, 11 February 2017 - 05:31 PM.

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