Is this calculation for radiation pressure correct?

  • #1
hewhorizons
1
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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 haven't 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
hewhorizons said:
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
hewhorizons said:
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 haven't 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
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
russ_watters said:
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
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.
 

1. What is radiation pressure?

Radiation pressure is the pressure exerted by electromagnetic radiation on any surface that absorbs or reflects it. It is a result of the transfer of momentum from the photons of the radiation to the surface. This pressure is usually very small and only becomes significant in extreme conditions such as near stars or in laser experiments.

2. How is radiation pressure calculated?

Radiation pressure is calculated using the formula P = I/c, where P is the pressure, I is the intensity of the radiation, and c is the speed of light. This formula assumes that the surface is perfectly absorbing and reflects all the photons back in the opposite direction.

3. Is this calculation for radiation pressure accurate?

The accuracy of the calculation for radiation pressure depends on the assumptions made and the conditions under which it is being applied. In most cases, it is a good approximation, but in some situations, such as near highly reflective surfaces, it may not give an accurate result.

4. Can radiation pressure be measured in real-life situations?

Yes, radiation pressure can be measured using various techniques, such as using a radiation pressure balance or a laser beam deflection setup. These measurements are important in understanding the effects of radiation pressure in different environments and for various applications.

5. What are some practical applications of radiation pressure?

Radiation pressure has various practical applications, such as in solar sails for space propulsion, in optical tweezers for manipulating particles, in laser cooling techniques, and in measuring the properties of materials through the pressure they exert on light. It is also an important factor in understanding the dynamics of celestial bodies and their interactions with each other.

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