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### Satellites and Dartboards

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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.
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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
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

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/
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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
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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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Associated Topics:
High School Probability

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