Would a giant satellite laser work as a weapon?

If someone built a satellite to float about in space, which used mirrors to focus and magnify energy from the sun directly into one direct spot (like in the James Bond film Die Another Day) would it actually work? Could it be powerful enough to destroy buildings and burn through the ground?

Or is it not possible? And if not, why?

Yes, it is possible so long as one has a large enough set of mirrors and the focal points are determined accurately enough (ignoring heat dissipation, which might be an issue in space). There is a solar furnace in France that does just this, though it is based on ground and has on the order of 10,000 mirrors, each of which appears to be quite large.

The Mega Watt Solar Furnace, for instance, has a total collecting area of 2835 m2 and it focuses the beam of light into a spot roughly 80 cm in diameter (i.e., ~0.503 m2)

Longer Answer with Background and Examples

The sun emits at Earth's orbit roughly 1380 W/m2, so you can determine the necessary collecting area based upon the target you wish to destroy. In the movie, the beam was shown to have a rather large diameter, but let's assume (to make the numbers easy to deal with) that area over which the death ray fires is 1 m2. Therefore, assuming no losses in the focusing or transmission through the atmosphere and no frequency conversion, one would have a gain factor in power of roughly 1400 for every square meter of collecting area on the spacecraft.

So what you want to do in this case is look at the heat of sublimation (also called the enthalpy of sublimation), $\Delta$Hsub, for various materials. This is the required energy necessary to turn a given solid into a gas without transitioning through the liquid phase, a process called sublimation.

There is an exhaustive list of $\Delta$Hsub values for numerous organic molecules from NIST in this PDF. You can look on most Wikipedia pages for the heat of vaporization, $\Delta$Hvap, which is always less than $\Delta$Hsub. Say, for example, we look at aluminum or Al, $\Delta$Hvap ~ 284 kJ/mol (kilojoules per mole) and its $\Delta$Hsub ~ 326 kJ/mol (called heat of atomization in this case). Al has a density of ~2.7 g/cm3 (or 2700 kg/m3) and an atomic mass of ~26.98154 g/mol. Thus, one mole of Al should have a mass of ~26.98154 g and occupy a volume of ~9.9932 $cm^{3}$ (recall that 1 cm3 is one milliliter).

Examples

Example 1

So now let's put those numbers to use.

• Let's say our spacecraft had a collecting area of 1000 m2, just to make the numbers easy, and perfect mirrors (i.e., no losses).
• That means the spacecraft would be collecting a total of ~1,380,000 watts of power (recall: 1 W = 1 J/s) or ~1.38 MW.
• Focusing that into a 1 m2 beam means our power density is ~1.38 MW/m2.
• So how much Al could we vaporize in one second (again, to make the numbers easy) if it were distributed over a nice circular disk with an area of 1 m2?

For now, let's ignore the angles of incidence and all those complications. Then we need only divide our power density, call it $\Delta$S, by the $\Delta$Hvap value and the area, $A$, to get the total number of moles, which we can use to determine the total mass and volume. This is given mathematically by: $$Moles \ of \ Al = \left( \frac{ \Delta S }{ \Delta H_{sub} * A } \right) \sim \left( \frac{ 1.38 \times 10^{6} \ J \ m^{-2} \ s^{-1} }{ 384 \times 10^{3} \ J \ mol^{-1} } \right) \sim 3.59 \ \text{moles}$$ which corresponds to ~96.965 g of Al or a ~35.913 cm3 volume. This is only ~$3.591 \times 10^{-5}$ m3 or a 1 m2 that is only ~35 $\mu$m thick (i.e., roughly half the thickness of a typical human hair).

Example 2

Perhaps a better way to ask this question is to start with a given amount of Al (you can do this with any material, just change the numbers) and work backwards to determine the amount of power density needed.

• Let's say we have a 1 m2 disk again, but now it's 1 foot thick (or ~0.3048 m), corresponding to a total volume of ~0.3048 m3.
• This means we have ~822.96 kg of Al or ~30500.854 moles.
• From the known value of $\Delta$Hsub, this means we need ~11.712 GJ of energy to sublime the Al disk (recall 1 GJ = $10^{9}$ J).
• To do this in one second with a 1 m2 beam means we need $\Delta$S ~ 11.712 GW/m2.
• We know the sun's power density output at Earth is ~1380 W/m2, so we just divide these two numbers to get the amount of collecting area needed to vaporize our Al disk. The area turns out to be ~$8.5 \times 10^{6}$ m2, which seems a little unreasonable, since one of the largest buildings by volume is the Boeing Everett Factory at only ~400,000 m2.

Side Note: The one advantage about having a mirror system like this in space is that one need not worry about inter-mirror stresses/strains (except during any orbital maneuver, obviously). However, that is hardly a significant enough advantage to actually make this worthwhile or feasible.

Example 3

• Let's try using the beam size of the Mega Watt Solar Furnace, i.e., ~0.503 m2 for our focusing point and again use a 1 ft thick piece of Al, i.e., a ~0.1532 m3 volume.
• This corresponds to ~413.66 kg of Al, or ~15,331.40 moles of Al.
• Thus, we need ~5 GJ of energy to sublime the Al disk.
• A 5 GJ beam covering ~0.503 m2 for a one second duration corresponds to $\Delta$S ~ 9.9 GW/m2.
• Therefore, to generate this energy density we would need a collecting area of ~$3.6 \times 10^{6}$ m2, which is still unreasonably large.