<p>Related: <a href="https://astronomy.stackexchange.com/questions/7778/is-it-possible-to-break-apart-a-neutron-star/7802#7802" title="Is it possible to break apart a neutron star?">Is it possible to break apart a neutron star?</a></p>

<p>The rotation must be powerful enough to overpower the force of gravity in order for anything to happen.  I'm not sure about the 'ripping apart,' but if the rotation does overpower gravity surface material will be ejected from the body, a process known as <strong>mass shedding</strong>.</p>

<p>As neutron stars are extremely dense they can have great amounts of angular velocity, like the pulsar class neutron star mentioned in your question. If you can quickly increase the angular velocity of the body without also increasing the mass and density, perhaps you could rip it apart, but that would be very difficult to do to a neutron star.</p>

CC BY-SA 3.0
2015-04-04T09:56:41.440Z
Is it like a ice skater doing a spin on the ice rink and when she tuck her arms inward she will go faster?
user6760
CC BY-SA 3.0
2015-04-04T10:03:51.330Z
This is one of the ways they achieve such great rotational speeds.  Generally neutron stars come from stars with about 10x the mass of our Sun. The resulting neutron star likely has a radius of less than 15 kilometers. Not all the mass becomes a part of the neutron star though - plenty is ejected (though this is generally credited to the supernova, not the rotational velocity).
Mitch Goshorn
CC BY-SA 3.0
2015-04-04T10:08:57.213Z
It would probably also be worth noting that as the strength of gravity increases as the distance from the center of gravity decreases, as the body becomes more dense the escape velocity at the surface will increase.
Mitch Goshorn
CC BY-SA 3.0
2015-04-04T21:09:41.763Z
Not really related, but young stars spin faster.   This makes sense if you think about it.  The coalescing of a cloud of gas into a star would greatly increase angular velocity, as angular momentum is conserved.   Stars tend to lose rotational speed over time.   http://www.space.com/28255-star-spin-age-kepler-spacecraft.html
userLTK
CC BY-SA 3.0
2015-04-04T09:53:10.187Z
2015-04-04T10:07:21.910Z
2017-04-13T12:59:07.017Z
Mitch Goshorn
<p>The structure (mass versus radius and density profile) <em>is</em> influenced by its rotation rate, but not by as much as you might think.</p>

<p>Even in Newtonian physics you can think of a mass element <span class="math-container">$m$</span> at the surface of a star of mass <span class="math-container">$M$</span> and radius <span class="math-container">$R$</span>, rotating with angular velocity <span class="math-container">$\omega$</span>.</p>

<p>A condition for stability would be that the surface gravity is strong enough to provide the centripetal acceleration of the test mass.
<span class="math-container">$$ \frac{GMm}{R^2} &gt; m R \omega^2$$</span>
If this is not satisfied then the object might break up (it is more complicated than this because the object will not stay spherical and the radius at the equator will increase etc., but these are small numerical factors).</p>

<p>Thus
<span class="math-container">$$ \omega &lt; \left(\frac{GM}{R^3}\right)^{1/2}$$</span>
or in terms of rotation period <span class="math-container">$P = 2\pi/\omega$</span> and so
<span class="math-container">$$ P &gt; 2\pi \left(\frac{GM}{R^3}\right)^{-1/2},$$</span>
is the condition for stability.</p>

<p>For a typical <span class="math-container">$1.4M_{\odot}$</span> neutron star with radius 10 km, then <span class="math-container">$P&gt;0.46$</span> milliseconds. A proper General Relativistic calculation of this limit would give a similar result, but depends to some extent on the equation of state of the neutron star.</p>

<p>Happily, this is easily satisfied for all observed neutron stars - they can spin extremely fast because of their <em>enormous</em> surface gravities and all are well below the instability limit. I believe the fastest known rotating pulsar has a period of 1.4 milliseconds.</p>

<p>You also ask how pulsars can attain these speeds. There are two classes of explanation for the two classes of pulsars.</p>

<p>Most pulsars are thought (at least initially) to be the product of a core-collapse supernova. The core collapses from something a little smaller than the radius of the Earth, to about 10km radius in a fraction of a second. Conservation of angular momentum demands that the rotation rate increases as the inverse of the radius <em>squared</em>. i.e. The spin rate increases by factors of a million or so.</p>

<p>Pulsars spin down with age because they turn their rotational kinetic energy into magnetic dipole radiation. However, the fastest rotating pulsars - the "millisecond pulsars" are "reborn", by accreting material from a binary companion. The accreted material has angular momentum and the accretion of this angular momentum is able to spin the neutron star up to very high rates because it has a relatively (for a stellar-mass object) small moment of inertia.</p>

CC BY-SA 4.0
2020-06-04T23:04:30.843Z
Also, there's the magnetic field to consider. In some magnetars, the magnetic field is so intense that the neutron star is actually a prolate ellipsoid! See https://arxiv.org/abs/0712.2162
PM 2Ring
CC BY-SA 4.0
2015-04-04T12:36:21.370Z
2020-06-04T22:56:50.693Z
2020-06-04T22:56:50.693Z
ProfRob
<p>Some stars such as a neutron star can spin very fast around 600 times a second, wouldn't the star be ripped apart? Although the gravitational pull is very strong but gravity is the weakest of all known forces and any good theory why neutron stars spin that fast does it have anything to do with the magnetic field and the surrounding dust clouds?</p>

<neutron-star>
What happens when star spins too fast?
user6760
# What happens when star spins too fast?

