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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/   
    
Associated Topics:
High School Probability

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