You argue that it isn’t, and then provide several examples where it is.
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.
themeatbridge@lemmy.world 1 year ago
raspberriesareyummy@lemmy.world 1 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.
themeatbridge@lemmy.world 1 year ago
Boy that’s a lot of words for “lol, you’re right. My mistake.”
jon@lemdro.id 1 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.
raspberriesareyummy@lemmy.world 1 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.
jon@lemdro.id 1 year ago
I was thinking of the Equivalence Principle:
equivalence principle is 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.
raspberriesareyummy@lemmy.world 1 year ago
okay, but that would be an accelerated frame of reference, not equivalent to one that is “gravity free”
otter@lemmy.ca 1 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@kbin.social 1 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.
raspberriesareyummy@lemmy.world 1 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”.
raspberriesareyummy@lemmy.world 1 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)
otter@lemmy.ca 1 year ago
Makes sense, thank you :)
cecilkorik@lemmy.ca 1 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.
raspberriesareyummy@lemmy.world 1 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.
cecilkorik@lemmy.ca 1 year ago
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.
raspberriesareyummy@lemmy.world 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, we’re talking 1000 astronomic units, so 1.5 * 10^14 meters.
The current value of its mass is 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 solutions make 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 * R_accretion_disk
<=> 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.
jon@lemdro.id 1 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.
raspberriesareyummy@lemmy.world 1 year ago
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.