this post was submitted on 03 Sep 2024
48 points (98.0% liked)

Spaceflight

647 readers
21 users here now

Your one-stop shop for spaceflight news and discussion.

All serious posts related to spaceflight are welcome! JAXA, ISRO, CNSA, Roscosmos, ULA, RocketLab, Firefly, Relativity, Blue Origin, etc. (Arca and Pythom, if you must).

Other related space communities:

Related meme community:

founded 1 year ago
MODERATORS
you are viewing a single comment's thread
view the rest of the comments
[–] threelonmusketeers@sh.itjust.works 5 points 2 months ago (4 children)
[–] Rangelus@lemmy.nz 2 points 2 months ago* (last edited 2 months ago) (1 children)

Cheers mate. Neat stuff!

I wonder how much force a sail of that size is expected to produce?

[–] nikaaa@lemmy.world 2 points 2 months ago* (last edited 2 months ago) (1 children)

I just did some quick maths (and assuming my formulas are correct), beaming at it with a 10 MW laser would cause it to speed up by 0.001 m/s² assuming the sail has 20 kg mass. So not very much, but also not nothing.

[–] 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.

load more comments (2 replies)