Question about space elevators:

Originally posted by joe shmo
Yeah that's what I meant,...Is the universes internal energy increased by the reaction?

No, it stays the same. With the possible exception of dark energy, which is not well understood and may not exist, the energy of the universe remains constant.

Originally posted by joe shmo
Yeah that's what I meant,...Is the universes internal energy increased by the reaction? I'm thinking that it would have to since the universe lost mass.

Energy can be converted to mass and mass can be converted to energy.
So if you measure all the energy and mass in the same energy unit, then the sum of it all is thought of being the same throughout the life of the universe.

Originally posted by DeepThought
I don't think the magnetic field idea is a goer. However you'd need a pretty hefty magnetic field to protect the crew from cosmic rays, whether they're in deep freeze or awake.

For a grain of dust weighing a milligram a collision at a relative speed of a tenth of the speed of light would release about 450 megajoules of energy, or about 125 kWH. I do ...[text shortened]... o greater problem than getting to that speed in the first place, but economically it's horrible.

I forgot to mention that we shouldn't just expect one collision with space dust but a great many such collisions during an interstellar journey. In fact, we should expect every square millimeter of any outer layer of a Whipple shield to be peppered with at least a few holes from dust impacts. Because of the way a Whipple shield works, especially if the dust strikes it at ~0.1c, each impact would create a tiny hole in the first layer it strikes (and the dust particle is vaporized into a kind of destructive fire ball in the process ) but that hole would be much larger further back and will get progressively larger each layer you go further back to create a kind of funnel-shaped volume of destruction of significant volume. The problem is that well before the journey is over, there would be no effective Whipple shield left as all its essential back layers would have been completely exploded off will before then. Then when the subsequent load of collision occurs, it would be spaceship that gets it.

I once tried to work out many workarounds this but each has its own big problems. The most obvious workaround would have many spare layers and just keep replacing or repairing the damaged layers ofter each collision. But that would require an awful lot of mass of spare layers and an awful lot of repair work during the journey!

Originally posted by humy
Trust me, you wouldn't want to use an antimatter shield against collisions with matter even if that matter is only dust size! Just one collision would be like a small atomic explosion that would blow up the shield and anything (like a spaceship ) attached to it.

At relatively 'low' velocities, a Whipple shield would do just fine:
http://en.wikipedia.org/wik ...[text shortened]... s pumped into them to generate that sort of magnetic field strength.
-Anyone else have an idea?

My idea would be to use huge magnetic fields to deflect dust, combined with extreme high voltage fields but instead of a craft that looks like a normal space craft, that is to say, a long cylinder, instead, it would be a giant donut shape, the idea here being the dust won't be guided around the ship, but through it, where there IS no ship! So the dust would just harmlessly pass through the center of the donut where the magnetic fields guide the stuff.

Also, a powerful laser could also be use to ionize dust particles before they get to the ship, allowing easier control of the dust particles.

Originally posted by sonhouse
My idea would be to use huge magnetic fields to deflect dust, combined with extreme high voltage fields but instead of a craft that looks like a normal space craft, that is to say, a long cylinder, instead, it would be a giant donut shape, the idea here being the dust won't be guided around the ship, but through it, where there IS no ship! So the dust would ...[text shortened]... onize dust particles before they get to the ship, allowing easier control of the dust particles.

Also, a powerful laser could also be use to ionize dust particles before they get to the ship, allowing easier control of the dust particles.

I had also thought of that. But then I also thought how would the dust particles be detected in time and also to allow the lasers time to ionize them if each dust particles is approaching at a massive speed relative to the spacecraft of ~0.1c!?

Originally posted by humy

Also, a powerful laser could also be use to ionize dust particles before they get to the ship, allowing easier control of the dust particles.

I had also thought of that. But then I also thought how would the dust particles be detected in time and also to allow the lasers time to ionize them if each dust particles is approaching at a massive speed relative to the spacecraft of ~0.1c!?

I was thinking using sufficiently strong magnetic fields, 100 tesla or more, future engineering project🙂 you might not need lasers, the fields, especially with super strong electric fields, 100 million volts or more, might just guide the particles into the center of the donut without needing the laser.

Lasers could be used just to trash larger particles that won't be guided by those huge fields, in that case you have milliseconds of time, 1 millisecond puts the particles at around 20 miles out when the laser hits it.

