Is this calculation for radiation pressure correct?

  • #1
I was thinking about how much propulsion you could get out of a solar sail, so I just did a rough calculation:
A solar sail 100m*100m in size (10'000m²), with 20grams of weight per m², an additional 100kg for probe and other stuff, so 300kg probe weight. Using the solar constant of 1400W/m², this should give us an 5256m/s of deltaV per year with ~50mN of thrust through the sail.
I know there are a lot of factors which i havent considered here, like increasing distance to the sun for interstellar travel, the weight is probably to optimistic too. Just wanted to know if 5256m/s is correct so I can continue from there.
 

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  • #2
russ_watters
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I was thinking about how much propulsion you could get out of a solar sail.... Using the solar constant of 1400W/m², this should give us an 5256m/s of deltaV per year with ~50mN of thrust through the sail....
No, that isn't how it works. Radiated energy is not mechanical kinetic energy. When absorbed, radiated energy just heats up an object. What you need is the momentum of the photons (and double it for reflection):
https://en.wikipedia.org/wiki/Solar_sail#Solar_radiation_pressure
http://www.georgedishman.f2s.com/solar/Calculator.html
 
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  • #3
jbriggs444
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I was thinking about how much propulsion you could get out of a solar sail, so I just did a rough calculation:
A solar sail 100m*100m in size (10'000m²), with 20grams of weight per m², an additional 100kg for probe and other stuff, so 300kg probe weight. Using the solar constant of 1400W/m², this should give us an 5256m/s of deltaV per year with ~50mN of thrust through the sail.
I know there are a lot of factors which i havent considered here, like increasing distance to the sun for interstellar travel, the weight is probably to optimistic too. Just wanted to know if 5256m/s is correct so I can continue from there.
Not many calculations there. Just inputs and results. Back of the envelope...

1400 Watts per square meter times 10000 square meters should give 14 megawatts.

14 megawatts times 365 days per year times 86400 seconds per day gives 441,504,000,000,000 = 442 terajoules per year.

E=pc. So divide that 442 terajoules by the speed of light (300,000,000 meters per second) to get the resulting momentum. 1.4 million kilogram meters/sec. [optional factor of two for reflection]

Divide by 300 kg and I get 4900 meters per second. Which is tolerably close to your result.
 
  • #4
russ_watters
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Hmmm, thanks for that - I had assumed based on the wording the OP was using the energy directly, but didn't do a calc to check.
 
  • #5
jbriggs444
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Hmmm, thanks for that - I had assumed based on the wording the OP was using the energy directly, but didn't do a calc to check.
I was tempted to the same conclusion. But then went "that should be a heck of a lot of energy for a puny 5 km/sec delta-V, maybe he's doing it right".
 
  • #6
sophiecentaur
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A nice mechanical approach is to calculate the momentum change when all the individual photons are absorbed or (better) reflected. Momentum of a photon is easy.
 

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