# Is it ever possible that the graviation from the mass of kinetic rotational energy will overcome centrifugal force?

what is gravitational force?

"Rotation speed can create centrifugal force opposing gravity and making things lighter," Rotation also stores energy, and energy is mass, per $E=mc^2$. A body spinning sufficiently fast will exert higher gravity, e.g. a slowly-spinning neutron star will have a weaker gravitational pull than equivalent neutron star that spins very fast."

Intuitively, my answer is "no way", added mass by rotational velocity I would think, could never be extensive enough to counteract the "flicking off" or centrifugal force of very fast rotation. Even with a Neutron Star I would think it's impossible, but I'm not 100% sure, so I thought I'd ask.

If we ignore relativity from motion but not $E=mc^2$, as that's the crux of the question

Centrifugal force = $m v^2/r$

Kinetic energy of rotation = $1/2 I w^2 = 1/5 m v^2$

Mass equivalent of kinetic energy = $1/5 m v^2/c^2$

gravitational force $g=\frac{Gm}{r^2}$

so if we apply the gravitational force of kinetic energy

$g = G(1/5 m * v^2/c^2) / r^2$, or, simplified, $G*m*v^2 / r^2 c^2$

and we compare the two equations Centrifugal $m v^2/r$ additional gravitational $G*m*v^2 / r^2 c^2$

we can remove $m*v^2$ from both on the top and 1 $r$ from the bottom

additional gravitation ratio to centrifugal force = $G/r * c^2$

$G$ and $c$ are numerical. $G$ is very small, $c$ is very big and the ratio grows smaller as the radius grows larger.

gravitational constant: $6.67408 × 10^{-11} m^3 kg^{-1} s^{-2} c = 2.998 x 10^8 m s-2 the ratio, unless my math is broken, centrifugal force to additional gravity from added mass by kinetic energy of motion =$1/r * 1.35 * 10^{27}$, so you'd need a hugely small r, almost a plank length or a singularity where the added gravity from kinetic energy of rotation would overcome the "flicking off the surface" or centrifugal force. When I work out the units, I get meters per kilogram, which I don't think is right. The units should cancel out with a ratio of two forces in opposite directions, so I think I made an error, but I don't see where I made it. My question is two fold. 1) is my math broken? and if so, where? and 2) is the added mass from kinetic rotational energy ever relevant, say in a very rapidly rotating Neutron star? Could it ever assist the Neutron star in collapse or add gravity?, I can see how it could add to flattening, as rotation flattens objects naturally and perhaps, add a speck of gravity on the polls where centrifugal force is zero, but logically, I think, rotational energy would end up spinning any non black hole object apart, long before the added energy of rotation had enough mass to make a measurable difference on gravity. Is my sense right or is there a situation where added mass from energy of rotation could overcome the centrifugal force? • This site supports TeX style maths, You could make your question prettier by TeX'ing the maths. – James K Feb 7 '16 at 11:43 • Yes, your maths is broken. Questions like these cannot be addressed without GR. But yes, rapid rotation does increase the radius of a neutron star and increase the maximum mass of a neutron star that can be supported by a given eqn of state. Centrifugal force always wins for lower mass neutron stars. – Rob Jeffries Feb 7 '16 at 21:33 • "added mass by rotational velocity I would think, could never be extensive enough to counteract the "flicking off" or centrifugal force of very fast rotation. Even with a Neutron Star I would think it's impossible" - You are right with the neutron star. It would increase gravity and as result reduce impact of centrifugal force but not enough to prevent escape with sufficient angular velocity. Only if you increase the density even more (and get a black hole) this ceases to be the case. (and how black hole, which is a singularity=point can rotate... that's a whole can of worms.) – SF. Feb 7 '16 at 21:34 • BTW, there's (probably) one body between a neutron star and black hole, density-wise: a quark star. But we haven't discovered any yet (conditions for one to form are exceedingly rare) so we don't exactly know if it would satisfy conditions of your question. – SF. Feb 8 '16 at 23:50 • I think you are misunderstanding the statement you quote about the centrifugal force making things lighter, it wasn't meant to be about kinetic energy being equivalent to mass and therefore exerting gravity, it's just a much simpler statement about fictitious forces in rotating coordinate systems in Newtonian physics. In a rotating coordinate system centered on the center of the Earth, any object experiences a centrifugal force which is in the opposite direction as gravity, so it can partly cancel the gravitational force from their perspective. – Hypnosifl Apr 14 '16 at 17:58 ## 1 Answer In some sense, the answer is trivially 'yes': Due to spacetime curvature, there exist no circular orbits below$\frac{3}{2}r_s$, with$r_s$the Schwarzschild radius. So whenever you get an object sufficiently dense, that its radius is below its$\frac{3}{2}r_s$, the mass equivalent to any nonzero rotational energy overcomes the centrifugal force of an orbit at$\frac{3}{2}r_s$. Below$\frac{3}{2}r_s\$ even zero rotational energy overcomes the centrifugal force, since the latter is then pointing inward.

This doesn't require a Black Hole.