27
$\begingroup$

The Sun moves around the Milky Way disk in the same direction as most of the other stars in our galaxy (prograde). But there are a number of older stars in the galactic halo that move in retrograde orbits, and some of them pass between the Sun and the center of the galaxy. What are the chances that the Sun will collide with another star, say in the next 5 billion years?

$\endgroup$
13
  • $\begingroup$ Fantastic question! (unfortunately I can't upvote) I wonder in particular if it's realistic enough stars collide with the Sun to reach the 8 solar masses required for a supernova, if I'm correct about this value. 5 billion years is an extremely long time, after all. $\endgroup$ – Plutos Loyer Jan 22 at 18:06
  • $\begingroup$ Do you mean a star entering the solar system or colliding and merging with the Sun? $\endgroup$ – fasterthanlight Jan 22 at 18:31
  • 4
    $\begingroup$ I seem to remember reading something about the possibilities of stars colliding being compared to two mosquitoes in the Grand Canyon colliding. $\endgroup$ – Arthur Kalliokoski Jan 22 at 18:42
  • 2
    $\begingroup$ @Plutos Loyer: It would seem much more likely that an actual collision would result in both stars losing considerable mass, as much material would be accelerated to more than escape velocity. And that doesn't even consider what blowing off the outer layers would do to internal nuclear reactions. I wonder if anyone's done a simulation of this? $\endgroup$ – jamesqf Jan 23 at 18:58
  • 1
    $\begingroup$ I remember an estimate of stars colliding when two milki-way size galaxies collide head-on. The estimate was something single-digit like 4 or 6. $\endgroup$ – fraxinus Jan 24 at 9:45
20
$\begingroup$

TL;DR: Virtually zero.

The distances between stars are HUGE and stars are tiny compared to the astronomical scales of distance between neighboring stars. The sun, is about 0.0000001 or one ten-millionth of a light year. The probability of a star (to be generous, say a $10 R_\odot$ star) colliding with the Sun is tiny.

Every star has a different velocity vector and different positions in space-time. The probability that their positions will intersect in a relatively short timescale is almost zero. Imagine throwing a tiny dart, and hitting a bullseye the size of an atom from several miles away. That still is larger than the probability of the Sun and some other star colliding in the near future (5 billion years is a tiny moment of time when it comes to events like these).

According to Wikipedia, it takes about 30 trillion years for a star to undergo a close encounter with another star. Even if the encounter is less than 1 AU, the stars would probably just pass by and not do anything special. For a complete collision and merge, you'd need the distances to be much, much smaller, about $0.1 R_\odot$.

Years from now Event
3×1013 (30 trillion) Estimated time for stars (including the Sun) to undergo a close encounter with another star in local stellar neighborhoods. Whenever two stars (or stellar remnants) pass close to each other, their planets' orbits can be disrupted, potentially ejecting them from the system entirely. On average, the closer a planet's orbit to its parent star the longer it takes to be ejected in this manner, because it is gravitationally more tightly bound to the star.

Source: https://en.wikipedia.org/wiki/Timeline_of_the_far_future

Also, according to Wikipedia:

While stellar collisions may occur very frequently in certain parts of the galaxy, the likelihood of a collision involving the Sun is very small. A probability calculation predicts the rate of stellar collisions involving the Sun is 1 in $10^{28}$ years.

Source: https://en.wikipedia.org/wiki/Stellar_collision#Stellar_collisions_and_the_Solar_System

Back to crude math, assuming an uncertainty of 2 AU, the probability given these parameters is equal to:

$$\dfrac{5 \cdot 10^9 \text{ years}}{3 \cdot 10^{13} \text{ years} \cdot 10^{24} \text{ stars in the universe}} \cdot \dfrac{0.1 R_\odot}{450 R_\odot} = \dfrac{1}{27,000,000,000,000,000,000,000,000,000,000} = \dfrac{1}{2.7 \cdot 10^{31}}$$

Warning: This is only a very crude estimate, don't take this as a "real, super-mathy" answer. But I still hope this helps.

