<p>Let's assume that what is falling onto the neutron star is "normal" material - i.e. a planet, an asteroid or something like that. As the material heads towards the neutron star it gains an enormous amount of kinetic energy. If we assume it starts from infinity, then the energy gained (and turned into kinetic energy) is approximately (ignoring GR)
$$ \frac{1}{2}m v^2 = \frac{GMm}{R}, $$
where $m$ is the mass of the object (which cancels) and $M$ and $R$ are the mass and radius of the neutron star (let's assume typical values of $1.4 M_{\odot}$ and 10 km respectively).</p>

<p>This results in a velocity as it approaches the neutron star surface of $1.9 \times 10^{8}$ m/s - i.e. big enough that you would have to do the calculation using relativistic mechanics actually.</p>

<p>However, I doubt that the object would get to the surface intact, due to tidal forces. The <a href="http://en.wikipedia.org/wiki/Roche_limit" rel="noreferrer">Roche limit</a> for the breakup of a rigid object occurs when the 
object is a distance 
$$d = 1.26 R \left(\frac{\rho_{NS}}{\rho_O}\right)^{1/3},$$
where $\rho_{NS}$ and $\rho_O$ are the average densities of our neutron star and object respectively. For rocky material, $\rho_O \simeq 5000$ kg/m$^{3}$. For our fiducial neutron star $\rho_{NS} \simeq 7\times10^{17}$ kg/m$^{3}$. Thus when the object gets closer than $d= 500,000$ km it will disintegrate into its constituent atoms.</p>

<p>It will thus arrive in the vicinity of the neutron star as an extremely hot, ionised gas. But if the material has even the slightest angular momentum it could not fall directly onto the neutron star surface without first shedding that angular momentum. It will therefore form (or join) an accretion disk. As angular momentum is transported outwards, material can move inwards until it is hooked onto the neutron star magnetic field and makes its final journey onto the neutron surface, probably passing through an accretion shock as it gets close to the magnetic pole, if the object is already accreting strongly. Roughly a few percent of the rest mass energy is converted into kinetic energy and then heat which is partly deposited in the neutron star crust along with matter (nuclei and electrons) and partly radiated away.</p>

<p>At the high densities in the outer crust the raw material (certainly if it contains many protons) will be burned in rapid nuclear reactions. If enough material is accreted in a short time this can lead to a runaway thermonuclear burst until all the light elements have been consumed. Subsequent electron captures make the material more and more neutron rich until it settles down to the equilibrium composition of the crust, which consists of neutron-rich nuclei and ultra-relativistically degenerate electrons (no free neutrons).</p>

CC BY-SA 3.0
2015-04-06T15:23:16.917Z
2015-04-06T15:23:16.917Z
ProfRob
<p>No detailed calculations, but a qualitative answer: Depending on the trajectory of the impactor, the results will vary a bit, but it's clear, that the potential energy of the impactor will be transformed into a high amount of kinetic energy before the impact happens. The kinetic energy will then be transformed mainly into heat during the impact, transforming a substantial part of the mass of the impactor into x-rays and gamma rays.</p>

<p>The remnants of the impactor will be transformed into a plasma, with most of the electrons moving independently of their former nuclei, and dispersed mainly into the atmosphere (a thin layer of a few millimeters) of the neutron star. The energies will be high enough to trigger nuclear fusion as well as fission, together with other high-energy particle reactions. Part of the energy will be transformed into magnetic fields, which can also be very strong on neutron stars.</p>

<p>No much intermixing with the interior of the neutron star is to be expected in the first instant for small impactors due to the high inertia and density of the inner parts of the neutron star.</p>

<p>In some cases the impact could trigger the collapse of the neutron star into a black hole, depending of the mass of the neutron star, and the mass of the impactor.</p>

<p><a href="http://en.wikipedia.org/wiki/Neutron_star" rel="noreferrer">More on the inner structure of neutron stars on Wikipedia</a>.
("Matter falling onto the surface of a neutron star would be accelerated to tremendous speed by the star's gravity. The force of impact would likely destroy the object's component atoms, rendering all its matter identical, in most respects, to the rest of the star.")</p>

<p><a href="http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/010607a.html" rel="noreferrer">More about the Chandrasekhar limit of neutron stars</a>.</p>

CC BY-SA 3.0
2014-02-08T14:51:03.933Z
2014-02-08T15:08:18.580Z
2014-02-08T15:08:18.580Z
Gerald
<p>We know neutron stars as a very massive object with extremely strong gravitational forces that composes mostly of neutrons.</p>

<p>I couldn't help but wonder, what would happen if an object fell into a neutron star, what would become of it? Will it turn it into neutrons as well? And will be any sort of accompanying radiation emission?</p>

CC BY-SA 3.0
2014-02-08T11:02:44.290Z
Neutron stars are nowhere near 'composed entirely of neutrons'. There's plenty of electrons in the crust, and most likely an outer envelope of fully ionized iron. So anything hitting that envelope will probably be fully ionized as well, but what fraction of it (if any) is likely to undergo fusion into heavier elements I'm not sure about, especially since it will be ripped apart by tidal forces first.
Stan Liou
CC BY-SA 3.0
2014-02-09T14:18:54.223Z
@StanLiou Thanks for pointing that out, I will edit the question accordingly.
Yoda
CC BY-SA 3.0
2014-02-09T15:38:44.810Z
@StanLiou: Given that the energy release per unit mass at the moment of impact is much larger than nuclear binding energy per unit mass of the impactor, ionisation and nuclear bonds are irrelevant. The outcome would be the same as if the object would be just a set of protons and neutrons.
Alexey Bobrick
<star><gravity><neutron-star>
What would happen if a body were to fall into a neutron star?
Yoda
# What would happen if a body were to fall into a neutron star?

