Like Fluoride or Oxygen.

        • @[email protected]OP
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          51 year ago

          Alle Dinge sind Gift, und nichts ist ohne Gift; allein die Dosis macht, dass ein Ding kein Gift ist.

          All things are poison, and nothing is without poison; the dosage alone makes it so a thing is not a poison.

          Paracelsus, 1538

          The word for poison in German is Gift?!

          The word has been used as a euphemism for “poison” since Old High German, a semantic loan from Late Latin dosis (“dose”), from Ancient Greek δόσις (dósis, “gift; dose of medicine”). The original meaning “gift” has disappeared in contemporary Standard German, but remains in some compounds (see Mitgift). Compare also Dutch gift (“gift”) alongside gif (“poison”).

          Well that’s dumb.

      • @[email protected]
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        41 year ago

        I’d argue gravitational force isn’t lethal. As long as you don’t arrive at whatever is pulling you & the gradient of gravity doesn’t change across your body length. You could be perfectly fine (for a while) orbiting a black hole at enormous speeds (assuming you don’t collide with matter in the accretion disc.

        • @[email protected]
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          181 year ago

          I’d argue against that. For one thing it is impossible to imagine a situation where there is no change in the gravitational gradient across your body over time. Your orbiting a black hole situation is a perfect example of a situation where the gradient alone would tear you apart. The conditions you’ve specified are tautological. There’s no way to maintain a zero gravitational gradient while also simultaneously having extremely high gravitational field. The two are mutually exclusive in any conceivable scenario.

          It’s like saying a human being in a hypersonic wind stream won’t necessarily hurt you, burn you alive and rip you to pieces (not necessarily in that order) as long as there is no turbulence and you have a sufficient boundary layer – but you’re a non-aerodynamic human body in a hypersonic wind stream, so of course there will be turbulence and the boundary layer will not protect you at all, you’re going to die, basically instantly.

          • @[email protected]
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            01 year ago

            Does the change in gravity gradient across your body kill you right now? No? You are currently orbiting the supermassive black hole in the center of the milky way. You and everything else in the milky way aside from a few intergalactic objects just traveling through.

            I am not an astrophysicist, but I do understand basic physics.

            • @[email protected]
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              11 year ago

              Does the change in gravity gradient across your body kill you right now? No? You are currently orbiting the supermassive black hole in the center of the milky way.

              It was implied by “accretion disc” and by the fact that we’re talking about gravitational gradients at all that we’re talking about a close orbit. Gravitational strength gets smaller with distance according to the inverse square law, so by the time you’re a few light years out from the galactic core the gravitational gradient is already extremely insignificant.

              • @[email protected]
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                1 year ago

                Accretion discs can be large enough that I am pretty sure a human body wouldn’t be torn apart at that distance (at least the outer bits) by the difference in gravity across it’s length. In the linked article about the supermassive black hole at the center of the Milky Way, we’re talking 1000 astronomic units, so 1.5 * 10^14 meters.

                The current value of this black hole’s mass is estimated at ca. 4.154±0.014 million solar masses.

                So let’s calculate the equivalent distance from the sun in terms of gravitational force on an object at the outer edge of the accretion disk:

                F_sun = C * (R_equivalent)^-2 * m_object

                F_black_hole = C * 4.15*10^6 * (R_accretion_disk)^-2 * m_object

                where C equals the gravity constant times the mass of our sun.

                ==> C * (R_equivalent)^-2 * m_object = C * 4.15*10^6 * (R_accretion_disk)^-2 * m_object

                divide by C and m_object:

                <=> (R_equivalent)^-2 = 4.15*10^6 * (R_accretion_disk)^-2

                invert:

                <=> R_equivalent^2 = (1/4.15) * 10^-6 * (R_accretion_disk)^2

                ==> R_equivalent^2 ~= 0.241 * 10^-6 * (R_accretion_disk)^2

                square root (only the positive solution makes sense here):

                ==> R_equivalent ~= 0.491 * 10^-3 * R_accretion_disk

                with R_accretion_disk = 1000 astronomic units = 10^3 AU

                <=> R_equivalent ~= 0.491 * 10^-3 * 10^3 AU

                <=> R_equivalent ~= 0.491 AU

                Unless I have a mistake in my math, I sincerely hope you will agree that the gravitational field (tidal forces) of the sun is very much survivable at a distance of 0.491 astronomical units - especially since the planet Mercury approaches the sun to about 0.307 AUs in its perihelion.

