I think the Mars series has a bunch of references to this, too. Basically if we ever had to go to war with the moon or mars colonies, it would just be kinetic weapons.
You might be able to detect acceleration at point of impact if it’s at an angle to the velocity at point of impact. This is assuming the mass driver is allowing the asteroid to fuel its own acceleration via mass drivers + a fuel source, not a separate frame of reference mass driving the asteroid into orbit.
On Earth it would absolutely show up. Because Earth is tectonically active, and has an active atmosphere that erodes things in relatively predictable ways, along with a number of other processes that can date things, we'd be able to tell that there's a whole bunch of impacts that seem to have taken place at suspiciously close to the same time, even millions of years after the fact. If there's more than a couple of them, it may even be the case that it becomes clear that they were specifically targeted; e.g., 2/3rds of the planet is water, if we found 8 impacts on land or coasts at the same time and could show that there were 0 in the middle of the water, that would be very strong evidence for intelligent action, for which "military" is the most likely conclusion.
In other environments, it may be much more difficult, but similar effects on more detailed scales may be in play.
It's possible the space war was won with One Big Rock, but for a variety of reasons, I'd expect space war is more likely to occur with a variety of smaller rocks. Primarily, One Big Rock is harder to steer at militarily useful speeds, and you have to be pretty committed to wiping out not just your enemy, but the entire ecosystem. A lot of much smaller rocks would be more militarily useful in all sorts of ways, speed of deployment and redundancy being among the most important. If there is a war between a space-faring civilization and the groundpounders, the space-farers win if they just degrade the ground civilization to the point that they are no longer space-faring and unable to interfere. Spending the considerably greater resources to massively overkill past that goal is possible, but would not be the most likely outcome in my opinion because of the outsize expenditures and much longer time to return on said expenditure vs. smaller, more nimble attacks.
Or maybe mutual annihilation? The dinosaurs where more ruthless and ended up wiping out the whole biome while martians limited themselves and ended up only killing the dinosaurs but letting Life survive.
Depends on how big the meteor. Large enough natural explosions would tend to lay down a layer rich in iridium in the sediments. But if it's just an explosion the size of the Tunguska event nobody looking a million years in the future will ever know it happened.
same as aurora 4x you can use the mass driver both to transfer minerals if the other end has one of equal strength to capture them or bombardment. heck it even simulates the bombardment dust cloud triggering a runaway cooling and subsequent ice age of you send enough mineral packets in short time!
> "In contrast to cargo-only chemical space-gun concepts, a mass driver could be any length, affordable, and with relatively smooth acceleration throughout, optionally even lengthy enough to reach target velocity without excessive g forces for passengers."
Wikipedia can generally be pretty good, but the above text from the subsection On Earth shows what can go wrong when deluded space enthusiasts go nuts on an article; it's pure wishful thinking. No major construction project of "any length" is "affordable," especially not one of the sophistication of a mass driver. At about 3g, about 30 m/s^2, it would take you about 270 seconds to get to 8 km/second, during which time you are traveling at an average speed of 4 km/second. This is a structure, with attendant control system and safeguards, about 1080 km, or 670 mi, in length. This is not affordable. By anyone.
What might be affordable (sticking my wet finger in the breeze) is getting 1 or 2 km/s at 5 or 6 g, or higher, and then doing the rest with a rocket, provided that you're firing continuously to amortize your initial cost. Skimming a couple km/s off the high end of the rocket equation brings your mass ratio way down, especially considering the reduced gravitational losses. (I'm assuming, in a fit of witless optimism, that we're launching from a high enough altitude that the increased aerodynamic losses are mitigated to the point that gravitational loss savings dominate, and that this adds no cost.)
Mass drivers are a fun idea because they seem to get around the rocket equation, but this affordability nonsense is just over the top.
I figure mass drivers are for non-atmosphere situations. In atmosphere it seems more likely that other exotic ideas are more feasible. The longer ones seem to be for a civilization that has a construction pipeline so automated that building a multi-hundred-mile structure isn't as insane as the idea is today.