(Asked by: username_0 on 2015-04-04)

Some stars such as a neutron star can spin very fast around 600 times a second, wouldn't the star be ripped apart? Although the gravitational pull is very strong but gravity is the weakest of all known forces and any good theory why neutron stars spin that fast does it have anything to do with the magnetic field and the surrounding dust clouds?

**Tags:** neutron-star


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## 2 answers

### Answer by username_3 on 2015-04-04. Score: 8 | Accepted answer

The structure (mass versus radius and density profile) *is* influenced by its rotation rate, but not by as much as you might think.

Even in Newtonian physics you can think of a mass element $m$ at the surface of a star of mass $M$ and radius $R$, rotating with angular velocity $\omega$.

A condition for stability would be that the surface gravity is strong enough to provide the centripetal acceleration of the test mass.
$$ \frac{GMm}{R^2} > m R \omega^2$$
If this is not satisfied then the object might break up (it is more complicated than this because the object will not stay spherical and the radius at the equator will increase etc., but these are small numerical factors).

Thus
$$ \omega < \left(\frac{GM}{R^3}\right)^{1/2}$$
or in terms of rotation period $P = 2\pi/\omega$ and so
$$ P > 2\pi \left(\frac{GM}{R^3}\right)^{-1/2},$$
is the condition for stability.

For a typical $1.4M\_{\odot}$ neutron star with radius 10 km, then $P>0.46$ milliseconds. A proper General Relativistic calculation of this limit would give a similar result, but depends to some extent on the equation of state of the neutron star.

Happily, this is easily satisfied for all observed neutron stars - they can spin extremely fast because of their *enormous* surface gravities and all are well below the instability limit. I believe the fastest known rotating pulsar has a period of 1.4 milliseconds.

You also ask how pulsars can attain these speeds. There are two classes of explanation for the two classes of pulsars.

Most pulsars are thought (at least initially) to be the product of a core-collapse supernova. The core collapses from something a little smaller than the radius of the Earth, to about 10km radius in a fraction of a second. Conservation of angular momentum demands that the rotation rate increases as the inverse of the radius *squared*. i.e. The spin rate increases by factors of a million or so.

Pulsars spin down with age because they turn their rotational kinetic energy into magnetic dipole radiation. However, the fastest rotating pulsars - the "millisecond pulsars" are "reborn", by accreting material from a binary companion. The accreted material has angular momentum and the accretion of this angular momentum is able to spin the neutron star up to very high rates because it has a relatively (for a stellar-mass object) small moment of inertia.

> **Comments:**
> **username_4**: Also, there's the magnetic field to consider. In some magnetars, the magnetic field is so intense that the neutron star is actually a prolate ellipsoid! See https://arxiv.org/abs/0712.2162  

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### Answer by username_1 on 2015-04-04. Score: 2

Related: [Is it possible to break apart a neutron star?](https://astronomy.stackexchange.com/questions/7778/is-it-possible-to-break-apart-a-neutron-star/7802#7802 "Is it possible to break apart a neutron star?")

The rotation must be powerful enough to overpower the force of gravity in order for anything to happen. I'm not sure about the 'ripping apart,' but if the rotation does overpower gravity surface material will be ejected from the body, a process known as **mass shedding**.

As neutron stars are extremely dense they can have great amounts of angular velocity, like the pulsar class neutron star mentioned in your question. If you can quickly increase the angular velocity of the body without also increasing the mass and density, perhaps you could rip it apart, but that would be very difficult to do to a neutron star.

> **Comments:**
> **username_0**: Is it like a ice skater doing a spin on the ice rink and when she tuck her arms inward she will go faster?  
> **username_1**: This is one of the ways they achieve such great rotational speeds.  Generally neutron stars come from stars with about 10x the mass of our Sun. The resulting neutron star likely has a radius of less than 15 kilometers. Not all the mass becomes a part of the neutron star though - plenty is ejected (though this is generally credited to the supernova, not the rotational velocity).  
> **username_1**: It would probably also be worth noting that as the strength of gravity increases as the distance from the center of gravity decreases, as the body becomes more dense the escape velocity at the surface will increase.  
> **username_2**: Not really related, but young stars spin faster.   This makes sense if you think about it.  The coalescing of a cloud of gas into a star would greatly increase angular velocity, as angular momentum is conserved.   Stars tend to lose rotational speed over time.   http://www.space.com/28255-star-spin-age-kepler-spacecraft.html  

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