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58

Other answers don't mention the fact that no single impulse (e.g, like being fired from a gun) can launch a projectile into orbit. A purely ballistic projectile fired from a gun must either crash back into the planet, or it must escape from the planet altogether. In order to achieve orbit, at least two impulses must be applied to the projectile. The first ...


30

Anything launched into orbit by such a gun needs to travel at orbital velocity (in fact above orbital velocity) in the lower atmosphere. That's generally undesirable, to put it mildly: there will be really serious heating.


16

Aside from the interior ballistic aspects of these various projects, it was quickly realized that any satellites launched by gun would have to withstand high g-loadings during firing of the gun and the size and mass of the satellite would be greatly constrained by the dimensions of the bore of the gun and the maximum impulse which could be provided by the ...


11

I think the heart of the question is whether one could arrange a continuous combustion of propellant along the length of the barrel. In that way the acceleration occurs along the length of the barrel in a more gentle way. Since the expanding gases from the propellant in a shell casing expand and the pressure of the expanding gases declines along the way it ...


6

I will try to answer as many questions as I can. I won't presume to give you complete exhaustive answers, but maybe they will be nonetheless useful to you. What variables determine the range of temperatures over which matter is liquid? My understanding of thermodynamics is that matter changes from a solid to a gas when some thermal vibrations create an ...


3

One more nasty factor: What is the expansion speed of your propellant. Take the Jules Verne approach and your spacecraft falls far short no matter how much powder you put in the gun because the expansion velocity is too low. Your craft will never exceed the expansion velocity of the propellant. Note, however, that you don't have to use explosives (or ...


2

Using Torricelli's law (https://en.wikipedia.org/wiki/Torricelli%27s_law) you get $v=\sqrt{2gh}$ irregardless of how big the opening is. Now you can calculate how much mass is leaving the tank at any time by multiplying the volume that is leaving the tank by its density: $$\Delta m=A\Delta s\cdot\rho$$ with $\Delta m$ the mass leaving the tank at any second, ...


2

To answer your specific question: absolutely none. The Millenium run is a "dark matter-only" simulation. In this sort of simulation gas physics is taken to play a negligible role. All the gas (and stars, indeed all "baryonic matter" as it's called in the jargon) is removed and replaced with additional dark matter. The extra dark matter is added just to keep ...


1

Orbital insertion is hard enough with ships where the initial stages can be aimed in directions favorable to the final orbit. This design calls for a fixed launch direction, which would be a terrible waste of on-board fuel for all but those satellites whose orbits coincide with the gun's trajectory. Even that can be overcome with fuel. The real issue is ...


1

Poiseuille's law will tell you that a pipe of 0.1 m diameter will achieve a flow velocity of around $v=0.1\mathrm{~m/s}$, which is actually more than I expected. For pipes a bit wider than this, the flow will be turbulent (Re>2200), for which you can't apply Poiseuille. For turbulent flow, you can look into the Darcy-Weisbach equation. The interesting part ...


1

I have discussed this problem with someone who solves this type of problems numerically and got the following response: The expression on the right-hand-side of (**) in my question is evaluated at $t_0+dt$ instead of $t_0$. This together with (*) provides two equations for $\mathbf{v}(t_0+dt)$ and $P(t_0+dt)$, \begin{align} ...


1

For a particular setup, the equations may get very simple: the tank should be massless ($M=0$) and the hole is all the way at the bottom of the container. Then $h$ is proportional to the mass of the water in the container: $m=m_0 h/h_0$, with $h_0$ the initial height. It's straightforward to derive that the force generated by the water jet is $F=2\rho g h ...


1

In this case, it is allowed. As a thought experiment, you can replace the top bucket with the siphon hose by an S-shaped tube that has a gradual change in diameter, like this: _ / \ |www| | | \www| | | \ww| | | \w| | | \|__/ | | | With this type of analysis, it is more ...


1

If the mouth of the bottle is small (e.g., a wine cork with drilled hole for a drinking straw), water indeed won't flow out. Would you also wonder why the water doesn't stay in an upside-down bucket (the limit case for a bottle with a very wide mouth)? What happens is that air enters through the mouth with a volume equal to the volume of water that leaves ...



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