The Spontaneous Beginning of Spacetime

[Anomalocaris]

I always though so long as our science still requires us to use rocket to get to space, we don't have good enough scientists.

Just be patient. The nano-wire space elevator is coming.

I never understood how space elevator can counteract without using rockets and fuel the fact that as it moves mass up, it's moment of inertia is increasing, while its angular momentum is not.      If it can't increase its angular momentum, it will gradually begin to lag the rotation of the earth until it eventually wrap itself around the earth as it goes flaccid and collapse.

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And when it does finally come, it'll be able to fire off huge loads up into the sky.

I'm pretty sure when they invented the theory of panspermia, that's not what they had in mind.

It's only spermia in the plane of the rotation of the earth, not panspermia.

[Alex K]

I think you're right, if we're bringing something up without taking sth of equal mass back down, the elevator needs to give the payload angular momentum - which it can't do by itself without slowly losing its orbit because it's not rigid and hence can't funnel the angular momentum from earths rotation. That can be done by a rocket firing sideways, and I'll do the calc over dinner how much power that would need, give me a minute

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Just be patient. The nano-wire space elevator is coming.

I never understood how space elevator can counteract without using rockets and fuel the fact that as it moves mass up, it's moment of inertia is increasing, while its angular momentum is not. If it can't increase its angular momentum, it will gradually begin to lag the rotation of the earth until it eventually wrap itself around the earth as it goes flaccid and collapse.

Allrighty!

As the resident rocket scientist (or so they tell me, I ain't never seen no rocket in my entire life) I shall deliver my space elevator calculation.

So, let's assume we want to move one metric tonne of material up all the way to geostationary orbit.

The geostationary orbit is roughly 36000 km away from the center of the earth. We are starting on the surface at the equator, which is roughly 6000 km away from the center. The earth rotates once every 24 h, or once every 86400 seconds. Hence, our metric tonne of stuff starts out with an linear momentum of

p0 = 1000kg * 6000 km / 86400 sec ~ 70 km kg/sec

up at the geostationary orbit, it needs to have an linear momentum of

p1 = 1000kg * 36000 km / 86400 sec ~ 420 km kg/sec

in order to still move with the correct speed and not slow the space elevator wire down.

In other words, we need a rocket engine which pushes sideways and gives our payload a momentum of

p1-p0 ~ 350 km kg/sec =350000 meter kg/sec = 350 Kilonewton*sec

Now we know what force we have to apply how long to give this linear momentum to our tonne of payload.

Now, a single engine of an F-14 has, without afterburner, a thrust of 60 Kilonewtons. In order to give our space lifted payload the necessary momentum, it therefore needs to fire for roughly 6 seconds at full power - a trivial feat, technologically.

In this calculation I have neglected that the direction of the space elevator changes when the earth turns, and hence have assumed that the journey up takes much less than 24 h, in order to allow for a simple calculation with absolute values of linear momenta instead of angular momenta and coriolis/centrifugal forces. The differences to the full calculation with a rotating frame should be minor, with some extra thrust to compensate for the centripetal force needed to lift the cargo. I think the above gives a very good estimate of the order of magnitude of jet engine power necessary to compensate for coriolis force. However, I should check that

[Alex K]

Another assumption here is that the radial momentum which you put on the counterweight in order to accelerate the payload is paid back upon braking. The momentum you put on the counterweight in order to *lift* it against gravity is, if I see it correctly, a function of the time you take to lift the object, and one should be able to reduce it by lifting quickly. But this is a bit messy and one possibly needs to do the full calculation after all to understand this.

[Anomalocaris]

I think you're right, if we're bringing something up without taking sth of equal mass back down, the elevator needs to give the payload angular momentum - which it can't do by itself without slowly losing its orbit because it's not rigid and hence can't funnel the angular momentum from earths rotation. That can be done by a rocket firing sideways, and I'll do the calc over dinner how much power that would need, give me a minute

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I never understood how space elevator can counteract without using rockets and fuel the fact that as it moves mass up, it's moment of inertia is increasing, while its angular momentum is not.      If it can't increase its angular momentum, it will gradually begin to lag the rotation of the earth until it eventually wrap itself around the earth as it goes flaccid and collapse.

Allrighty!

As the resident rocket scientist (or so they tell me, I ain't never seen no rocket in my entire life) I shall deliver my space elevator calculation.

So, let's assume we want to move one metric tonne of material up all the way to geostationary orbit.

The geostationary orbit is roughly 36000 km away from the center of the earth. We are starting on the surface at the equator, which is roughly 6000 km away from the center. The earth rotates once every 24 h, or once every 86400 seconds. Hence, our metric tonne of stuff starts out with an linear momentum of

p0 = 1000kg * 6000 km / 86400 sec  ~ 70 km kg/sec

up at the geostationary orbit, it needs to have an linear momentum of

p1 = 1000kg * 36000 km / 86400 sec ~ 420 km kg/sec

in order to still move with the correct speed and not slow the space elevator wire down.

In other words, we need a rocket engine which pushes sideways and gives our payload a momentum of

p1-p0 ~ 350 km kg/sec  =350000  meter kg/sec = 350 Kilonewton*sec

Now we know what force we have to apply how long to give this linear momentum to our tonne of payload.