Originally posted by sonhouse
I was thinking using sufficiently strong magnetic fields, 100 tesla or more, future engineering project🙂 you might not need lasers, the fields, especially with super strong electric fields, 100 million volts or more, might just guide the particles into the center of the donut without needing the laser.

Lasers could be used just to trash larger particle ...[text shortened]... seconds of time, 1 millisecond puts the particles at around 20 miles out when the laser hits it.

But getting back to the space elevator, I calculated the escape velocity of the tip of the cable, at 106,378 Km from the center of Earth, given the cable is stable and stretched out straight, the velocity there is about 27800 meters per second.

So using the escape velocity formula, I find escape velocity at that altitude at 1/10th that of what is already there, 2737 meters per second. Verify?

So that means just launching off the end gives you a 27000 meter per second velocity for free to launch anywhere in the solar system since you have 360 degrees of freedom in that circular path. It would probably not be totally 360 given the plane of the ecliptic but would be a wide choice of launch angles.

I also calculated the escape velocity of the sun at 1 au and that came out at about 34340 meters per second. So all you need is to add about 7000 meters per second to the cable tip velocity and you have passed the escape velocity of the sun!

Can anyone verify these numbers? I used V=(2GM/r)^1/2 where V escape velocity in meters per second, mass in kg, r in meters. Earth mass at 5.976 E24 kg and the sun at ~2E30 Kg and r for Earth at 1.06 E6 meters and r for the sun at 150 E9 meters and G at 6.67 E -11.

If you need slower launch velocity you just launch from lower down the cable.

Originally posted by sonhouse
But getting back to the space elevator, I calculated the escape velocity of the tip of the cable, at 106,378 Km from the center of Earth, given the cable is stable and stretched out straight, the velocity there is about 27800 meters per second.

So using the escape velocity formula, I find escape velocity at that altitude at 1/10th that of what is alread ...[text shortened]... at 6.67 E -11.

If you need slower launch velocity you just launch from lower down the cable.

It is clear that the end of the cable is already over the value of the escape velocity. Because if it wasn't it would fall back to the earth surface.

If you want to launch anything from the end of the cable - just let go at the right moment and you will be flung out at the peripheral velocity.

What about the sun...? You want to attach a cable to the surface of the sun and use a space lift from there...?

Originally posted by FabianFnas
It is clear that the end of the cable is already over the value of the escape velocity. Because if it wasn't it would fall back to the earth surface.

If you want to launch anything from the end of the cable - just let go at the right moment and you will be flung out at the peripheral velocity.

What about the sun...? You want to attach a cable to the surface of the sun and use a space lift from there...?

No🙂 Just that at the end of the cable, if my numbers are right, a small increase of 7 km/second added to the already 27+Km/second tip velocity, you have surpassed the escape velocity of the sun, which I calculated to be about 34Km/second at 1 AU. Verify my numbers?

That is not even counting the velocity of Earth itself in its orbit around the sun. Not sure what that velocity is but any orbit times the square root of 2 is escape velocity so the orbital velocity seems to be around 26Km/second so together it can add up to 53Km/second so depending on which direction you let go you can already be doing much more than solar escape velocity. Going in the opposite direction you can have almost zero km/second with respect to the sun which could help get to the inner planets.

Originally posted by sonhouse
No🙂 Just that at the end of the cable, if my numbers are right, a small increase of 7 km/second added to the already 27+Km/second tip velocity, you have surpassed the escape velocity of the sun, which I calculated to be about 34Km/second at 1 AU. Verify my numbers?

That is not even counting the velocity of Earth itself in its orbit around the sun. Not s ...[text shortened]... an have almost zero km/second with respect to the sun which could help get to the inner planets.

Well, we forget about the sun, and just concern ourselves about a space elevator from the earth, shall we?

If I ask you the following questions:
Have you calculated how long this elevator wire should be if it is fastened at a site at the equator? I.e. how far from the surface is the operational platform at the end of the cable be? How much further is it from the geostationary point?
What mass per meter has the cable? Do you need different mass/meter at different heights? What is it at the geostationary point?
What mass does the platform at the end of the cable have?
What margin do you calculate would be necessary to compensate the transportation of mass from surface to the platform? What would be its lift capacity?
Will the transportation unit use its own motor, or is it lifted from the platform from the outer side?
... are these questions taken into consideration in you calculation?

