• Professorozone@lemmy.world
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    20 days 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.

    • Bosht@lemmy.world
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      20 days ago

      Gotta love Tumblr. Just massive amounts of disinformation and bullshit all the time.

    • SwordInStone@lemmy.world
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      20 days ago

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

    • merc@sh.itjust.works
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      20 days 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
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        17 hours 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.

      • druidjaidan@lemmy.world
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        19 days 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
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          19 days 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
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            19 days 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
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              18 days 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
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            18 days 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

    • CellarRat@sh.itjust.works
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      20 days 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)

    • stephen01king@lemmy.zip
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      19 days 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.

        • stephen01king@lemmy.zip
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          17 days 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.