• Captain Aggravated
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    267 months ago

    The calculation of its speed was made by high speed camera, as you’ve probably seen the Mythbusters do. In this case the manhole cover was seen in flight in precisely one frame of high speed camera footage, and for it to go “installed, in flight, gone” in three frames means it would have had to be moving at mach jesus.

    It likely didn’t make it to space intact; it would have had ultrasonic compression heating on one side and a nuclear explosion on the other. It’s probably still here in the form of iron oxide dust scattered about the Northwestern hemisphere.

    • @[email protected]
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      17 months ago

      If the event was near the equatorial near midday then there’s a very very (very) slim chance it was pointed directly at the sun.

      • atocci
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        147 months ago

        If it was pointed directly at the sun, it would miss. Not that this would make the odds any better, but aiming straight at the sun doesn’t work either.

    • @[email protected]
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      177 months ago

      Saying this with only an understanding of orbital mechanics learned from Kerbal Space Program, I’d say the chances are damn near 0%. Hitting the sun is actually pretty difficult and requires a precise amount of Δv (change in velocity). This thing had such a huge Δv that it would have left the solar system.

  • @[email protected]
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    7 months ago

    Nope, it would just have bursted due to thermal schock and pressure. Escape velocity, what are you dreaming, is the lid made of tungsten?

    • @[email protected]
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      7 months ago

      This is the origin apparently.

      RRB: “My calculations are irrelevant on this point. They are only valid in speaking of the shock reflection.” Ogle: “How fast did it go?” RRB: “Those numbers are meaningless. I have only a vacuum above the cap. No air, no gravity, no real material strengths in the iron cap. Effectively the cap is just loose, traveling through meaningless space.” Ogle: And how fast is it going?" This last question was more of a shout. Bill liked to have a direct answer to each one of his questions. RRB: “Six times the escape velocity from the earth.”

    • @[email protected]
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      27 months ago

      Hypersonic heating is really weird. We only have data going to about mach 17 (the HTV-2 was the fastest solidly atmospheric vehicle I found) but as we go from subsonic, to supersonic, to hypersonic regimes air becomes pretty much incompressible, and forms a really solid shockwave in front of a fast-moving object. Air is a pretty good thermal insulator, so for very fast, blunt objects they actually heat much slower than you might expect.

      Tl;Dr it absolutely vaporized, but it likely lasted longer than you might expect.

  • @[email protected]
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    187 months ago

    I love the idea that our first message to aliens might be “FRESH WATER ONLY. NO WASTE.”

  • @[email protected]
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    7 months ago

    One thing that no one ever talks about with this is the massive air resistance on it going Mach 164 through the atmosphere would incur (albeit for a very brief period)…I bet that would knock 25-50 kmph off it easily.

  • I Cast Fist
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    147 months ago

    We should test this again, but with a fridge and someone inside it for the nuclear blast. I bet that would work out great

  • linuxgator
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    97 months ago

    The foundry that made that manhole cover has some great potential advertising claims.

  • @[email protected]
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    87 months ago

    Ok, tin foil hats for this one, our universe isn’t exactly infinite in the way people traditionally think like numbers. The edges of the universe bend and form a large shape, say a sphere for simplicity. That cover speeds through and circles back eventually, but do to it’s speed and travelling along the edges of everything and relativity, when it returns it’s not at the same point or even at the same speed. It arrives before it initially left, quite a bit before it left… So much so that it kills off the dinosaurs.

    • LostXOR
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      97 months ago

      Sadly, the escape velocity of our galaxy is an order of magnitude higher than the manhole cover’s velocity. And even at that speed it wouldn’t hit with nearly enough energy to cause a mass extinction. Still a fun idea though. :)

  • @[email protected]
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    667 months ago

    Ummm, not sure where they got these numbers from but Earth’s escape velocity is not 7000mph and escaping the sun’s gravitational pull (leaving the solar system from Earth) is not 30,000mph. Respectively the numbers are approximately 25,000mph and 94,000mph. You’re welcome.

    • @[email protected]
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      27 months ago

      94000mph is relative to the sun’s surface. Relative to the Earth’s surface, it is around 37000mph, which means they were still wrong.

        • @[email protected]
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          6 months ago

          42.1 km/s is the speed required relative to the sun’s surface for objects launching from Earth’s surface. You need to look at the value labelled V_te, which is the speed relative to the minor body the object is launching from. In this case, it is 16.6 km/s.

    • @[email protected]
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      97 months ago

      yeah, and it is not “research” to check it. They literally teach it in primary school physics.

    • @[email protected]
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      37 months ago

      I like how they are implying the speed of light is only 500000mph (as opposed to 671,000,000 mph or 1,080,000,000kph)

    • @[email protected]
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      37 months 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.

      • @[email protected]
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        16 months 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.

        • @[email protected]
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          16 months 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.

      • @[email protected]
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        27 months 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.

        • @[email protected]
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          27 months ago

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

          • @[email protected]
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            17 months 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

          • @[email protected]
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            27 months 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.

            • @[email protected]
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              27 months 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

  • @[email protected]
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    267 months ago

    Ive seen this claim a dozen times. It’s a disc shape. How this thing isn’t going to start flipping and curving its trajectory, or just plain old running out of energy due to air resistance, and not making it out of earth’s atmosphere is beyond me.

      • @[email protected]
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        37 months ago

        If it’s like a frisbee, yeah, but it still curves. Now start it spinning like spinning a coin on edge. The curving will be much more dramatic.

      • @[email protected]
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        17 months ago

        Throw it into water or gelatin. At thousands of metres per second the air is going to seem much more dense.

            • @[email protected]
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              47 months ago

              tbf the calculated speed is actually roughly the minnimum based on its starting position and the frame it appeared in. it could have actually been going even faster.

              • Victor
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                27 months ago

                I don’t count having no visual indication of the object as “tracking” it, if we’re talking semantics. One frame could equal an even faster speed than what it would minimally take to cross the entire width of the image at some trajectory vector. For other vectors, it could be (much) less (like not passing straight through the image from on side to the opposite side, e.g.).

                It’s important to not hang too hard on this as the escape speed is dependent on air resistance, or rather lack thereof. Those escape speed numbers are defined along with the assumption of zero air resistance or other forces acting on the object.

                • @[email protected]
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                  27 months ago

                  You can use the frame from before to calculate the MINIMUM speed. It could have been going even faster.