The atmosphere is nice in a lot of ways, but it is not helpful in getting to space today.
The major problem with mass drivers in Earth atmosphere is that anything like escape velocity combined with atmosphere introduces a lot of G forces and a ton of heat. (For intuition, imagine reentry, except even faster at lower altitudes.)
It might be feasible to use a mass driver to substitute for a first stage, especially if the mass driver mouth is high enough up to substantially reduce air resistance on the things coming out - the top of Mt. Kilimanjaro say. But that would be at most a few km/s of the 8 you need to get into low Earth orbit.
Yeah, you need a 2nd stage to make it remotely viable. A single stage mass driver would need to have escape velocity minus drag loss at the exit mouth, which is vastly faster than any real chemical rocket is going at ~10-15 km elevation, where the air is still reasonably dense. Also, 10-15 km is around the range of Max Q for real-world rockets and given the equation, q = 0.5 p v², a higher velocity is going to result in some extreme structural stress.
Run it in a tunnel (cut and cover to reduce costs) with a reduced atmospheric pressure that goes up the side of an equatorial mountain. Mt. Kenya is the best prospect from what I remember the last time I looked at this. It is just south of the equator so you get max angular momentum from the spin of the earth, it is about 5km above sea level so you are at half the atmospheric pressure as sea level, and it is relatively close to a downrange ocean in case the launch does not go as planned.
The Moon is a Harsh Mistress by Robert Heinlein speaks about this. Rebels on the moon list it as a demand from the Earthside governments to build a mass driver to lower costs of shipping and make moonside life more equitable.
They even list a couple of ideal locations including I believe Kenya.
The tunnel can't just be cut and cover, because you can't take sharp turns of any kind and there's no straight-line gradient up a mountain. Yes, a termination at a high equatorial peak helps, but there's still a lot of air there that you're immediately buffeted by as soon as you leave the tunnel.
"This is a structure, with attendant control system and safeguards, about 1080 km, or 670 mi, in length. This is not affordable. By anyone."
The Hyperloop[1] is a proposal for a 350 mile structure that actually sounds remarkably similar to a mass driver, at a projected cost of $6 billion -- quite affordable for governments, corporations, and even some individuals.
A train is 150+ year old technology that could not be simpler and yet the cost of building out high speed rail is 10 times the number you quoted at least. You're delusional if you think experimental technology is going to be cheaper.
Reading this made me wonder a few physics questions.
- If you shoot a gun on the moon, will it return to your location from the behind, no matter what angle you shoot it?
- The escape velocity of the moon is 2.38km/s. Is this the velocity required for a bullet to leave orbit no matter what angle you shoot it? Or is the required velocity higher at smaller angles?
2) * Escape velocity is the velocity required to counter the force of gravity, ie 'away' or 'out' from the center of the planet, opposite the force of gravity.
* Edit: Looks like I'm wrong, guess I slept that day in physics
Escape velocity is actually the same in any direction, not just directly upward. This is true because in a two-body system, whether or not a trajectory escapes to infinity depends only on whether the sign of specific orbital energy[1] is positive or negative. Specific orbital energy is a function of kinetic energy and potential energy, but is independent of the direction of motion.
> If you shoot a gun on the moon, will it return to your location from the behind, no matter what angle you shoot it?
In most cases no. Although the orbit of the bullet is cyclical (below escape velocity), unless you shoot perfectly level with the surface, its orbit will intersect the moon. Obviously if you shoot slightly down, it'll hit the ground. If you shoot slightly up, it'll hit the ground behind you on its return.
> Is this the velocity required for a bullet to leave orbit no matter what angle you shoot it?
Yes. Although if it hits the moon first, it'll slow down and not escape (unless you were to shoot through the moon).
Bullets are sealed. They don't use the atmosphere as an oxidizer. Doesn't mean they fire well in a vacuum, because there are other factors involved. But you could certainly design a firing system to work perfectly fine in a vacuum.