            • @[email protected]
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              11 year ago

              If the gravity were strong enough and the source close enough then the tidal force would absolutely be strong enough to simultaneously crush you and rip you apart. The same effect gives rise to tides on this planet, hence the name.

              • @[email protected]
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                01 year ago

                Your orbiting a black hole situation is a perfect example of a situation where the gradient alone would tear you apart.

                I just proved this claim of yours wrong, and then you move the goalposts. I said from the very beginning that a gravity gradient is a problem.

                • @[email protected]
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                  1 year ago

                  I studied Relativity at university as part of combined Physics/Maths degree, but please feel free to continue entertaining us with your popular magazine-based learnings.

        • themeatbridge
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          31 year ago

          You argue that it isn’t, and then provide several examples where it is.

          • @[email protected]
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            -21 year ago

            Can’t help you if you don’t understand what “ideal cases” are, when the real world examples are not practical to describe the underlying principle. The point is: gravity doesn’t kill you, no matter how high the absolute. Arguably, in a perfect gravitational field, you could even be accelerated at insane speeds without experiencing discomfort, because each atom of your body would be experiencing the same acceleration.

        • @[email protected]
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          21 year ago

          I think General Relativity is based on the idea that a frame of reference that’s in freefall is equivalent to one that in a gravity free region of space (at least that was one of Einstein’s Gedankenexperiments that led him to his theory of GR).

          Having said that, in reality a sufficiently strong gravitational field will cause a tidal effect, which will crush you along one axis and pull you apart along another.

          • @[email protected]
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            01 year ago

            There was definitely something like that - I am not sure if free-fall and being accelerated in a gravitational field are the same though. It may be that GR is talking about moving along lines in space-time that have the same gravitational potential (orbits), and moving across potential lines counts as an accelerated frame of reference in which you wouldn’t observe the same as in a reference frame moving at constant speed.

            • @[email protected]
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              1 year ago

              I was thinking of the Equivalence Principle:

              the equivalence of gravitational and inertial mass, and Albert Einstein’s observation that the gravitational “force” as experienced locally while standing on a massive body (such as the Earth) is the same as the pseudo-force experienced by an observer in a non-inertial (accelerated) frame of reference.

              • @[email protected]
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                01 year ago

                okay, but that would be an accelerated frame of reference, not equivalent to one that is “gravity free”

        • Otter
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          11 year ago

          Wouldn’t a high enough force cause the gradient of gravity to differ?

          Unless I misunderstood how that works. I’m picturing a downed powerline that causes large differences in voltage across the ground, which is why you are supposed to shuffle instead of taking a normal step. Would a high enough gravity cause a harmful gradient across the length of a human body?

          • Bizarroland
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            121 year ago

            The term spaghettification comes into mind.

            Like if you were free falling into a black hole, the gravity forces would rip you to shreds long before you ever actually impacted anything because the difference in the force of gravity on the parts of your body that are closer to the black hole and the parts of your body that are farther away are enough to shred you like lettuce.

            • @[email protected]
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              01 year ago

              I have read popular scientific articles however according to which in a large enough black hole, it may be possible to fall through the event horizon before being inconvenienced by the gravity gradient, and even the smartest physicists do not know for sure what will happen beyond the event horizon. In theory, there could be the beginning of another universe there :) Like - the singularity at the center of the black hole could expand as a big bang into a brand new universe “on the other side”.

          • @[email protected]
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            01 year ago

            Gradient: the change of a value (here: gravitational force, or rather: potential) over a reference variable (here e.g. the length of the body)

            No, the absolute value of the gravitational force does not matter for the gradient. Gravitational force (potential) is proportional to the inverse distance squared from the center of mass that exerts the gravitational potential. If your distance from the object R is large enough, then the gradient of gravity across the length of your body is negligible: In the worst case, with your body length being s, the gravity at the part of your body closest to the center of mass pulling you would be: F_max = F_min * ( R^2 / (R-s)^2 ), and with s << R, this becomes F_min, the force at the part of your body furthest away from the mass pulling you in.

            This becomes problematic when you get “too close” to the body in question - and where too close begins, depends indeed on the absolute force. But for each black hole, there’s a safe distance at which you could fall around it, assuming no other factors killing you (like intersteller particles, or an accretion disc)

      • @[email protected]
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        21 year ago

        I thought about this a bit and concluded that it only applies to physical materials and forces.

        For example: There certainly are lethal ideas, but most of them are not, and much like bosons they can overlap, so filling a person with multiple copies of the same (benign) thought has a diminishing effect.

        But yeah, anything physical has a lethal concentration.