Satellites and Dartboards
Date: 02/21/2000 at 11:10:44 From: Brian Borden Subject: Probability of Accidentally Hitting a Satellite Assume we model our satellites in two groups. One group is at 500 km altitude, and the other group at geosynchronous altitude. Consequently, I can determine the surface area of the two spheres. I also assume a certain number of satellites in each sphere with a given average surface area for each. This yields a total surface area of satellites in each sphere. I also assume the satellites move randomly on each sphere for the sake of modeling simplicity. I realize this assumption is more valid for the lower satellites than for those at geosynchronous altitude. If I fire a laser beam with a known divergence, I can calculate the area it will cover on each of the two spheres. If I fire an instantaneous burst, what is the probability of hitting a satellite? I figure the probability must be proportional to the area of the satellites, proportional to the area of the beam, and inversely proportional to the area of the spherical cap above the local horizon. As the area of the beam approaches the area of the spherical cap, the probability should go to 1. Similarly, as the area covered by the satellites approaches the area of the cap, the probability should also go to 1. Obviously the probability can never exceed 1, thus as the area covered by the satellites and the area covered by the beam approach the area of the cap, the probability still must approach only 1. Also, when the area of the beam is greater than the area of the cap minus the area of the satellites, the probability of hitting a satellite must equal 1. I've tried (Asat + Abeam)/Acap, and (Asat/Acap)*(Abeam/Acap), but neither of these quite fit all the criteria that must apply. Also, I'm not certain if this is a conditional or an unconditional probability. The satellites could be anywhere, and the beam could be fired anywhere, but given that the beam is in a given spot, I need to know the probability that a satellite will also be there at the same time. Can you help? Thanks.
Date: 02/21/2000 at 22:57:53 From: Doctor Peterson Subject: Re: Probability of Accidentally Hitting a Satellite Hi, Brian. Since no expert on probability has taken your question, I'll try it. I think you have a couple of red herrings here. The "sky area" (your spherical cap) is irrelevant; if you think of a fixed laser and ask the probability that a satellite will be within your beam at a given time, the area of sky that you can aim at doesn't matter. Also, I suspect that the total surface area of the satellites will not be the determining factor; you could imagine a given total area either being covered just by one enormous (but still smaller than the moon) satellite, giving a small probability, or at the other extreme being divided into billions of micro-satellites scattered as dust, so that one will be hit with probability one! The actual sizes and number of satellites will be important. I'm going to proceed on the assumption that there are N satellites of about the same size, which is sufficiently smaller than the laser beam so that they can be thought of as points. (If their size is significant, I think you could adjust your measure of the beam diameter to make a larger effective diameter that would take into account that the center of the satellite could be some distance outside the beam and it would still count as a strike.) Now the probability that any one particular satellite will be hit is P_1 = Abeam / Asky where Asky is the area of the ENTIRE SPHERE on which the satellite is assumed to reside. The probability that ONE of N satellites will be hit is one minus the probability that NO satellite will be hit, or P_N = 1 - (1 - P_1)^N which may be approximated for large N and small P_1 as P_N = 1 - (1 - N*P_1) = N*P_1 = N * Abeam / Asky Whether this is a good approximation depends strongly on the values of N and P_1. Is Abeam much larger or smaller than Asky/N, the area assigned to any one satellite? I can't really proceed without knowing that. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Date: 02/28/2000 at 13:56:08 From: Brian Borden Subject: Re: Probability of Accidentally Hitting a Satellite Dr. Peterson, I think a similar way to approach the problem would be to determine the probability of hitting a particular sector on a dartboard given three things: 1) Your dart will definitely hit SOMEWHERE on the board, but your aim isn't great so your chances of hitting any one spot on the board are the same as hitting any other spot. 2) You know the sizes of the sector and the board. 3) You're playing suction cup darts, and you know the size of the suction cup - you score a hit as long as any part of the suction cup is inside the sector. What's the probability of scoring a hit if you throw one dart? Brian Borden
Date: 02/28/2000 at 16:58:44 From: Doctor Peterson Subject: Re: Probability of Accidentally Hitting a Satellite Hi, Brian. Thanks for writing back. If there's one sector we're trying to hit, this will certainly be equivalent to my N = 1 case (with one, possibly huge, satellite in the sky). We'll have to extend it if we want to model a large number of satellites. Suppose the target "sector" is a circle with radius r_t, and the dart has radius r_d, while the whole dart board has radius r_b. (If I used an actual sector, then as it became smaller, it would approach a line segment rather than a point, which doesn't seem realistic for a satellite.) Then we score a hit as long as the dart is within (r_t + r_d) of the center of the target spot. Therefore the probability of a hit is the ratio of the "hit" area to the total area: pi (r_t + r_d)^2 (r_t + r_d)^2 P_1 = ---------------- = ------------- pi r_b^2 r_b^2 Now let's move on to the N satellite case, by imagining N identical target spots, and assuming that the "hit" regions don't overlap. Then the probability of a hit will again be the ratio of the total area of all the "hit" regions to the total area: N pi (r_t + r_d)^2 N pi r_t^2 (1 + r_d/r_t)^2 P_N = ------------------ = -------------------------- pi r_b^2 pi r_b^2 Notice that N pi r_t^2 is the total area A_t of all the target spots, which you originally felt would determine the probability. If I let N increase and r_t decrease so as to keep this area constant, the r_d/r_t term will increase without bounds; the only limit on the probability of a hit will be that we assumed the hit regions would not overlap, and since they have a minimum area of pi r_d^2, N can't really increase forever without changing the conditions of the problem. As I said last time, if N is large enough, making the satellites "dust," the probability of a hit will be essentially one, because almost certainly there will be no part of the sky (dart board) that is not within r_d of a satellite (target spot). Rather than work with the area of the targets, we can focus on the area of the dart: N pi (r_t + r_d)^2 N pi r_d^2 (r_t/r_d + 1)^2 P_N = ------------------ = -------------------------- pi r_b^2 pi r_b^2 This way, when N gets large, r_t/r_d gets small and we are left with P_N = N A_d/A_b which is what I got last time (though I still haven't dealt with the overlap problem, so I can't really take r_t to zero). All of this essentially does what I did before, except that we are starting with the area of the target and adjusting it for the size of the dart, rather than starting with the size of the dart (treating the target as a point) and adjusting that for the actual size of the target spot. You can repeat the P_N = 1 - (1 - P_1)^N analysis I did last time if you wish, to remove the non-overlap assumption. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
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