# Rogue neutron star

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1. Apr 28, 2015

### Stephanus

Dear PF forum.
Folks, what I have here may be an unimportant question. Still I'm curious to know the answer.
When iron fusion happens in a massive star, it will undergo supernova.
What I want to know is...
1. There are some binary (or perhaps ternary) neutron star systems.
Neturon star is the product of a supernova (the other product may be a black hole). If the star explodes, will it scatter its solar system (or simply "its system," "SOL" is the name of our sun) away?
Why does it still have companion?
2. During a supernova explosion, can the star be ejected through its trajectory orbit with respect to the center of the galaxy?

I'm not trying to throw a Fermi Problem here, but...
If there are approximately 100 billions of stars in a galaxy, take Milky Way for an example.
And the rate of star born is -- 13.5 billions years (age of the universe) / 100 billions (stars in Milky Way) ≈ 1 star per month is born.
And 10% of those stars can undergo supernova according to its mass (our star can't and its just an average star).
And the lifespans of a massive star is perhaps 30 millions years. But 30 millions years should'nt be in the equation, right?.
Thus, we have only 1 month / 10% ≈ 1 year. So, every year a supernova explodes in our Milky Way. Up until now, there should be 13.5 billion supernovas in our galaxy alone. Let's say my approximation is incorrect. Let's say it's only 1 billion supernovas. And.. about 1% of them if question number 2 is correct is a rogue neutron star. So that would be... 10 millions rogue neutron stars or black holes in our galaxy alone!
Is that correct?
Other calculation
Our Milky way is 80 000 ly by 1000 ly thick. So... In a cylinder of 40000 x 40000 x 1000 x π = 5 trillion ly cube. (calculator here...)
So in a cyllinder of 5 trillion ly cube there are 10 million of these rogue stars. Or 1 star in every 500 000 ly cube.
It's "only" a cube of 80 ly length. And our nearest star is 4.3 ly away. So our Milky Way are actualy roamed by these rogue stars.
Again, if this is correct, and it's just an idea.

But, what I really want to know is, how is the chance of a supernova causing the star to deviate from its trajectory orbit to the center of the galaxy and if somehow our star (which can't) undergoes a supernova, will it scatter Jupiter or Neptune away?.

Steven

2. Apr 28, 2015

### D H

Staff Emeritus
First off, your numbers are off. The current star formation rate in the Milky Way is on the order of one star per year, not one per month. Estimates vary widely, but not that wide. You made the mistake of assuming star formation rate is constant. It isn't. Another mistake is your 10% figure for the fraction stars that go supernova. It's less than one percent. The end result: The supernova rate in the Milky Way is on the order of one per century, not one per year.

However (and this is a big however), pulsars tend to have a high velocity relative to the interstellar medium. Some are moving faster than 1000 km/s. (Compare with the ~500 km/s escape velocity at the Sun's distance from the galactic center.) Pulsars are rotating neutron stars. The pulsar velocity profile should be indicative of neutron stars in general. Since most stars are moving more or less with the interstellar medium, something must have given those neutron stars a big kick at some point. The dominant hypothesis (but it is just a hypothesis) is that the supernovae that created those fast-moving neutron stars was a bit asymmetric. It doesn't take much of an asymmetry to give a big kick.

There is a conundrum here, but it's not the question you asked. The conundrum is explaining how astronomers see neutron stars in clusters, where the escape velocity is on the order of 50 km/s, and how astronomers see neutron stars paired with other stars. This is the neutron star retention problem.

3. Apr 29, 2015

### |Glitch|

The star formation rate for the present day Milky Way Galaxy is between 0.68 M and 1.45 M per year.
A spectral type G main sequence star, like our sun, is not "average" but rather only accounts for 7.5% of all the stars in the Milky Way galaxy. Spectral types O, B, A, and F main sequence stars are even rarer than spectral type G main sequence stars. Spectral type K main sequence stars comprise 12% of the stars in the Milky Way galaxy. The most common star, by far, are the spectral type M main sequence stars, totaling as much as 76% of all the stars in the Milky Way galaxy.
It is estimated that a core collapse supernova (SN types Ib/c and II) will occur in the Milky Way galaxy 1.9 ± 1.1 times every century, or 1 every ~50 years.
Furthermore, when the universe was very young, there would have been a preponderance of massive Population III stars, which have very short lives. As the universe aged and the less massive Population II stars evolved, the rate at which SNe would have occurred would also slow down accordingly. Finally, we are left with primarily Population I stars (like our sun) today, which are even less massive than the Population II stars and therefore have longer lives. In other words, the rate of SNe in the Milky Way galaxy would have been significantly higher ~13.2 billion years ago, and that rate will steadily decline as the Milky Way ages.

Lastly, your premise that a binary companion of the SN progenitor will "be ejected through its trajectory orbit with respect to the center of the galaxy" is incorrect. With the exception of Type Iax SNe, the ejecta of all SNe is |v| > ~10,000 km/s. Depending on the mass of the progenitor, the ejecta would obliterate planets and any companion star for dozens of AU. Main sequence stars, when compared to rocky-type planets like Earth, have a relatively low density. Earth's density is 5.51 g/cm3, and the sun's density is 1.4 g/cm3. A SN would not eject a companion star, it would destroy the companion star if it was close enough.

