Simple enough:
Solar surface temperature: 5000K
Power emission at solar surface: 50 Kw/cm^2
Radiation pressure: 300 Mw per Newton
1. Calculate power emission for 100Mk blackbody:
Temp ratio: 1 x 10^8 / 5 x 10^3 = 2 x 10^4
Black body emission proportional to T^4
Power ratio: 16 x 10^16 = 1.6 x 10^17
Power emission: 1.6 x 10^17 x 5 x 10^4 = 8 x 10^22 watts/cm^2
Power emission of 1 square meter in megawatts:
8 x 10^22 x 10^4 x 10^-6 = 8 x 10^20 megawatts/m^2
2. Calculate power at 300 meters:
Ratio = 300/10 = 30. By 1/r^2, ratio is 1/900 or about 10^-3
Power incident on 1 m^2 at 300 meters is 1 x 10^17 megawatts.
3. Calculate force:
10^17 / 300 = 3 x 10^14 newtons or about 3 x 10^12 Kg.
This will accelerate your 100 kg steel plate at 10^12
gravities....
That's enough to get started. If you assume 1 microsecond before
the plate is obliterated, it will be completely crushed before
then by the radiation pressure. Knowing the force and time (and
mass of plate) it is easy to compute the velocity of the plate
when the fireball arrives. As an exercise, one can also compute
the temperature of the plate. (Specific heat of Fe is about 1/10
of water, btw).
> Are you really proposing that the peak forces generated are
> sufficient to deform a macroscopic object made of, say, 1 cm sheet
> steel?
surely
> If we assume above energy flux to persist for 0.5 s, what would the
> final travel velocity of a 1 x 1 x 0.01 m steel sheet be?
You got the numbers, now apply your high school physics.
> I'd say negligeably small, few 10 cm/s at best.
Need I say more?
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| Hara Ra <harara@shamanics.com> |
| Box 8334 Santa Cruz, CA 95061 |
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