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NASA's solar sail successfully spreads its wings in space (www.space.com)

submitted 2 months ago by threelonmusketeers@sh.itjust.works to c/spaceflight@sh.itjust.works

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[–] intensely_human@lemm.ee 1 points 2 months ago (1 children)

Can that calculation be done assuming conservation of momentum and the laser’s power?

How’d you calculate it? Does the area of the sail figure into your calculations?

[–] nikaaa@lemmy.world 3 points 2 months ago

Oh yes, of course:

I guess the formula to calculate the momentum p of light is p = E/c where c is the speed of light and E is the energy.

So, if the Energy is 10 MW, or alternatively, 10 MJ/sec, then E = 1e7 J. c = 3e8 m/s. p = E/c = 1e7/3e8 J/(m/s) = 1e7•1e-17/3e8 kg•(m/s)²/(m/s) = 3e-2 kg•(m/s).

(Since E = m•c², 1 kg • c² = 1 kg • (3e8 m/s)² = 1e17 kg•(m/s)².)

Now, if we assume that that momentum gets transferred to the sail, which weighs approximately 20 kg, then we get the velocity of the sail: v = p/20 kg = 3e-2 kg•(m/s)/20 kg ≈ 1e-4 m/s. And this transfer of momentum happens every second, so the acceleration is roughly 1e-4 m/s², or 0.001 m/s².