Last edited by a moderator: May 7, 2017
4. Apr 29, 2015

### D H

Staff Emeritus
I can cite other, more recent articles that say something else. One article is rarely definitive, and that's the case here. Astronomers still agree to disagree on the exact value of the Milky Way star formation rate. Saying that it's on the order of one solar mass per year -- that's something most can agree with. There's a lot of slop with "on the order of". The same goes for your supernova rate article. There's a lot of slop in that 1.9± 1.1 per century, and this is consistent with a sloppy on the order of one per century.

That's the one part of the OP's premise that had some truth to it. Astronomers have discovered a number of hypervelocity stars. What causes that? Possibly a close encounter with an intermediate mass black hole? That explanation runs into a number of problems. The most widely accepted hypothesis is that these hypervelocity stars were ejected when their binary pair went supernova. I'll try to add some references later, but for now I'm off to work.

5. Apr 29, 2015

### Stephanus

Thanks a lot Glitch for the answer.
But why there are binary neutron stars?
Or... are binary neutron stars "rare"? I don't know! What the word "rare" means If it's about astronomy!

6. Apr 29, 2015

### |Glitch|

Stars with a mass of between 5 M and 10 M are uncommon in the first place. Such stars will lose ~70% of their mass during the SN process so they end up with a mass greater than 1.44 M and less than ~3 M in order to form a neutron star. There have been ~2,000 neutron stars discovered thus far, but NASA estimates that there are approximately one billion neutron stars in the Milky Way, or ~0.25% of all the stars in the Milky Way. Out of all the neutron stars discovered, only 40 binary neutron pairs have been discovered. It is estimated that only ~5% of all neutron stars are in binary pairs. Which would make neutron binary pairs only ~0.0125% of all the stars, or one neutron binary pair for every ~8,000 stars in the Milky Way. I think the adjective "rare" would definitely apply in this case.
Supernova ejecta (except for Type Iax SNe) will be traveling between 10,000 km/s and 30,000 km/s and carrying with it 1044 Joules [J/(s·m²)] of energy. If the amount of energy from the SN ejecta is able to overcome the gravitational binding energy of a planet or another companion star, then it will destroy the hydrostatic equilibrium of that planet or star, and blow it apart. For a spherical mass of uniform density (which planets are not, but it is close enough for an order-of-magnitude approximation) the gravitational binding energy can be calculated as follows:

U = 3GM2 / 5R​

Where:
G = Gravitational Constant = 6.6742 × 10−11 m3kg-1s-2;
M = Mass of the planet, in kilograms; and
R = Radius of the planet, in meters.​

Using Earth as an example, the gravitational binding energy (U) would be 2.25 × 1032 Joules [J/(s·m²)] of energy. Using Jupiter as an example, the amount of gravitational binding energy increases to 2 × 1036 Joules [J/(s·m²)] of energy. Stars, on the other hand, have a gravitational binding energy equivalent of about twice their internal thermal (kinetic) energy. For example, the sun would have a gravitational binding energy of 6.9 × 1041 Joules [J/(s·m²)] of energy.
• https://ia700806.us.archive.org/12/items/AnIntroductionToTheStudyOfStellarStructure/Chandrasekhar-AnIntroductionToTheStudyOfStellarStructure.pdf [Broken] - Chandrasekhar, S. (1939), Section 9, Equations 90-92, p. 51 (Dover edition) [PDF]
Therefore, if the planet or a companion star were close enough to the SN progenitor, there would be nothing left of either. However, the further away the planet or companion star is from the SN progenitor, the less energy will impact them, using the inverse-square law. If the sun were to supernova and produce 1044 Joules [J/(s·m²)] of energy, it would obliterate the inner planets, and strip off the atmospheres of Saturn and Jupiter leaving only their cores behind. A companion star at a distance of 5 AU from the SN progenitor will certainly lose some mass, but otherwise remain intact. But that is only the beginning of the problem. Since the progenitor has lost ~70% of its mass, all the objects orbiting the progenitor, including any stellar companions, will change their orbits.

The shifting orbits could easily explain how companion stars could be ejected from their binary system, or how a companion star might migrate closer to the neutron star. It is not the SN ejecta "kicking" a companion star out of the system, but rather the loss of ~70% of the progenitor's mass and the resulting changes in the orbit of the companion star that causes the star to be ejected from the binary system. If a companion star has not been ejected from the binary system, and still has between 5 M and ~10 M remaining, then it will also become a neutron star. The kinetic energy of a SN is not strong enough to overcome the gravitational binding energy of a neutron star, thus resulting in a neutron binary system.

It should also be noted that we have discovered exoplanets orbiting neutron stars. It has been suggested that these exoplanets formed after the SN event, during a second round of planetary formation.
• http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?db_key=AST&bibcode=1993ASPC...36..149P&letter=.&classic=YES&defaultprint=YES&whole_paper=YES&page=149&epage=149&send=Send+PDF&filetype=.pdf [Broken] - Proceedings of the Conference, California Institute of Technology, Pasadena, Apr. 30-May 1, 1992 (A93-36426 14-90), p. 149-165 [PDF]

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