Edit: That equation above is still probably too big. Asked a question about this: If two stars collide, what is the probability that they merge to form a single star?

$\endgroup$
10
  • $\begingroup$ The estimate cited in your answer is for "another star in local stellar neighborhoods." Are you sure this estimate would also apply for retrograde orbiting stars in the Galactic Halo? $\endgroup$ – Connor Garcia Jan 22 at 19:47
  • 1
    $\begingroup$ @nick012000 Each star's gravity will alter the course of the other star slightly as it swings by. $\endgroup$ – HiddenWindshield Jan 23 at 17:17
  • 9
    $\begingroup$ While having a "close" encounter with another star will likely not much affect the Sun, it will certainly disturb the orbits of the planets. $\endgroup$ – Paŭlo Ebermann Jan 23 at 17:34
  • 2
    $\begingroup$ @PaŭloEbermann Yup, I was thinking that having a second sun pass within 1 AU of us would be pretty spectacular $\endgroup$ – JollyJoker Jan 23 at 18:50
  • 2
    $\begingroup$ @nick012000 conservation of energy says no. Gravity will alter the trajectories but it doesn't "suck things in". If the stars weren't gravitationally bound before the encounter, they won't be afterwards, unless they come close enough to literally smack into each other. $\endgroup$ – hobbs Jan 23 at 19:05
9
$\begingroup$

As fasterthanlight says, the probability of our Sun colliding with another star in the galaxy is virtually nil. In fact, the probability of any star in the galaxy colliding with another (unrelated) star is very small.

Stars can and do collide, but they're stars that are already gravitationally bound to each other in binary or multiple star systems. And they generally don't collide until they're quite old. A star can pull material from a companion star, and in some cases that upsets the gravitational balance sufficiently that the stars merge, which can be rather explosive. Another option is that the stars' orbits decay via gravitational wave radiation. Even for neutron stars in tight orbits, which radiate gravitational energy at a relatively fast pace, this process takes a long time.

There's a lot of space between the stars in the galaxy. And even when whole galaxies collide, star collisions have a very low probability. Wikipedia says that when the Milky Way and Andromeda merge "the stars involved are sufficiently far apart that it is improbable that any of them will individually collide".


The following may give you an idea of how low the star density of our galaxy is. According to Wikipedia, the mass of the Milky Way is somewhere between 0.8 and 1.5 trillion solar masses. That includes dark matter, which is a significant majority of the total mass, as well as the mass of loose interstellar gas and dust, so the actual mass of stars is quite a bit less. But let's go with the $1.5×10^{12} M_\odot$ figure, just to be generous. ;)

That article also says the diameter of the Milky Way's stellar disk is around 185 ± 15 thousand lightyears, depending how you define the outer edge. So let's use a radius of 85,000 lightyears. Also, the mean thickness of the galactic disk is around 1,000 lightyears.

Stars are made of hot plasma, so they're fairly tenuous in their outer regions. But they get quite dense in their cores, where most of the nuclear reactions happen. The mean density of the Sun is 1.41 g/cm³ (1,410 kg/m³), i.e., 1.41 times the standard density of water, but the density in its core is around 150 g/cm³, many times greater than the densest metal in terrestrial conditions.

Let's imagine that we could somehow homogenise and compress the Milky Way into a uniform disc, with its current radius, and with a density of 1g/cm³. So we're squishing that 1000 lightyear thickness of stars and everything else, down into a thin disk of radius 85,000 lightyears, with the same density as water.

Take a guess at how thick that disc would be.

1.469 millimetres.

