Post by: simplified on 28/01/2012 15:48:10
Post by: CliffordK on 28/01/2012 20:53:18
The advantages of working on the moon is access to materials, and thin atmosphere minimizing atmospheric friction.
If one chose to limit the centrifugal/centripetal force of the system to 10G, or even 100G, one still ends up with an abysmally slow speed, at least when compared to the speed of light.
However, a lunar base and accelerator would be excellent for building ships in space, and moving around our solar system.
Post by: simplified on 29/01/2012 07:37:16
A while ago I had thought about the advantages of building a ring shaped track around the equator of the moon (tangent to the Earth to minimize the chance of accidents).Thanks. I forgotten about the lunar attraction. What there is the maximal safe speed of leaving of the moon by spacecraft with robots?(your opinion)
The advantages of working on the moon is access to materials, and thin atmosphere minimizing atmospheric friction.
If one chose to limit the centrifugal/centripetal force of the system to 10G, or even 100G, one still ends up with an abysmally slow speed, at least when compared to the speed of light.
However, a lunar base and accelerator would be excellent for building ships in space, and moving around our solar system.
Post by: CliffordK on 29/01/2012 10:10:46
For people, 10g's (say 100m/s2) is a pretty intense acceleration. But, just for the sake of argument, let's say one can pull 100g's (1000m/s2) with good positioning and a good pressure suit, for a short period, and computer controlled navigation.
The two primary forces one would consider would be rearward acceleration, and centrifugal/centripetal acceleration if a circular path is chosen. I had originally only calculated for a circular path... But, let's see.
Moon:
1,737 km radius.
3,474 km diameter
0.1654 g Gravity. Low enough that this can be ignored, especially since I'm estimating 1g=9.8m/s2 --> 10m/s2.
Two possible launchers. One might be to tunnel through the middle of the moon. This would give a straight track of about 3,474 km. There would be a slight Coriolis effect, dependent on your actual acceleration, but let's assume a straight track.
So, at 100g acceleration, 1000 m/s2, or 1km/s2
In about 83 seconds, one reaches a speed of 83km/s, and a distance of about 3486km, and blasts out the other side of the moon.
If, on the other hand, one chose to build a circular track around the equator of the moon (tangential to Earth to avoid collisions),
The centripetal acceleration is a=v2/r
Set a = 1km/s2
r = 1,737 km
Solve for v2 = a*r, or v = sqrt(a*r)
v = sqrt(1,737 km * 1km/s2) = 42 km/s
So, assuming one can tolerate the 100g acceleration, one actually does better with a tunnel through the middle of the moon, reaching about 83km/s, rather than trying a loop around the outside of the moon.
The speed of light is 300,000 km/s
So, with a brutal launch acceleration, you are not even at 1/1000 the speed of light.
I think I had made a similar calculation for a circular track around Jupiter (a track through the middle wouldn't be practical).
Equatorial radius: 71,492 km (I am assuming above atmosphere).
Gravity: 2.528 g
Let's give it 102.5g's * 9.80 m/s = 1004.5 m/s = 1.0045 km/s. Not a lot of help from the intense Jupiter gravity.
v = sqrt(71,492 km * 1.0045 km/s2)
And, I get up to about 268 km/s. Whew... just under 1/1000 c!!!!!!!!!!!!!!!!!!!!!!
But, perhaps one could design a straighter track, the equivalent length of the diameter of Jupiter, or 143,000 km long, with a 100g rearward acceleration. That would give about 535 seconds of acceleration at 100g, and a speed of 535 km/s, just under 2/1000c. It wouldn't have to be perfectly straight, just a tolerable curve radius.
Post by: CliffordK on 29/01/2012 10:20:55
Should I repeat the calcs with 1000g forces? Is that tolerable?
For the straight shot through the middle of the moon, 1000g (10km/s2) acceleration, one gets up to 260km/s, in 26 seconds for 3510 km.
That's getting close to 1/1000c!!!
For the circular centrifugal, at 1000g, one gets.
132km/s. Still better to do high acceleration through the middle of the moon.
Post by: simplified on 29/01/2012 12:09:13
Oh, you said robots... (or perhaps eggs)?If mass of spacecraft=100,000kg
Should I repeat the calcs with 1000g forces? Is that tolerable?
For the straight shot through the middle of the moon, 1000g (10km/s2) acceleration, one gets up to 260km/s, in 26 seconds for 3510 km.
That's getting close to 1/1000c!!!
For the circular centrifugal, at 1000g, one gets.
132km/s. Still better to do high acceleration through the middle of the moon.
v=260 km/s
We can not use energy of photons,we can use only their momentum for acceleration of spacecraft.
p=26,000,000,000 kg*m/s
Therefore we should turn about 80 kg of mass into photons.
Post by: CliffordK on 31/01/2012 09:42:10
The advantage of magnetic acceleration is that the mass of the spacecraft remains more or less constant through the acceleration, unlike the chemical rockets where the initial rocket size is huge compared to the final payload put into orbit.
Voyager 1 was about 722 Kilos (for the deep space probe).
The average communication satellite today is 2,000 to 5,000 kilos, with estimates that they will be on the order of 10,000 kilos in the near future.
Let me try to work out some units (and try out this TEX stuff too)
= 
= 
= 
= 
= 
= 
= 
= 
Ok, so I had estimated a maximum of 1000g of acceleration, over 26 seconds, and 3510 km, and a final velocity of final velocity is 260,000m/s.
Hmmm, how does that all fit in?