- Ah, I see. Does there always exist a velocity `v(theta)` that will make the bullet hit you in the back after one orbit? A slightly different question, but if you shoot a bullet perpendicular to the moon's surface at any speed (as long as it won't hit the ground), it will always hit you in the back after one orbit, right?
- I suppose if you think about it in terms of kinetic energy, this makes sense.
I realize that I made an assumption in my first answer, that the moon is not rotating. A rotation may cause you to move out of the orbital path of the bullet before it gets back to you. If you shot from the poles though, you would remain in the bullet's orbit.
That said, in ideal conditions, even with rotation, the bullet will hit you eventually, but not for a long time on average, given the scale of the moon. In reality it's orbit would probably destabilise due to collisions with dust and perturbations from other gravitational bodies (the Earth, Sun, etc. create an n-body situation which slowly changes the orbit over time).
> Does there always exist a velocity `v(theta)` that will make the bullet hit you in the back after one orbit?
The velocity must be great enough to stay above the surface, and not high enough to escape. Additionally, you must shoot horizontally, like I mentioned before, and not rotate out of the orbit (e.g. by being at the poles, or at the rotational equator, shooting along the equator).
Fragile cargo at least. For some otherwise fragile things, you could submerge them in an incompressible fluid of similar density to reduce the difference in force throughout them. Your brain's tissue is very soft, yet by being buoyant in its cerebrospinal fluid it becomes capable of withstanding much higher acceleration.
You're correct but... the effect cushioning the brain is insufficient to withstand the rapid deceleration of a car accident. It would be hard to imagine a mass driver with low enough force that a human could live through it.
What defines precious cargo? Satellites and astronauts seem to do alright going into space attached to a giant rocket, but I will admit I'm not well versed on the forces created from a electromagnetic launcher.
So here's one way to think about the accelerations. To a first approximation, you need the same total change in velocity to go from standing on the surface of an object to be in orbit at a given altitude. You can choose to spread that change in velocity over any given time and distance interval (those two terms are linked). So if your mass driver accelerates your payload to final velocity in 1/10th the time (and more or less distance), you will get 10x the acceleration.
So the forces you would be subject to will be more or less directly related to how large of a mass driver (in terms of track length) it is relative to the mass of the object you are leaving.
Rockets are actually relatively gentle in comparison, generally <3G for humans. This is because you can apply constant thrust the entire way to orbit (>~400 km on Earth)
With a Mass Driver, you can only apply acceleration for however long your driver is, so you have to get up to speed in a much shorter distance. If you build a 4km driver (length of Heathrow Runway), you have to get up to speed in 1/100th the distance causing drastically higher accelerations.
(note, some poor assumptions here because horizontal distance =/= vertical distance, but the general idea is correct).
"On November 18, 1966 the Yuma gun fired a 400 lb (180 kg) Martlet 2 projectile at 7,000 ft/s (2,100 m/s)[4] sending it briefly into space and setting an altitude record of 180 km (590,000 ft; 110 mi)"
If my math napkin correct the projectile lost less than 1.5 Mach of the speed when it reached 30km height, ie. where air is just 2% density of the sea level.
"The second supergun, "Big Babylon", of which a pair were planned (one to be mounted horizontally, at least for test purposes), was much larger. The barrel was to be 156 metres (512 feet) long, with a bore of 1 metre (3.3 feet).[1] Originally intended to be suspended by cables from a steel framework, it would have been over 100 metres (300 feet) high at the tip. The complete device weighed about 2,100 tonnes (the barrel alone weighed 1,655 tons). It was a space gun intended to shoot projectiles into orbit, a theme of Bull's work since Project HARP."
what a pity :
"The project began in 1988; it was halted in 1990 after Bull was assassinated, and parts of the superguns were seized in transit around Europe. The components that remained in Iraq were destroyed by the United Nations after the 1991 Persian Gulf War. "
one can dream - put such a gun on plateau at 4km altitude, half sea level pressure, somewhere in Bolivia or Peru (it is even close to equator) or may be Tibet, and shoot the fuel, food, water, etc. into the orbit to build up the Moon, Mars, etc, missions. If only we could find a few tens of millions dollars :). I'm waiting for Musk to hopefully get to it as he has a business case and money.