You can check my calculation by plugging this formula into Google's search bar, which invokes the Google Calculator: (1.5e12 solar masses) / ((1 g/cm^3) * pi * (85000 lightyears)^2)

In contrast, if we squash the Sun down to a 1 g/cm³ disc of radius 30.1 au, the orbital radius of Neptune, we get a thickness of

31.2 metres

(1 solar mass) / ((1 g/cm^3) * pi * (30.1 au)^2)

$\endgroup$
6
  • 1
    $\begingroup$ I am looking for something more specific than "very small." Do you agree with the other answer of 1 in 2.7 million? $\endgroup$ – Connor Garcia Jan 23 at 0:41
  • $\begingroup$ @ConnorGarcia I think it's actually an overestimate, based on the probabilities associated with the Milky Way / Andromeda merger. None of the trillion stars in the 2 colliding galaxies are likely to collide. But the odds for crossing trajectories are presumably greater than what they are in a stable galaxy. Years ago, when I first read about the galaxy merger, the article mentioned that collisions were very unlikely, but if they do happen, it could only involve a couple of stars, less than a handful. But we have better stats on star numbers, distributions, and the galaxy mass profiles now. $\endgroup$ – PM 2Ring Jan 23 at 1:03
  • 1
    $\begingroup$ @ConnorGarcia - How about 0 in 5B? Keith's answer $\endgroup$ – Mazura Jan 23 at 4:27
  • $\begingroup$ @Mazura I don't know what "0 in 5B" is supposed to mean, but 1 collision per 10²⁸ years sounds reasonable. So if the Sun ever does collide with another star, it's likely to be a white (or black) dwarf by then. On that timescale, the conditions will have changed significantly: as I said here the entire Local Group will merge into a single giant elliptical galaxy in 450 billion years or so. So probability calculations on that timescale should be taken with a grain of salt. $\endgroup$ – PM 2Ring Jan 23 at 14:07
  • 1
    $\begingroup$ @ConnorGarcia It's not easy to estimate the probability of events that have never been observed, like unrelated star collisions. In several answers & comments on a recent Physics SE question, there are some great points made regarding the difficulties of doing valid probability calculations for events with very low probability. The basic idea is that with tiny probabilities, the calculated value may be smaller than the probability that your model is wrong, or at least the probability gets swamped by the errors in the model. $\endgroup$ – PM 2Ring Jan 23 at 14:23
4
$\begingroup$

Imagine that that there are 2 pigeons on LSD flying around the world. What are the chances that they would collide?

Colliding stars are less likely because if the stars were as big as pigeons, their average distance would be 200,000 kilometers.

Stars travel at 500.000 mph on average, so if they were birds, the birds would be flying at 0.000005 mph.

Stars are 10^11 times bigger than pigeons.

Here is an article about close encounters of our galaxy with other stars, i.e. Scholz'Star which passed about 50,000 AU from the sun, through the Oort cloud. https://www.discovermagazine.com/the-sciences/wandering-stars-pass-near-our-solar-system-surprisingly-often

That's like a pigeon passing 500 kilometers from another pigeon, being called a "close call".

The closest fly-by currently predicted is Gliese 710 will pass 5 times closer than Scholz star in 1.2 million years at 0.17 light years distance. It's currently 63 ly away.

I just checked this again, Wolfram sais that if the milky way was shrunk so it fits in the earth's orbit around the sun, there would be 100,000,000,000 birds(stars) flying at 1mm per minute, and 200,000 km in between the birds on average.

$\endgroup$
2
  • $\begingroup$ On the other hand, if there were a hundred billion drugged up pigeons flying around, what are the chances of some of them crashing? The chances of any two individual stars colliding is low, but the combined chances of any stars in the same galaxy colliding with us are... well, still very low, but we have to run all the numbers to be sure of that. $\endgroup$ – user3153372 Jan 24 at 10:31
  • 2
    $\begingroup$ @user3153372, Yes indeed. I checked other information... It said: if astronomers were fleas riding on the pigeons, one astronomer would say to the other: You know what? 70,000 years ago, another pigeon came as close as 500 kilometers from us. That was a close call! he was called Sholz's pigeon, he would have changed the night sky! wow! It kindof means that we can star-hop every 50,000 years without having to travel light years away, i.e. in 50,000 years, we can perhaps fly to another star system. en.wikipedia.org/wiki/Scholz%27s_Star#Solar_System_flyby $\endgroup$ – aliential Jan 24 at 17:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.