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submitted 3 weeks ago by fossilesque@mander.xyz to c/science_memes@mander.xyz

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[-] merc@sh.itjust.works 3 points 3 weeks ago

That's 11.2 km/s and 42.1 km/s.

Also, even if the manhole cover was going at above 12 km/s the trajectory has to be right for that to result in orbit. Most paths it would take would result in it going up and then coming back down again. Similarly, if somehow it did manage more than 50 km/s and wasn't destroyed in the atmosphere, it might have the velocity to escape the sun's gravity, but probably wouldn't be on the right path to do it. Most likely it would fall into the sun.

So, assuming the 125,000 mph (55 km/s) velocity is correct, the most likely outcome is that it was a reverse-meteor, something that burned up going up through the atmosphere, not down. And even if it did have enough speed to get out of the atmosphere, and there was enough of it left, it most likely fell right back down through the atmosphere somewhere else, either burning up on re-entry or hitting the ground (or the water) somewhere else.

[-] Swedneck@discuss.tchncs.de 1 points 5 days ago

correction to your correction: it would not fall into the sun, falling into the sun is basically impossible, it would just end up in a highly eccentric orbit around the sun.

[-] merc@sh.itjust.works 1 points 4 days ago

Yeah, "fall into the sun" was sort of hyperbole. If it truly got out into space and was going fast enough to escape Earth's gravity, it would start orbiting with earth's orbit plus some delta. Out of all the possible angles it could leave the earth, there are probably 2 angles where it would directly hit the sun One is the angle that cancels out all the orbital velocity of the earth and sends it directly at the sun, the other is the one that does the same but sends it directly away from the sun. Of all the possible trajectories on the surface of a sphere, only those two tiny solutions would end up with it contacting the sun, everything else would result in an orbit.

Of course, given enough time, it's pretty likely that if it isn't collected by a planet, it will eventually end up in the sun. There isn't much friction in space, but there's a tiny bit: solar wind, micrometeoroids, etc. Eventually its orbit would decay and it would stray too close to the sun.

[-] druidjaidan@lemmy.world 2 points 3 weeks ago

Ignoring that it burned up and ignoring losses due to drag if it somehow didn't. Isn't the point of escape velocity that it explicitly won't come back down.iar least not on earth. Your trajectory won't matter as you have enough velocity to escape the gravity of earth and will orbit the sun. Further if you managed the solar system escape velocity you will end up orbiting the galactic core. Trajectory doesn't matter if you have escape velocity. Correct trajectory just minimizes the delta v needed to reach that escape velocity.

At least that's all my recollection.

[-] Maggoty@lemmy.world 2 points 3 weeks ago

Escape velocity means you could stay in orbit. It doesn't guarantee anything if you launch at the wrong angle.

[-] merc@sh.itjust.works 2 points 3 weeks ago

Exactly. It's the minimum speed required to get into orbit assuming you get the direction correct. If you launch vertically, you'll almost certainly come back down, no matter how far out into space you go. The only consideration is that if you go far enough out you might be influenced by the gravity of something else like the moon which could change your trajectory.

[-] druidjaidan@lemmy.world 2 points 3 weeks ago

That is not the definition of escape velocity. Escape velocity is the minimum velocity to escape a body's gravity well entirely. Orbital is much lower

[-] druidjaidan@lemmy.world 1 points 3 weeks ago

That is not the definition of escape velocity. Escape velocity is the minimum velocity to escape a body's gravity well entirely. Orbital is much lower