Re: FT-Airless bodies

bif smith wrote:
> Would beam weapons have any range modifiers firing from a airless
body?, as
> opposed to firing through a atmosphere with it`s attendent difusion
effects?

	Short answer: Yes.

	Long answer:

	An atmosphere does reduce the range of beam weapons, but
	even in airless space such weapons are subject to the
	dreaded Inverse Square law.

	This is because a beam *spreads* as it travels.

	The law says that if you double the distance, the
	beam strength drops by one quarter.  Doubling means
	a factor of two, two squared is four.
	In other words, the beam strength is 1/(d^2)
	
	The original law was derived for searchlights and things
	like that.  One would think that coherent sources of light
	like lasers would be immune.  Alas, lasers are subject
	to something called diffraction, which also spreads the
	beam.  Not quite as bad as the inverse square law, but
	bad enough.  The legendary Erik Max Francis said that
	the practical limit of a visible light laser is a few
	hundred thousand kilometer before they start suffering
	serious range losses.

--------start of quote---------------------
> Is the lightspeed delay always the limiting factor?  Could you hit
> anything
> you could see--assuming that it didn't move (that is, didn't move
> unpredictably)? Surely a target could be visible but a laser weapon
> not
> precise enough to be able to reliably hit it?  Wouldn't you, at
> extreme
> ranges, be forced to rely on a Gaussian sort of range?

You also have diffractions -- even lasers will diffract.  The extent of
the diffraction (the halfangle of the beam) depends on how large your
aperture is and what frequency of light you're using.

So no, in the general case, a distant, predictably-moving target will
_not_ always be vulnerable.  At extreme distances, you can bathe them
with bright light, but it probably won't cause much damage.

> But suppose you detected an object near Pluto
> that was moving in a steady, predictable course--would you expect to
> be able
> to hit it with a laser from Earth orbit?

At any given time, Earth and Pluto will be somewhere between 38.4 au
(5.76 Tm) and 40.4 au (6.06 Tm) apart.	The median is, naturally, the
Earth-Pluto distance, which is 39.4 au (5.91 Tm).

Let's say you're firing a 1 GW beam with wavelength 500 nm
(greenish-blue) -- or a frequency of 600 THz -- from a 1 m diameter
aperture.

Diffraction halfangle theta, wavelength lambda, and aperture diameter l
are related by 

    sin theta = 2.44 lambda/l.

Here, lambda = 500 nm and l = 1 m, so theta = 1.22 urad.

The beam radius r starts at l/2 (half the width of the aperture), and
increases as a function of distance R as 

    r = l/2 + R tan theta.

At the median distance to Pluto -- 5.91 Tm -- the beam radius has
increased to 7.21 Mm, or a little over one Earth radius (6.37 Mm).

The initial beam intensity is 1.27 GW/m^2.  At Pluto's orbit, the beam
radius has increased to 7.21 Mm, so the cross-section beam area is 1.63
x 10^14 m^2.  That gives a beam intensity at Pluto's orbit of 6.12
uW/m^2.

You can use as a metric for your laser losing effectiveness when the
beam radius doubles (and thus the beam area quadruples, meaning that the
intensity drops by a factor of four).  This happens when r = l (twice
the initial beam radius):

    l = l/2 + R tan theta

    l/2 = R tan theta

    R = l/(2 tan theta).

For our hypothetical beam with theta = 1.22 urad, this means that the