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While listening to the latest updates it seems like there is a limited 60 hour battery capacity left for Philae. There seems to be some information on the web on wireless power transfer using lasers.

Wikipedia: ...thereby delivering almost all emitted power at long ranges.

Since Rosetta is still in sampling sight of solar power, could it possibly transfer this power using the wireless method to Philae, or would the loss in transfer be too severe for any kind of practical solution? Seems like this could be an option for future mission to explore locations while blind to sun light.

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This is indeed possible. The US spent a fair amount of money in the 60s and 70s on beamed power. In recent years it has gotten some more interest for lunar rovers that can survive the 14 day lunar night (without RTGs). The losses are not negligible, and the pointing requirements are rather extreme but in practice there's no reason why this couldn't happen.

Losses: Initially the losses are driven by the solar panels on the satellite, if we assume they are 25% efficient then you've already lost a big lump of your 1300W/m2 (at 1 AU). Then you have the internal storage and transfer loses, maybe a few more percent ~2%. Then the beaming losses, creating the beam might be as inefficient as 10% or 20%, lets be optimistic and say 20%. There are potential losses based on the footprint size of the beam, but we'll (very optimistically) assume we can point just right. The receiver at the lander might manage another 25% efficiency, plus a few extra % for inboard storage etc ~ 2%. Total power efficiency: 0.25*0.97*0.20*0.25*0.98 = 0.012 = 1.2%.

So about 1% of the power you get in orbit will make it to the lander. Of course all of these efficiencies can be improved upon, but it gives you an idea of the sort of problems. For 100W on the surface you would need 100W/0.012/1300W/m2 = 6.4m2 of solar panels in orbit.

Once I get on my laptop (and off my mobile) I'll pepper this with links and correct typos.

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  • $\begingroup$ Why would there be 80% loss in beaming the power? Microwave beaming can be 95% efficient. $\endgroup$ – kim holder wants Monica back Jan 7 '16 at 16:42
  • $\begingroup$ @kimholder Creating laser light from electric power is inefficient. Creating microwaves is much more efficient. $\endgroup$ – Wirewrap Jan 7 '16 at 16:49
  • $\begingroup$ He's not really talking about beaming loss, but about converting electrical power to laser light. $\endgroup$ – Hobbes Jan 7 '16 at 16:51
  • $\begingroup$ The question mentions using a laser, but i don't think meant to restrict it to that. $\endgroup$ – kim holder wants Monica back Jan 7 '16 at 16:52
  • $\begingroup$ An interesting way to look at microwave transmission is using link budgets. There you can simply calculate the power at the receiver (just ignore all the details about noise). You'd still have to included solar panel/microwave emitter efficencies and the reconverstion efficencies on the surface, but it gives you a nice way to look at the problem. $\endgroup$ – ThePlanMan Jan 7 '16 at 16:56
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Doing this efficiently requires knowing exactly where the lander is. A laser will generate a tiny beam over the distances we're talking about, so you have to aim the laser accurately.
Rosetta doesn't have enough equipment on board to find where Philae is. You could use a radio direction finding system, for example.
An additional problem: the orbiter will be in sight of the lander for only a fraction of its orbit, so you need to transfer enough power for an entire orbit in that small window.

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  • $\begingroup$ Would microwave beaming help? $\endgroup$ – kim holder wants Monica back Jan 7 '16 at 16:43
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    $\begingroup$ Perhaps, but the wider the beam is, the lower your efficiency. If 90% of your beam doesn't illuminate the lander, you have to scale up the system by a factor of 10. $\endgroup$ – Hobbes Jan 7 '16 at 16:53

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