That doesn’t seem particularly fast of a spin and I’m surprised it’s the fastest one observed.
There is a moment in 2001: A Space Odyssey that I’ve always liked. It’s at about 1:13:15 or so, and is just a distant shot of the Discovery One. Suddenly two space rocks silently tumble past. I like the scene because it gives a sense of the scope and silence of outer space, but I suppose it would be incredibly rare to be so close to even one rock, let alone a pair. And they are tumbling at about the pace measured by the one in that article, so perhaps even rarer still.
Well as the article states the bigger ones would tear themselves apart if they spun too fast so that is one reason. In the end they are just big rocks and thus not very good at tensile loads...
I’m surprised to learn that one rotation every 2.6 seconds is the fastest we’ve ever seen. That actually seems kind of slow. I can think of lots of things that rotate faster than that, including me if I want to. Is this mainly because the solar system is old and energy has dissipated over time?
I am not sure about the exact numbers but I would think objects of a certain size would fall apart if they spin too fast. I think I read that this is 1m. Spinning that fast creates some significant centrifugal force.
There are two main mechanisms that cause asteroids to spin. The first is as a result of collisions, which is fairly intuitive. But even in the absence of any collisions asteroids can acquire spin due to the YORP Effect (named after Yarkovsky, O'Keefe, Radzievskii, and Paddock). The idea is that if the surface is anisotropic (which is true of asteroids), then as the asteroid radiates heat, the photon pressure will be anisotropic and will start to impart a torque onto the body.
Not to nitpick (I only stumbled upon this recently myself), but the mechanism driving a Crookes radiometer (which I presume is the type you're talking about) is not photon pressure, but interactions with the low pressure gas in the bulb [1].
At some point gravitational forces start counteracting centrifugal forces again, so density means a lot. Apparently the fastest-spinning neutron star known so far spins at about 716 rotations per second. (see https://en.wikipedia.org/wiki/Neutron_star)
It's approximately only a problem of density. For large objects, you can neglect tensile strength, so "How fast until it's torn apart" becomes "how fast would it have to spin for an object on the equator to be in orbit?"
F_c = m_1 r ω^2 = F_g = G m_1 m_2 / r^2
cancel out the orbiting mass (m_1), rearrange it a bit
ω^2 = G (m_2 / r^3)
where you'll note that term in parenthesis is directly proportional to the density.
"how fast would it have to spin for an object on the equator to be in orbit"
I think this is correct, but then it can be simply calculated knowing the escape velocity on Earth is 11.2 km/sec and the circumference is 40e3 km:
40e3 / 11.2 = 3571 seconds
Therefore it's the Earth span at one revolution per hour, objects at the equator would escape gravity... I'm not sure where is your error in your formula that leads you to a different response.
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[ 3.5 ms ] story [ 52.0 ms ] threadThere is a moment in 2001: A Space Odyssey that I’ve always liked. It’s at about 1:13:15 or so, and is just a distant shot of the Discovery One. Suddenly two space rocks silently tumble past. I like the scene because it gives a sense of the scope and silence of outer space, but I suppose it would be incredibly rare to be so close to even one rock, let alone a pair. And they are tumbling at about the pace measured by the one in that article, so perhaps even rarer still.
That's a toy i had as a kid, essentially!
https://youtu.be/t-JN2U4jHgk?feature=shared
[1]: https://en.wikipedia.org/wiki/Crookes_radiometer#Explanation...
Curious, what is that number for a space rock the size of Earth?
F_c = m_1 r ω^2 = F_g = G m_1 m_2 / r^2
cancel out the orbiting mass (m_1), rearrange it a bit
ω^2 = G (m_2 / r^3)
where you'll note that term in parenthesis is directly proportional to the density.
So the answer would probably be ~a 2.2 hour day.
I think this is correct, but then it can be simply calculated knowing the escape velocity on Earth is 11.2 km/sec and the circumference is 40e3 km:
40e3 / 11.2 = 3571 seconds
Therefore it's the Earth span at one revolution per hour, objects at the equator would escape gravity... I'm not sure where is your error in your formula that leads you to a different response.