As I suppose, your escape velocity is from a stationary point, rather from an orbiting point. Do you really take the orbit parameters in consideration in your calculations?

Originally posted by FabianFnas
Well, we forget about the sun, and just concern ourselves about a space elevator from the earth, shall we?

If I ask you the following questions:
Have you calculated how long this elevator wire should be if it is fastened at a site at the equator? I.e. how far from the surface is the operational platform at the end of the cable be? How much further is ...[text shortened]... n orbiting point. Do you really take the orbit parameters in consideration in your calculations?

The real calculations would of course be a lot more complex for travel to a given planet or interstellar journey, I just did the calculation of the velocity relative to Earth. I did include the orbital velocity of Earth when I calculated the velocity at the top of the cable and the angle of release of a craft at the top can be way over the escape velocity of the sun or way under it depending on the angle of release.

If you released in a direction going backwards from the direction of travel of Earth in its obit around the sun, the velocity would be reduced a lot from that if you chose an angle of release that was in the same direction as the Earth obits the sun.

I think all the mass, mass per meter and such has all been worked out for theoretical cables, probably 20 years ago or more. It pretty much HAS to be fastened to Earth close to the Equator, since the tilt of the Earth at 22.5 degrees does not allow one from the poles as we already worked out.

The geo point is about 22,500 miles or 36,000 kilometers high and the top end counterbalance is supposed to be at 100,000 kilometers up or 62,500 miles.

It is clear you have to add velocity to the top mass if you want it to not fall behind and create havoc.

I don't think it much matters what the geo point mass is, I think the mass at the top of the cable would be more important to know since it is stretching the cable through centripetal forces and the more mass the more stress there would be on the cable so there would be a limit to that mass determined by the ultimate strength of the cable.

The geo point platform doesn't even have to add stress to the cable, just be fastened at the geo stationary point where the cable and the platform would both be basically in the same orbit so the stress on the cable due to the platform would be minimal.

It would be at the end of the cable where mass Vs length and cable strength would come into play.

One version of the lifter goes like this: There is a massive laser pointing up and the laser energy is concentrated below the lifer and that energy powers wheels attached to the cable by friction forces.

The time to get to the geostationary platform was worked out to be in the order of 2 weeks to go from the surface of Earth to the Geo point.

Originally posted by FabianFnas
Energy can be converted to mass and mass can be converted to energy.
So if you measure all the energy and mass in the same energy unit, then the sum of it all is thought of being the same throughout the life of the universe.

With the caveat that at the end of the universes life we will not be able to do any useful work (entropy generation for any real process). That basically states that (in time) the universe energy per unit volume will be constant over the entirety of space and the universe will be at a uniform temperature? Is this a correct interpretation of the Second Law of Thermodynamics as it applies to the universe?

Originally posted by joe shmo
With the caveat that at the end of the universes life we will not be able to do any useful work (entropy generation for any real process). That basically states that (in time) the universe energy per unit volume will be constant over the entirety of space and the universe will be at a uniform temperature? Is this a correct interpretation of the Second Law of Thermodynamics as it applies to the universe?

There are a number of scenarios for the end of the universe that would change that. If the universe was not expanding then yes, it would become homogenous. If the universe stopped expanding and re-contracted into a big crunch then the end of the universe would end as a singularity (modulo quantum gravity effects). As it is the universe is expected to end as an anti-deSitter space assuming they are right about the accelerated expansion of the universe. This would mean that on average the universe would be empty, with the occasional isolated particle at a to all intents and purposes infinite distance from any other particle. It's a little meaningless to assign temperatures to systems like that.

Originally posted by DeepThought
As it is the universe is expected to end as an anti-deSitter space assuming they are right about the accelerated expansion of the universe. This would mean that on average the universe would be empty, with the occasional isolated particle at a to all intents and purposes infinite distance from any other particle.

I propose we call that scenario the Big Bummer.

Originally posted by joe shmo
With the caveat that at the end of the universes life we will not be able to do any useful work (entropy generation for any real process). That basically states that (in time) the universe energy per unit volume will be constant over the entirety of space and the universe will be at a uniform temperature? Is this a correct interpretation of the Second Law of Thermodynamics as it applies to the universe?

"... end of the universes life we will not be able ..."

We? Who we? When 'we' cannot do any useful work anymore, then 'we' has seized to exist long time ago. At the extreme conditions at this time, the laws of physics doesn't act in 'our' favor exactly...