I like the ice ball idea. Macgyver solution. Good one.
I'm not so sure about this one as a target though. It does not spin. We kinda need that I think. Sure we could make do without. but...
Also, don't forget relativistic effects: if we managed to achieve a probe that could do 0.9c (or even a manned craft, at much lower accelerations)... Most importantly, time dilation on board and length contraction from our observational frame.
To make it simple, let's assume our probe/ship does 0.9c for 100% of the journey (overly simplistic for ease of calculation, neglecting a passenger appropriate acceleration/deceleration curve - heinous math):
Let's see what would the time be dilated to from our frame for such an abused passenger?
[mathb]\Delta T^{\prime}\;=\;\frac{\Delta T}{\sqrt{1\;-\;\frac{v^{2}}{c^{2}}}}[/mathb]Where [math]\Delta T\;[/math] is the time interval between events for a local observer and [math]\Delta T^{\prime}\;[/math] is the interval between those events as measured by an observer in motion (our space traveller) with respect to the first (family at home).
Using the Lorentz factor:
[math]\gamma\;=\;\frac{1}{\sqrt{1\;-\;\frac{v^{2}}{c^{2}}}}[/math][math]\Delta T^{\prime}\;=\;\gamma\Delta T\;=\;\frac{\Delta T}{\sqrt{1\;-\;\frac{v^{2}}{c^{2}}}}[/math][math]\Delta T\;=\;\frac{\Delta T^{\prime}}{\gamma}\;=\;\frac{\Delta T^{\prime}}{\frac{1}{\sqrt{1\;-\;\frac{v^{2}}{c^{2}}}}}[/math]sub in our values...
[math]\Delta T\;=\;\frac{22.23\;y\;\times\;(365\;d/y)\;\times\;(24\;h/d)\;\times\;(3600\;s/h)}{\frac{1}{\sqrt{1\;-\;\frac{(269813212.2 (m/s))^2}{(299792458\;(m/s))^2}}}}[/math]
[math]\Delta T\;=\;\frac{701045280\;s}{\frac{1}{\sqrt{1\;-\;0.81}}}[/math]
[math]\Delta T\;=\;\frac{701045280\;s}{\frac{1}{0.435889}}[/math]
[math]\Delta T\;=\;\scriptsize 305577926\;s\;=\;9.69\;y[/math]Edit: fixed - so the journey at 0.9c that the "stationary" observer sees as 22.23 years is clocked at 9.69 years by the "moving" observer. (use of quotes as it is all relative
)
Hmm. Did I screw that up somewhere? Math layout is not easy I'm discovering. Lets check by the Lorentz factor for 0.9c (2.294 x 22.23 y = 50.995) - yup, looks right - neglecting gravity effects along the way for now of course. So I guess the special relativistic effects of such a journey are significant. That complicates matters. Or does it really? Still, I'm not sure I have it quite right.If I do have it right (someone please check?), then the 20 light year journey at 0.9c that we'd expect to take 22.23 years for the traveling clock would be observed as 50.99 years from Earth.Hmmm. I wonder how the trade-off in velocity works out for time considerations and cost? What's the best velocity to shoot for (mean over the acceleration curve of course - for now, in this exercise)
Wakey, wakey everybody... I'm making my second coffee!
I think we should aim for
.
I bet we could have at least an unmanned 0.9c probe within eight years if we really put ourselves on it.
I need to build an excel super-science spreadsheet, keeping it OO.o and GDocs compatible. What would be really cool would be a LaTeX solution that spans all three seamlessly to provide pretty equations for the cell formulas. Actually, looks like MathML would be the way to go for that, as it seems new office versions support it.
(grr.. seems neither jsMath or MathJax handle units properly. I might have to fix that.)
I have decided 0.9c makes no sense at all really other than to demonstrate the possible magnitude of the effect. One must consider the acceleration curve.
The fun part is that the speed of light (EM - i.e. radio) is constant in all frames, that will magically simplify comms (somewhat).
Have people built algorithms to work all this crap out? I expect they have, but if not, the time is now.
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