(Asked by: username_0 on 2014-02-08)

We know neutron stars as a very massive object with extremely strong gravitational forces that composes mostly of neutrons.

I couldn't help but wonder, what would happen if an object fell into a neutron star, what would become of it? Will it turn it into neutrons as well? And will be any sort of accompanying radiation emission?

**Tags:** star, gravity, neutron-star

> **Comments:**
> **username_1**: Neutron stars are nowhere near 'composed entirely of neutrons'. There's plenty of electrons in the crust, and most likely an outer envelope of fully ionized iron. So anything hitting that envelope will probably be fully ionized as well, but what fraction of it (if any) is likely to undergo fusion into heavier elements I'm not sure about, especially since it will be ripped apart by tidal forces first.  
> **username_0**: @username_1 Thanks for pointing that out, I will edit the question accordingly.  
> **username_2**: @username_1: Given that the energy release per unit mass at the moment of impact is much larger than nuclear binding energy per unit mass of the impactor, ionisation and nuclear bonds are irrelevant. The outcome would be the same as if the object would be just a set of protons and neutrons.  

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## 2 answers

### Answer by username_4 on 2014-02-08. Score: 7 | Accepted answer

No detailed calculations, but a qualitative answer: Depending on the trajectory of the impactor, the results will vary a bit, but it's clear, that the potential energy of the impactor will be transformed into a high amount of kinetic energy before the impact happens. The kinetic energy will then be transformed mainly into heat during the impact, transforming a substantial part of the mass of the impactor into x-rays and gamma rays.

The remnants of the impactor will be transformed into a plasma, with most of the electrons moving independently of their former nuclei, and dispersed mainly into the atmosphere (a thin layer of a few millimeters) of the neutron star. The energies will be high enough to trigger nuclear fusion as well as fission, together with other high-energy particle reactions. Part of the energy will be transformed into magnetic fields, which can also be very strong on neutron stars.

No much intermixing with the interior of the neutron star is to be expected in the first instant for small impactors due to the high inertia and density of the inner parts of the neutron star.

In some cases the impact could trigger the collapse of the neutron star into a black hole, depending of the mass of the neutron star, and the mass of the impactor.

[More on the inner structure of neutron stars on Wikipedia](http://en.wikipedia.org/wiki/Neutron_star).
("Matter falling onto the surface of a neutron star would be accelerated to tremendous speed by the star's gravity. The force of impact would likely destroy the object's component atoms, rendering all its matter identical, in most respects, to the rest of the star.")

[More about the Chandrasekhar limit of neutron stars](http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/010607a.html).


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### Answer by username_3 on 2015-04-06. Score: 6

Let's assume that what is falling onto the neutron star is "normal" material - i.e. a planet, an asteroid or something like that. As the material heads towards the neutron star it gains an enormous amount of kinetic energy. If we assume it starts from infinity, then the energy gained (and turned into kinetic energy) is approximately (ignoring GR)
$$ \frac{1}{2}m v^2 = \frac{GMm}{R}, $$
where $m$ is the mass of the object (which cancels) and $M$ and $R$ are the mass and radius of the neutron star (let's assume typical values of $1.4 M\_{\odot}$ and 10 km respectively).

This results in a velocity as it approaches the neutron star surface of $1.9 \times 10^{8}$ m/s - i.e. big enough that you would have to do the calculation using relativistic mechanics actually.

However, I doubt that the object would get to the surface intact, due to tidal forces. The [Roche limit](http://en.wikipedia.org/wiki/Roche_limit) for the breakup of a rigid object occurs when the
object is a distance
$$d = 1.26 R \left(\frac{\rho\_{NS}}{\rho\_O}\right)^{1/3},$$
where $\rho\_{NS}$ and $\rho\_O$ are the average densities of our neutron star and object respectively. For rocky material, $\rho\_O \simeq 5000$ kg/m$^{3}$. For our fiducial neutron star $\rho\_{NS} \simeq 7\times10^{17}$ kg/m$^{3}$. Thus when the object gets closer than $d= 500,000$ km it will disintegrate into its constituent atoms.

It will thus arrive in the vicinity of the neutron star as an extremely hot, ionised gas. But if the material has even the slightest angular momentum it could not fall directly onto the neutron star surface without first shedding that angular momentum. It will therefore form (or join) an accretion disk. As angular momentum is transported outwards, material can move inwards until it is hooked onto the neutron star magnetic field and makes its final journey onto the neutron surface, probably passing through an accretion shock as it gets close to the magnetic pole, if the object is already accreting strongly. Roughly a few percent of the rest mass energy is converted into kinetic energy and then heat which is partly deposited in the neutron star crust along with matter (nuclei and electrons) and partly radiated away.

At the high densities in the outer crust the raw material (certainly if it contains many protons) will be burned in rapid nuclear reactions. If enough material is accreted in a short time this can lead to a runaway thermonuclear burst until all the light elements have been consumed. Subsequent electron captures make the material more and more neutron rich until it settles down to the equilibrium composition of the crust, which consists of neutron-rich nuclei and ultra-relativistically degenerate electrons (no free neutrons).


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