Over the first second, one uses the power:
x
And, I think one gets
x 
= x 
I am having a little troubles with my units conversions, and why the power requirements appear to be velocity dependent (and thus would increase over time).
But, anyway, it would take about 50MW per kilo to send the payload off at a 1000g acceleration.
If my final velocity is 260,000m/s, then does that mean that my power requirements ramps up from 50MW per kilo to 2.6GW per kilo?
I guess I'm having troubles understanding why I need to add more energy to maintain the constant acceleration the faster I go [:-\]
Hmmm, for Kinetic Energy:
= 
So, my final Kinetic Energy is:
= x 
= x 
Ok... getting close.
I should be able to divide that by my 26 seconds of acceleration, and get my power requirements as:
Average Power =
x 
= x 
Hmmm, so maybe I was right with the increasing power requirements.
That is going to take a lot of solar panels, although it is only a 26 second launch period. I can imagine a 26 second blackout for my entire lunar colony for every satellite launch!!
I could buffer the power drain somewhat. If a Lead Acid battery can put out 12V x 1000A, or 12KW for a few seconds. I would only need about 217,000 car batteries per kilo to power the satellite launch [::)]
Post by: wolfekeeper on 01/02/2012 18:54:57
Quote
I guess I'm having troubles understanding why I need to add more energy to maintain the constant acceleration the faster I go
Yes, for a given acceleration power is proportional to speed.
Proof:
work = force x distance
differentiating with respect to time:
power = force x speed
A corollary of the first equation is that you need a fixed amount of energy per unit length to give constant acceleration.
(That's true of rockets as well, but there's something sneaky going on, so I'll leave with you the puzzle of how rockets are able to give the same acceleration independent of speed, with the same propellant and the same propellant burn rate.)
Post by: CliffordK on 01/02/2012 20:18:23
(That's true of rockets as well, but there's something sneaky going on, so I'll leave with you the puzzle of how rockets are able to give the same acceleration independent of speed, with the same propellant and the same propellant burn rate.)I saw a discussion about moving reference frames with rockets, however, I'm not convinced it is different.
One of the things about "conventional rockets" is that they have an extraordinary change in mass.
The Saturn V takeoff weight was about 7 million pounds, with about 70 thousand pound spacecraft modules (or about 1%, which is rather large).
Post by: wolfekeeper on 01/02/2012 22:10:12
But, perhaps one could design a straighter track, the equivalent length of the diameter of JupiterPerhaps you could stick it in solar orbit and make it the radius of Earth's orbit (or Jupiter but Earth's is big enough to start with.)
If so, I make it about 380km/s.
I think it would have to be about 72,000 km long, which isn't too bad.
Post by: CliffordK on 01/02/2012 22:47:29
An exercise for another day.
What would it take to mine all the high density material out of Jupiter, its moons, and all the Trojans and Greeks, and make it into a ring or partial ring in its own orbit.
Or, perhaps doing the conversion with Saturn or Neptune.
Also, for a circular track, is the length of track dependent on acceleration (10G, 100G, 1000G), assuming maximum forward acceleration and maximum centrifugal/centripetal acceleration?
Would one wish to combine rearward and centrifugal acceleration vectors? Or just realize that if the centrifugal and rearward acceleration vectors were equal, the true vector would follow a right triangle, 1,1,sqrt(2).
I suppose another thing to consider.
Say one put a track in orbit around the sun. If it was in a tangential orbit, one would get a curve. However, would it be possible to put a long track into its own perpendicular orbit around the sun (somewhat like the space elevator idea). Or would it always rotate around its center of mass back into a tangential orbit? Maybe consider it like a tidally locked planet, with the greater mass towards the sun. Or, maybe one could put it in a very slow spin, so periodically it would become perpendicular. Could one achieve a straighter track at some point in its spin?
Post by: wolfekeeper on 01/02/2012 22:55:09
Say one put a track in orbit around the sun. If it was in a tangential orbit, one would get a curve. However, would it be possible to put a long track into its own perpendicular orbit around the sun (somewhat like the space elevator idea). Or would it always rotate around its center of mass back into a tangential orbit? Maybe consider it like a tidally locked planet, with the greater mass towards the sun. Or, maybe one could put it in a very slow spin, so periodically it would become perpendicular. Could one achieve a straighter track at some point in its spin?Yes, you could put it at a solar L1 or L2 lagrange point perhaps (or possibly L4/5).
Post by: wolfekeeper on 01/02/2012 23:04:45
The other trick is to design it to not vaporise the payload; if you have even a teeny tiny fraction of energy being coupled into the payload, the payload won't make it to the end of the track, so essentially all of the waste heat has to be in the track, where it can be dealt with.
So the track has to be really smooth.
Post by: CliffordK on 02/02/2012 00:11:29
I'm not sure I'd hire you as an engineer...However, would it be possible to put a long track into its own perpendicular orbit around the sun (somewhat like the space elevator idea).Yes, you could put it at a solar L1 or L2 Lagrange point
A perpendicular orbit in L1 would blast the probe either into the sun, or into the Earth... So, I'd probably choose a different position for it. I suppose it wouldn't take more than a degree or so to miss the Earth & Sun, but I think I'd choose a different location to put the accelerator.
If the goal was to power the launches with solar energy, L2 does receive some solar energy, but it is significantly less than other locations.
L4&L5 might be a good place to put it in a slow spin. Could one power the rotation with a non-particle energy such as a solar sail, or solar magnetic power?
Post by: wolfekeeper on 02/02/2012 06:19:55