• jorge
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    10 months ago

    The US FCC defines ionizing radiation as wavelengths smaller than 124 nm (which corresponds more or less to the ionization energy of both oxygen and hydrogen, so it is a sensible definition).

    The “most starlight” part is a bit trickier. Stars emit light in a wide spectrum (approximately black body radiation) depending on their temperature: hotter stars emit bluer light and are more luminous, but very rare, while cooler stars are redder and fainter, and much more common. Yellow stars (spectral type G), like the Sun, emit mostly between 400 nm and 750 nm (visible spectrum), while red stars (spectral type M) mostly emit from 700 nm to 1000 nm,

    So let’s say that you want all the light with wavelengths of 1000 nm or smaller turned into ionizing radiation. That gives us a blue-shift of 1+z = lambda_obs/lambda_em = 124 nm/ 1000 nm = 0.124.

    The relation between speed and blue(/red)-shift is given by the relativistic Doppler effect:

    1+z = sqrt((1+beta)/(1-beta))

    where beta=v/c and c is the speed of light. Solving for beta

    beta = ((1+z)^2 -1)/((1+z)^2 +1)

    And plugging the numbers, you get beta = -0.970, where the minus sign means that you are moving towards the star. At 97% of the speed of light.

    If you only wanted to turn most of the sunlight into ionizing radiation, you would need “just” 94.7% of the speed of light.

    • Troy@lemmy.caOP
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      10 months ago

      This is a fantastic answer that clearly spells out the assumptions made and the effects of those assumptions. Well done :)

      The sci fi fan in me now wonders: assuming you get to 0.95c before it becomes truly problematic, what kind of shielding would you need at the front of your probe(/spacecraft/etc.) to prevent the effect from totally destroying the probe. Like you can’t just blast electronics with x-rays and harsher and expect it to run. So there is probably a mass tradeoff here with a big ole block of ablating lead at the front that ablates from radiation spalling?

      There’s probably a cost optimization curve there – mass of shielding, travel time, energy cost to accelerate the probe.

      Furthermore, there would be a pressure exerted by that radiation that should slow the probe down over time, but the rate of slowing will depend on the mass/cross section/etc.

      And of course running into an interstellar proton or such will be somewhat of a high energy event… More shielding, or hypothetical defenses like the bussard ramscoop…

      • HeckGazer@programming.dev
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        10 months ago

        A couple non-quantitative thoughts on this:

        • Space is big and empty. In order to have significant amounts of radiation energy hitting you you’d have to be going quite close to stars thanks to the cube energy drop-off.

        • Space probes do have this kind of shielding on their chips! They also use multiple independent identical processors that vote on results as it’s unlikely for a high energy particle to scramble a majority of the chips in an identical way. This is also why space tech tends to use old manufacturing nodes that have huge (by earth bound comparison) transistors.

        • Troy@lemmy.caOP
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          10 months ago

          Good points.

          I wonder what the dose would look like to a square metre, framed in human reference terms. Better or worse than medical imaging. That would require figuring out photon flux, above 750nm. But there is also relativistic cross section changes happening, so does that affect the flux?

          I’m reminded of a first year physics prof that suggested we figure out how fast we’d need to go to fit through the head of a needle (in a vacuum).

          The redundant computing this is a fantastic invention. I’m aware that SpaceX is using off-the-shelf computers for this instead of the longstanding tradition to use only “rad hardened hardware”, preferring to rely on multiple redundancy for weight and cost savings. Without knowing the flux at 0.95c though, it’d be hard to estimate how well the strategy would work :)