Mass Driver
en.wikipedia.org
en.wikipedia.org
"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.
- 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?
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).
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
https://en.wikipedia.org/wiki/Andrei_Sakharov#Magneto-implos...
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.
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 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.
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.
Mike, the protagonist of this book, runs the catapult on the moon, among other duties.
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).
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...