Wouldn't that be a solution to the never-ending objections saying that Jupiter cannot be used as a resources for various gases because that gravity sink would use too much fuel to leave?
As the spacecraft launcher can be station-powered the whole time of the acceleration, there is no more any worry about fuel mass, isn't it...
I don't think so. wikipedia.org used to have a mild downweight on it, the same one HN has on most major media websites. A few months ago a reader emailed and asked us to try removing that. There seemed to be enough randomly interesting Wikipedia submissions that we agreed to try it. Randomly interesting is sort of the highest good on HN, so on the whole I think it's working. We tend to moderate so there's no more than one on the front page at a time.
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[ 3.6 ms ] story [ 27.9 ms ] threadhttps://childrenofadeadearth.wordpress.com/tag/railguns/
In other environments, it may be much more difficult, but similar effects on more detailed scales may be in play.
It's possible the space war was won with One Big Rock, but for a variety of reasons, I'd expect space war is more likely to occur with a variety of smaller rocks. Primarily, One Big Rock is harder to steer at militarily useful speeds, and you have to be pretty committed to wiping out not just your enemy, but the entire ecosystem. A lot of much smaller rocks would be more militarily useful in all sorts of ways, speed of deployment and redundancy being among the most important. If there is a war between a space-faring civilization and the groundpounders, the space-farers win if they just degrade the ground civilization to the point that they are no longer space-faring and unable to interfere. Spending the considerably greater resources to massively overkill past that goal is possible, but would not be the most likely outcome in my opinion because of the outsize expenditures and much longer time to return on said expenditure vs. smaller, more nimble attacks.
Thank you, interplanetary brothers.
https://en.wikipedia.org/wiki/Launch_loop
https://en.wikipedia.org/wiki/Space_fountain
https://en.wikipedia.org/wiki/Andrei_Sakharov#Magneto-implos...
Wikipedia can generally be pretty good, but the above text from the subsection On Earth shows what can go wrong when deluded space enthusiasts go nuts on an article; it's pure wishful thinking. No major construction project of "any length" is "affordable," especially not one of the sophistication of a mass driver. At about 3g, about 30 m/s^2, it would take you about 270 seconds to get to 8 km/second, during which time you are traveling at an average speed of 4 km/second. This is a structure, with attendant control system and safeguards, about 1080 km, or 670 mi, in length. This is not affordable. By anyone.
What might be affordable (sticking my wet finger in the breeze) is getting 1 or 2 km/s at 5 or 6 g, or higher, and then doing the rest with a rocket, provided that you're firing continuously to amortize your initial cost. Skimming a couple km/s off the high end of the rocket equation brings your mass ratio way down, especially considering the reduced gravitational losses. (I'm assuming, in a fit of witless optimism, that we're launching from a high enough altitude that the increased aerodynamic losses are mitigated to the point that gravitational loss savings dominate, and that this adds no cost.)
Mass drivers are a fun idea because they seem to get around the rocket equation, but this affordability nonsense is just over the top.
The atmosphere is nice in a lot of ways, but it is not helpful in getting to space today.
They even list a couple of ideal locations including I believe Kenya.
The Hyperloop[1] is a proposal for a 350 mile structure that actually sounds remarkably similar to a mass driver, at a projected cost of $6 billion -- quite affordable for governments, corporations, and even some individuals.
[1] - https://en.wikipedia.org/wiki/Hyperloop
- If you shoot a gun on the moon, will it return to your location from the behind, no matter what angle you shoot it?
- The escape velocity of the moon is 2.38km/s. Is this the velocity required for a bullet to leave orbit no matter what angle you shoot it? Or is the required velocity higher at smaller angles?
