cubinator

Assuming a reflectivity of 49% in the visible light range for the stainless steel Starship hull, and a 55 m long orbiter, I'm going to wave my hands and say the part of the ship that's reflecting the sun directly towards you is nominally a strip about 0.5 m wide along the whole length of the ship. That's an area of about 27.5 m, we'll round down to 27 to be conservative. So, what we want is the area that takes up in the sky. At a distance of 400 km (the height of the ISS) our Starship spans 0.004 degrees. As far as an area in the sky, since we said the sun-direct part of that reflection was 0.5 m wide, we'll cut that value in half for a degree-area of 0.0019 deg^2. The sun's apparent magnitude is -26.74, and its area is pi * 0.5^2 deg^2 = .785 deg^2. We're dealing with small enough areas that I'm going to pretend the sky is flat for this. So, we're dealing with an area of 0.00246 of the sun's area, reflecting 49% of its light. That means that .49 * .00246 = .0012 of the Sun's brightness is coming off the ship. I will use the apparent magnitude equation m2-m1 = -2.50 log(B2/B1) m2 = magnitude of the sun = -26.74 m1 = magnitude of Starship = ? B2 = brightness of the Sun = 1 B1 = brightness of Starship relative to the sun = .0012 -26.74 - m1 = -2.5*log(828.8909) -26.74 - m1 = -7.296 m1 = -19.44? That seems very high. Although this was something of a best-case scenario like an Iridium flare but much bigger. If it does get this bright, it would likely only be for a fraction of a second. I'll try for an area of just a spot instead of a whole line. 0.5 m by 0.5 m, we'll say. Using the methods above, we get an area of 0.0000358 deg^2, which is .0000456 of the Sun's area. In this case we get m1 = -15.89. Still very high. I doubt this is accurate.

@CatastrophicFailure Heh, I'm glad you think so highly of me. I actually did dig around in some old papers but couldn't find a really solid reflectivity percentage for stainless steel, and I had some other stuff to do so I dropped it for a bit. Since you insist, I'll look around some more and try to math-whiz up some magnitude numbers for various distances for you. I think it's around 90%, maybe a little over like 93% so I'll probably go with that. Though this site says 49%: https://www.azahner.com/materials/stainless-steel

The bulk of what we see from the station is reflected off dark solar panels. I think it's quite unlikely the station will be able to outshine an actual mirror.

4/10 a shallow reentry is no fun, you've got to go steep for real excitement!

Error 12: (

Hellas Basin [Mars]

Wow, that's cool.

Well it's not the solar storm, that has an ETA tomorrow.

Last night I dreamt I was in the ISS cupola controlling a spaceship to dock. Then, when I finished that, I went around doing lots of front flips.

Is it bolted down, though? If not, it might be a hop by default...

KORAAAAAHHHH MATAAAAAAHHH KORAAAAAAHHHHHH RAAATAAMAAAAAHHH

I don't know why I didn't upload this sooner, here is a theme for Mercury that has been sitting on my hard drive for a while: Also Venus:

"Adios, amigos!" Buzz Aldrin's last words on the Moon's surface

Hello there!

E

A big honking skyscraper flies away.

N o y ' t

Now that's unexpected. Every place we visit reinforces for me the idea that space is every bit as exciting as science fiction makes it seem.

A full sized prototype with full fuel tanks could very nearly make orbit...I suppose they don't have the cooling system installed on this prototype, though.

Installed MATLAB today. I will likely soon become veeeery familiar with it...

Tycho Crater [Moon]

When was Jupiter named, and why was the name chosen? It seems weirdly coincidental that the planet was named after the king of the gods before we had any idea it was actually the largest planet. And Venus is brighter most of the time.

Strange twilight lighting and bright exposure is my guess.

Good choice!