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#42
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Solar sailing DOESN"T break laws of physics'
In sci.space.policy John Ordover wrote:
Someone correct me if I'm wrong, but in general an analogy to solar sailing would be to move a terrestrial sailboat by hitting the sail with a whole lot of high-speed ping-pong balls. Light is massless but not energy-less, natch, so it winds up having a similar effect on objects to the ping-pong balls. Just as some of the momementum of the ping-pong balls would be transfered to the sails, some of the energy of the light is transfered to the sail. Just as the ping-pong balls would therefore lose some energy, the photons in the light lose some energy, but light does that all the time. So what's the problem? There isn't one. -- Sander +++ Out of cheese error +++ |
#43
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Solar sailing DOESN"T break laws of physics'
In article , Laurel Amberdine writes:
On Sun, 06 Jul 2003 22:35:55 GMT, wrote: ... Before you even start getting to relativistic physics, the point missed by many is that even in classical physics momentum *is not* defined as mv (yes, there is quite a lot of physics beyond high school physics). Momentum is defined as a gradient of the Lagrangian (yes, I know this doesn't mean much to whoever didn't study it but, as I said, there is lots of physics beyond high school physics). In the particular case of a classical massive particle, this *evaluates* to mv but that's a result, not a definition. For other entities you get different result. Thus, even within classical physics electromagnetic waves carry momentum even though they're massless. When you're not so busy you need to write a book, Mati. Something like "All the Physics you Learned Wrong" or something. After all this practice you could probably write it half asleep with your eyes closed. Probably so, this ng is good practice for such purpose:-) Mati Meron | "When you argue with a fool, | chances are he is doing just the same" |
#44
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Solar sailing DOESN"T break laws of physics'
"Alex Terrell" wrote:
Please don't misintepret me. I do "believe" in solar sails. I'm just trying to figure out how it works, because, in the normal world: Momentum = mass * velocity If mass = 0 and velocity = 3E8, then momentum = 0 Please tell me what I'm missing? This is one of the cases where you have to remember that "mass" is really "rest-mass" but in this case rest-mass is the wrong mass, so you want something more like the old "relativistic mass". In other words, you're missing that e=mc^2, and that since photons have energy they also have mass (and momentum). |
#45
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Solar sailing DOESN"T break laws of physics'
In article , (Gregory L. Hansen) writes:
In article , Laurel Amberdine wrote: On 6 Jul 2003 17:43:30 -0700, Edward Green wrote: "Steve Harris" wrote in message ... If there's no change in photon energy to first order, then obviously that's a breaking of Carnot's law to first order, since Carnot requires an decrease in photon temperature (photon energy) for work to be extracted. True, I snipped context and all, but I love the way "to first order" sometimes takes on a meaning all of its own -- as if it meant something without even specifying "with respect to X". Ignorance time: people are saying "to first (second, zeroth) order" etc, quite often lately. What does it mean, anyway? If you have some function f(x), perhaps an unknown that you're trying to solve equations of motion to find, it's often possible to approximate it and wind up with a polynomial expansion. So you might get f(x) ~= a + bx + cx^2 + ... First order would be linear in x, second order would be quadratic in x, etc. In this particular discussion, I don't think "first order" really means anything, it's just something that people are saying. Actually, it does mean just what it is supposed to. First order in delta_p (momentum transfer). Mati Meron | "When you argue with a fool, | chances are he is doing just the same" |
#46
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Solar sailing DOESN"T break laws of physics'
In article ,
wrote: In article , (Gregory L. Hansen) writes: In article , Laurel Amberdine wrote: On 6 Jul 2003 17:43:30 -0700, Edward Green wrote: "Steve Harris" wrote in message .. . If there's no change in photon energy to first order, then obviously that's a breaking of Carnot's law to first order, since Carnot requires an decrease in photon temperature (photon energy) for work to be extracted. True, I snipped context and all, but I love the way "to first order" sometimes takes on a meaning all of its own -- as if it meant something without even specifying "with respect to X". Ignorance time: people are saying "to first (second, zeroth) order" etc, quite often lately. What does it mean, anyway? If you have some function f(x), perhaps an unknown that you're trying to solve equations of motion to find, it's often possible to approximate it and wind up with a polynomial expansion. So you might get f(x) ~= a + bx + cx^2 + ... First order would be linear in x, second order would be quadratic in x, etc. In this particular discussion, I don't think "first order" really means anything, it's just something that people are saying. Actually, it does mean just what it is supposed to. First order in delta_p (momentum transfer). But how does that relate to that Carnot business? -- "Is that plutonium on your gums?" "Shut up and kiss me!" -- Marge and Homer Simpson |
#47
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Solar sailing DOESN"T break laws of physics'
In article ,