2) * Escape velocity is the velocity required to counter the force of gravity, ie 'away' or 'out' from the center of the planet, opposite the force of gravity.
* Edit: Looks like I'm wrong, guess I slept that day in physics
[1]: https://en.wikipedia.org/wiki/Specific_orbital_energy
In most cases no. Although the orbit of the bullet is cyclical (below escape velocity), unless you shoot perfectly level with the surface, its orbit will intersect the moon. Obviously if you shoot slightly down, it'll hit the ground. If you shoot slightly up, it'll hit the ground behind you on its return.
> Is this the velocity required for a bullet to leave orbit no matter what angle you shoot it?
Yes. Although if it hits the moon first, it'll slow down and not escape (unless you were to shoot through the moon).
- I suppose if you think about it in terms of kinetic energy, this makes sense.
That said, in ideal conditions, even with rotation, the bullet will hit you eventually, but not for a long time on average, given the scale of the moon. In reality it's orbit would probably destabilise due to collisions with dust and perturbations from other gravitational bodies (the Earth, Sun, etc. create an n-body situation which slowly changes the orbit over time).
> Does there always exist a velocity `v(theta)` that will make the bullet hit you in the back after one orbit?
The velocity must be great enough to stay above the surface, and not high enough to escape. Additionally, you must shoot horizontally, like I mentioned before, and not rotate out of the orbit (e.g. by being at the poles, or at the rotational equator, shooting along the equator).
So here's one way to think about the accelerations. To a first approximation, you need the same total change in velocity to go from standing on the surface of an object to be in orbit at a given altitude. You can choose to spread that change in velocity over any given time and distance interval (those two terms are linked). So if your mass driver accelerates your payload to final velocity in 1/10th the time (and more or less distance), you will get 10x the acceleration.
So the forces you would be subject to will be more or less directly related to how large of a mass driver (in terms of track length) it is relative to the mass of the object you are leaving.
With a Mass Driver, you can only apply acceleration for however long your driver is, so you have to get up to speed in a much shorter distance. If you build a 4km driver (length of Heathrow Runway), you have to get up to speed in 1/100th the distance causing drastically higher accelerations.
(note, some poor assumptions here because horizontal distance =/= vertical distance, but the general idea is correct).
"On November 18, 1966 the Yuma gun fired a 400 lb (180 kg) Martlet 2 projectile at 7,000 ft/s (2,100 m/s)[4] sending it briefly into space and setting an altitude record of 180 km (590,000 ft; 110 mi)"
If my math napkin correct the projectile lost less than 1.5 Mach of the speed when it reached 30km height, ie. where air is just 2% density of the sea level.
https://en.wikipedia.org/wiki/Project_Babylon
"The second supergun, "Big Babylon", of which a pair were planned (one to be mounted horizontally, at least for test purposes), was much larger. The barrel was to be 156 metres (512 feet) long, with a bore of 1 metre (3.3 feet).[1] Originally intended to be suspended by cables from a steel framework, it would have been over 100 metres (300 feet) high at the tip. The complete device weighed about 2,100 tonnes (the barrel alone weighed 1,655 tons). It was a space gun intended to shoot projectiles into orbit, a theme of Bull's work since Project HARP."
what a pity :
"The project began in 1988; it was halted in 1990 after Bull was assassinated, and parts of the superguns were seized in transit around Europe. The components that remained in Iraq were destroyed by the United Nations after the 1991 Persian Gulf War. "
one can dream - put such a gun on plateau at 4km altitude, half sea level pressure, somewhere in Bolivia or Peru (it is even close to equator) or may be Tibet, and shoot the fuel, food, water, etc. into the orbit to build up the Moon, Mars, etc, missions. If only we could find a few tens of millions dollars :). I'm waiting for Musk to hopefully get to it as he has a business case and money.
As the spacecraft launcher can be station-powered the whole time of the acceleration, there is no more any worry about fuel mass, isn't it...
Mike, the protagonist of this book, runs the catapult on the moon, among other duties.