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Accord. to my Prof. a 92 astroblaster would be sufficient to shot a ball to the moon. I have done some research on the internet, but could not find any sufficient source.

Does anyone know about astroblasters as alternatives to rocket propulsion systems? Has NASA/ESA ever built a system? If so, what are the dimensions of it?

to get the idea: http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=1738.0

CoilKid
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Randy Welt
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  • What's the speed of sound in your balls? That's probably pretty close to the velocity limit that the device is really capable of, but maybe I am wrong and with the right construction one can go additive without the whole thing destroying itself. If anything, I would go with NASA hydrogen gas guns... those come pretty close to getting things into orbit: http://www.nasa.gov/centers/wstf/laboratories/hypervelocity/gasguns.html. Thats a proven technology. – CuriousOne Jan 01 '15 at 00:27
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    What, exactly, is a "92 astroblaster"? Questions posted here should stand on their own. I am not about to enable java for an unknown site. – David Hammen Jan 01 '15 at 16:37
  • drop a perpendicular and let 92 balls fall this way. the momentum (caused by gravity) of the 91 balls will be completely absorbed by the ball on top. (ellastic collision, if I remember correctly) – Randy Welt Jan 02 '15 at 20:51

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In the best case scenario in which all the momentum from the lower balls is transferred to the to ball, and in absence of atmospheric friction, you will have:

$Mv_M=mv_m$, where M and m are the masses of the bottom balls and top ball respectively, $v_M$ is the speed of the balls immediately before touching the floor and $v_m$ is the speed of the top ball after touching the floor.

If you want to get to the moon, you need to at least reach Earth's escape velocity, $v_e=10^4$m/s. Using $v_M=\sqrt{2gh_0}$, where $h_0$ is the initial height of the balls, you get:

1) For $h_0=10$m: $\space m/M=1000$

2) For $h_0=1000$m: $\space m/M=100$

Which means, briefly: To reach escape velocity, if you drop the astroblaster from a height of $10$m, you need the bottom balls to be at least 1000 times heavier than the top one. If you drop it from a height of $1000$m, you will need the bottom balls to be only 100 times heavier than the top one.