Now, a single engine of an F-14 has, without afterburner, a thrust of 60 Kilonewtons. In order to give our space lifted payload the necessary momentum, it therefore needs to fire for roughly 6 seconds at full power - a trivial feat, technologically.

In this calculation I have neglected that the direction of the space elevator changes when the earth turns, and hence have assumed that the journey up takes much less than 24 h, in order to allow for a simple calculation with absolute values of linear momenta instead of angular momenta and coriolis/centrifugal forces. The differences to the full calculation with a rotating frame should be minor, with some extra thrust to compensate for the centripetal force needed to lift the cargo. I think the above  gives a very good estimate of the order of magnitude of jet engine power necessary to compensate for coriolis force. However, I should check that

If I am not mistaken, you underestimate the momentum and changes in momentum by the ratio between circumference and radius.  So the total increase in momentum ought to be about 2,000,000 Newtons.   So an F-14 engine needs to burn for about half minute.   Since an elevator cable is not rigid, one could not simply fire the engine for a 36 second burst after the load gets to the top.   It has to fire continuously all the way up to keep the cable from deflecting under the angular momentum of the payload.   So a light, high efficiency, long duration, low thrust engine like an ion engine would work better than a rocket, at least once the payload is out of the atmosphere.

[pocaracas]

Dudes... you're forgetting one very important way to make work without any effort: MAGNETS!
Magnetize the nano-wire, like a maglev.... and just see it go!

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[Spooky]

That fact that we don't know doesn't make guesses that much more credible.

Shouldn't*

[Alex K]

If I am not mistaken, you underestimate the momentum and changes in momentum by the ratio between circumference and radius.  

Crap. Where?

[Anomalocaris]

If I am not mistaken, you underestimate the momentum and changes in momentum by the ratio between circumference and radius.  

Crap. Where?

Veolcity is distance / time.  Distance is the circumference of the earth or of the geosynchronous orbit, not the radius, if the time is 24 hours.

[Alex K]

Crap. Where?

Veolcity is distance / time.  Distance is the circumference of the earth or of the geosynchronous orbit, not the radius, if the time is 24 hours.

Oh noes, I did in fact miss a factor of 2 pi, didn't I. Embarrassing, but to my defense, it is really late and I was half asleep when I typed it.

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I think you're right, if we're bringing something up without taking sth of equal mass back down, the elevator needs to give the payload angular momentum - which it can't do by itself without slowly losing its orbit because it's not rigid and hence can't funnel the angular momentum from earths rotation. That can be done by a rocket firing sideways, and I'll do the calc over dinner how much power that would need, give me a minute

---

I never understood how space elevator can counteract without using rockets and fuel the fact that as it moves mass up, it's moment of inertia is increasing, while its angular momentum is not.      If it can't increase its angular momentum, it will gradually begin to lag the rotation of the earth until it eventually wrap itself around the earth as it goes flaccid and collapse.

Allrighty!

As the resident rocket scientist (or so they tell me, I ain't never seen no rocket in my entire life) I shall deliver my space elevator calculation.

So, let's assume we want to move one metric tonne of material up all the way to geostationary orbit.

The geostationary orbit is roughly 36000 km away from the center of the earth. We are starting on the surface at the equator, which is roughly 6000 km away from the center. The earth rotates once every 24 h, or once every 86400 seconds. Hence, our metric tonne of stuff starts out with an linear momentum of

p0 = 1000kg * 2 pi * 6000 km / 86400 sec  ~ 440 km kg/sec

up at the geostationary orbit, it needs to have an linear momentum of

p1 = 1000kg * 2 pi * 36000 km / 86400 sec ~ 2600 km kg/sec

in order to still move with the correct speed and not slow the space elevator wire down.

In other words, we need a rocket engine which pushes sideways and gives our payload a momentum of

p1-p0 ~ 2200 km kg/sec  =2200000  meter kg/sec = 2200 Kilonewton*sec

Now we know what force we have to apply how long to give this linear momentum to our tonne of payload.

Now, a single engine of an F-14 has, without afterburner, a thrust of 60 Kilonewtons. In order to give our space lifted payload the necessary momentum, it therefore needs to fire for roughly 37 seconds at full power - a relatively trivial feat, technologically.

In this calculation I have neglected that the direction of the space elevator changes when the earth turns, and hence have assumed that the journey up takes much less than 24 h, in order to allow for a simple calculation with absolute values of linear momenta instead of angular momenta and coriolis/centrifugal forces. The differences to the full calculation with a rotating frame should be minor, with some extra thrust to compensate for the centripetal force needed to lift the cargo. I think the above  gives a very good estimate of the order of magnitude of jet engine power necessary to compensate for coriolis force. However, I should check that

[Pyrrho]

...

As the resident rocket scientist (or so they tell me, I ain't never seen no rocket in my entire life) ...

Seriously? Have you not seen one in a museum? How about a bottle rocket, or other tiny fireworks rocket?

I am not familiar with the museums in Germany that might have rockets, but you can see them in the U.S., at places like either building of the Smithsonian Air & Space Museum (in the Washington, D.C. area; one in town, one near Dulles Airport), at the Air Force Museum in Dayton, Ohio, and at the places where they actually launched rockets, like the Kennedy Space Center in Cape Canaveral, Florida. I am sure that there are other places where you can see rockets, which you should be able to find with a little online searching.