# Is this calculation for radiation pressure correct?

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.

russ_watters
Mentor
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):
http://www.georgedishman.f2s.com/solar/Calculator.html

• Lynda Jhon
jbriggs444
Homework Helper
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.

russ_watters
Mentor
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.

jbriggs444