3 ms·
You'd see the "bow wave" regardless - the energy imparted as the impactor hits interstellar medium. To a first approximation, that's on the order of 67kw / m^2
by TheLoneWolfling 12y ago
You'd see the "bow wave" regardless - the energy imparted as the impactor hits interstellar medium. To a first approximation, that's on the order of 67kw / m^2 at 0.9c ([1]), which is... non-negligible. And in practice, you'd notice the occasional flashes when you hit a grain of dust.
This cannot be solved by a shroud, as you'd then see the interstellar medium hitting the shroud instead.
The best response to something like this would be a whipple shield. Except you don't need an actual shield. Just drop a couple stages of a bunch of chaff in its trajectory, or a bunch of spin-stabilized mylar disks. Even ball bearings would be too much. You don't need to stop it, "just" diffuse it enough so that the atmosphere can handle it.
[1]: http://www.wolframalpha.com/input/?i=%28%281+-+0.9^2%29^-0.5+-+1%29+*+c^2+*++0.9+*+c+*++%281*0.9+%2B+4*0.1%29+amu%2Fcm^3 http://www.wolframalpha.com/input/?i=%28%281+-+0.9^2%29^-0.5..., Assuming ~1 atom/cc, 90% hydrogen 10% helium
- deeviant 12y agoTrue, but I think a lot of people here are assuming something like .9C or the like when the reality is you nothing nearly that high. 1000 tons at .1c is 9818 megatons, which would be roughly the size of a US air craft carrier. The bow shock would be much less and the I'm assuming it would be using some sort of magnetic scoop ramjet would would channel interstellar medium to the propulsion system rather than striking the bow. With a stealthy design, the thing would be very difficult if not impossible to detect, barring some sort of entirely need detection method.
- TheLoneWolfling 12y agoYou use a magnetic scoop ramjet, great. Now you have cyclotron radiation. Congratulations, now you're even more detectable! And even at 0.1c, you encounter a grain of sand, that's, what, ~23MJ of energy released.
- deeviant 12y agoYou got me there, good comments. Thanks.
- TheLoneWolfling 12y agoOh yeah, and two other things. Your 1000 tons at 0.1c is going to have a whole lot more cross-sectional area than, say, 4 tons at 0.9c. At least the cube root of 250 times, or ~6.3x. You'll also have, what, 11x more time to react to it at a given distance. Neither of these compensate entirely, but both of these are compensatory effects.
- deeviant 12y agoMore cross-sectional area because of lorentz contraction or just looking at the mass-surface area / kinetic energy ratio?
- TheLoneWolfling 12y agoWhoops, I messed up. Try 250^(2/3) times, instead. (I was intending to assume constant density and shape, but forgot to turn that into cross-sectional area as opposed to radius.)