Laurel Amberdine wrote: Ignorance time: people are saying "to first (second, zeroth) order" etc, quite often lately. What does it mean, anyway? While those terms have precise meanings, they are often applied somewhat more loosely and generally. Speaking loosely and generally... "To first order" means "considering only the biggest and most obvious effects, neglecting details which don't change things very much". "To second order" means "including the most significant of the details, but neglecting really small ones which have rather smaller effects". And so forth. And by analogy, "to zeroth order" means "making drastic simplifications to get an answer which will be somewhere in the right ballpark". For example, if you wanted to know how much my little nephew weighed... To zeroth order, you might try to pick him up (cautiously :-)). To first order, you'd put him on a bathroom scale. To second order, you'd have him strip to his undershorts, put him on a doctor's scale, and tell him to stand still. To third order, you'd need a research-grade scale, you'd have him strip completely, you'd offer him an advance copy of the next Harry Potter book if he stood very, very still, and you'd pay attention to issues like when he last ate, whether his bladder is full or empty, etc. -- MOST launched 1015 EDT 30 June, separated 1046, | Henry Spencer first ground-station pass 1651, all nominal! | |
#48
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Solar sailing DOESN"T break laws of physics'
On Mon, 7 Jul 2003 18:45:12 +0000 (UTC), Gregory L. Hansen wrote:
In article , Laurel Amberdine wrote: snip Ignorance time: people are saying "to first (second, zeroth) order" etc, quite often lately. What does it mean, anyway? If you have some function f(x), perhaps an unknown that you're trying to solve equations of motion to find, it's often possible to approximate it and wind up with a polynomial expansion. So you might get f(x) ~= a + bx + cx^2 + ... First order would be linear in x, second order would be quadratic in x, etc. Okay, thanks! It sounded familiar but I couldn't recall enough to piece a meaning together. In this particular discussion, I don't think "first order" really means anything, it's just something that people are saying. I don't know if the phrase is being used technically now or not, but it does seem to have nearly become slang. -Laurel |
#49
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Solar sailing DOESN"T break laws of physics'
On 7 Jul 2003 20:57:47 GMT, Laurel Amberdine wrote:
On Mon, 7 Jul 2003 18:45:12 +0000 (UTC), Gregory L. Hansen wrote: In article , Laurel Amberdine wrote: I so like responding to myself. snip Ignorance time: people are saying "to first (second, zeroth) order" etc, quite often lately. What does it mean, anyway? If you have some function f(x), perhaps an unknown that you're trying to solve equations of motion to find, it's often possible to approximate it and wind up with a polynomial expansion. So you might get f(x) ~= a + bx + cx^2 + ... First order would be linear in x, second order would be quadratic in x, etc. Okay, thanks! It sounded familiar but I couldn't recall enough to piece a meaning together. And now I have come across "second order" and "third order" (etc) determinants, and I don't know if (or how) that relates to the same phrasing above. (I do know what it means in this context. Kinda obvious.) That isn't really a question. I'm just babbling as a break from really excessive quantities of multiplying, adding, and subtracting. Zeroes. I just wish there were more zeroes...! -Laurel |
#50
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Solar sailing DOESN"T break laws of physics'
wrote in message ... In article , "Steve Harris" If there's no change in photon energy to first order, then obviously that's a breaking of Carnot's law to first order, since Carnot requires an decrease in photon temperature (photon energy) for work to be extracted. Two points: 1) Photon energy is *not* "photon temperature". Photon (or any particle) does not have a temperature). Temperature is an ensemble property. Yes, yes, but the point is you can have a thermalized bath of photons, and if you do, then it has to follow the same thermodynamic laws as any heat source. to wit, if you turn some of the photon energy of such a collection into free energy (charging a battery on the rocket or pushing the rocket faster), then you have to pay some entropy price for destroying that much heat. In simple reflective sails that means creating another bunch of (reflected) photons that are effectively thermalized at a lower temperature (heat going into a lower temp reservoir). Or else you have to heat up some sail material which was previously at a lower temp than the illuminating radiation, and dump your entropy that way. If you have a monochromatic source of photons like laser or a microwave beam, then (so far as I can tell) nothing prevents you from converting such a beam *entirely* into free energy (charging a battery on board the rocket), and destroying the EM radiation completely, so that no photons are left at all, and sail heating is minimized (perhaps doesn't occur at all). All the energy could be extracted from the beam and used to charge a battery or make chemical fuel, or something. Which could then be used indirectly for propulsion, subject only to a rocket's ability to extract fuel energy and turn it into rocket kinetic energy (not a thermodynamic problem, but certainly a practical one which I supposed must be related to the thermodynamic one). For reasons that are obscure to me, there are additional limitations for using the energy in photons that come in a blackbody spectrum. They act like heat, and only some or their energy is available for conversion into free energy at the target, and thence into rocket kinetic energy. Gold, I'm sure, specified a "heat" driven sail (thermalized light from the sun), because thermodynamics limits in that case what fraction of light energy can be extracted from the heat beam as a first step. But I agree it does just serve to complicate the problem, since the major shortcoming in using EM radiation to power a rocket, is in converting all that energy into something useful for propulsion. If you just reflect it, you waste most of your energy, and that has nothing to do with thermodynamics. SBH |
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