A simulation carried out using supercomputers led scientists to expect that stellar black holes found in globular clusters may merge, fly out of the cluster and float in the galaxy * There are about 200 globular clusters in the Milky Way, and it is possible that hundreds of stealthy black holes fly in the galaxy and swallow dust and stars
"It's very difficult to detect exclusionary black holes," says astronomer Kelly Holly-Bockelman of Vanderbilt University, who presented the results of a supercomputer simulation at the American Astronomical Union's winter meeting on January 9 in Austin, Texas. Most of the research was done at Pennsylvania State University in collaboration with two researchers from there and a researcher from the University of Michigan before Holly-Beckelman moved to Vanderbilt.
"Until it swallows a lot of gas, the only way we can detect such a black hole is to watch the way its strong gravitational field causes gravitational lensing and distorts the light from stars behind it." saying.
The research focused on building models for "medium black holes", whose very existence is controversial. Astronomers have managed to locate black holes less than 100 solar masses in size, predicting that they will be the remains of a large star that collapsed. Black holes with millions of solar masses to billions of solar masses are found in the centers of most galaxies, including the Milky Way. In addition, the theorists predict that globular clusters, which are groups of one hundred thousand to one million stars bound together by gravity within a galaxy, may contain at their center a third type of black holes - the so-called medium black holes weighing several thousand solar masses. However, so far the astronomers have managed to discover two such and that too only according to the Hicks.
In the last two years, scientists have been able to show through simulation that merger processes between black holes can take place, without violating Einstein's theory of relativity. One of the big surprises to come from the prediction efforts is that two black holes spinning at different speeds or sizes will give the fused black hole a big kick to conserve momentum, and the black hole is thrown in a random direction at over 4,000 kilometers per second. Therefore, this means that any black hole created in such a merger will be thrown out of the globular cluster, because the escape velocity is about 100 kilometers per second.
The group of researchers led by Bockelman performed a number of simulations predicting the growth process of medium-sized black holes when they merge from several stellar black holes, of which there are many in globular clusters, and they placed particular emphasis on the momentum they receive after each merger.
"We used different estimates for the sizes of the black holes, starting with stellar black holes within the globular clusters and also assumed a range of random rotation speeds and directions. We found that even if all globular clusters start with a medium-sized black hole, only about 30% of them remain in the cluster after the merger. According to conservative estimates, less than two percent of globular clusters should contain medium-sized black holes today. If all 200 globular clusters in the Milky Way have ejected medium-sized black holes into space, this means that hundreds of such black holes are wandering around the Milky Way, seeking to bite into the unlucky nebulae, stars, and planets that get in their way.
Fortunately, even if there are a few black holes roaming around in our area, they shouldn't cause us any harm. "These black holes will not harm us during the entire lifetime of the universe," Beckelman said. "Their risk zone - Schwarzschild radius, extremely tiny, only a few hundred kilometers. There are much more dangerous things near the solar system.
For information on the Vanderbilt University website
More news on the same topic:
48 תגובות
Addendum to Judah:
There is especially no point in you demanding explanations from me for false claims you make.
Yehuda:
You got off the ants because you don't know how to prove.
No matter how many explanations you require of me for anything else.
To Michael
So if you think I was so wrong then how do you explain the fact that I always arrive at correct answers?, I'm very lucky to have?, a positive Hugin cosmological effect?
You know what?, so be it!, I'm getting off the ants.
good afternoon
Sabdarmish Yehuda
Yehuda:
When it comes to feeding me garbage - not even a spoon will help you!
You write: "After one second, everyone who is in the first cm, goes out of the first cm. Either he goes out to the second cm or he goes out of the rod at all." And you don't understand that this is simply exactly the original problem only with a length of a centimeter instead meter and it requires proof just as easily.
The truth is that you are not accurate and your original wording hid the fact that there is another complication that you ignored that arises from collisions between ants in the first centimeter and those in the second but this, as I said in the previous response - only makes the claim that you are relying on without proving it more complicated than the original problem and not the same as the original problem .
I see Michael that you really need to be spoon-fed:-
After one second, everyone who is in the first cm, goes out of the first cm. Either they go out to the second cm or they go out of the pole at all.
After a second, several years, everyone who is in cm number two leaves it, either towards the first cm and from there out or towards the third cm.
And so on and so forth.
After fifty seconds everyone who is at the fiftieth cm goes out to our right either towards the 49th cm and out from there, or towards the fifty-first cm and from there out the other side.
The same on the other side from the 99th cm and then 98, 97 etc.
That is, after a maximum of fifty-one seconds we will have an ant that has at most another 49 seconds to go out, that is, a maximum of one hundred seconds and the rod will be empty!!!.
And look, Mr. Michael, I'll explain to you again, without claiming that you're lying, simply with the understanding that sometimes there are people who...
Not important
Good night Michael
Sabdarmish Yehuda
Hugin:
I don't know if you really follow the discussion.
It's hard for me to believe that you're really following, but since you didn't say anything about it, I'll give you the benefit of the doubt.
What is clear to me is that Yehuda does not follow.
Yehuda:
You simply do not understand that your claim "after one second everyone who is in the first cm is on their way out" is a more difficult claim to prove than the original riddle.
In fact, the only way I see to prove this claim is based on the proof of the original riddle, so maybe you happen upon some kind of "spoon feeding" proof, but then you would have to skip the entire proof and say that the riddle itself (which, as mentioned - is simpler to prove than your above claim) does not require proof at all .
In short - as usual - you are lying and what you call spoon feeding is simply a cover for the fact that you do not know how to prove it. point.
And another word of Yehuda, in order to add specifically to your words/and your father Shmuel..
Everything is built on simple principles! When a child/person understands the fundamental principle of the person. All the solutions turn around and are easily solved this way.
So in mathematics and so in everything. This is how pure intelligence works.
And all the rest is sometimes a needless refusal to twists and turns of brain drains and a refusal of the main point and the principle.
Hugin
Yehuda and Michael.
Since our Magnibush hates grandmother's stories.. (and I haven't had a day for that yet) he is challenging me to get on Yehuda's associative wave and tell about the calculus teacher from the XNUMXth grade, who was really and truly an honest Eisen.
In one of the classes he brought up a problem and its solution, and I jumped up and said... Yossi (that's his name), but there is a mistake here... and the answer is this... The whole class was outraged, especially one, his name is Morris, who defiantly closed his notebook and shouted, "What is she thinking to herself?" This "spoiled" me .. the teacher says he knows!! And I am the nice one.. still stubborn.. done
The next day, the teacher came and confessed, I checked the problem at home... and indeed she (my name) was right!! and he
Put it on the board, all over again. That's it.
A small example of the tenacity of a student, and the pure honesty of a good teacher-educator... and of the crowd that to this day
He will be jealous and jealous of geniuses and geniuses for alternating shifts at times.
Bye Yehuda, refer to Hugin sometimes...
To Michael
The well-remembered teacher Shmuel from the XNUMXth grade told me to limit my response, it is true that there are those who need to be fed everything, but since the result showed that I had reached a solution, which shows an above average healthy intuition, I will be content with that and regarding your words I will not comment because that is who you are and there is nothing to do.
But maybe I didn't understand you well because according to Hugin, you will be different for a long life, you are currently in the section of a joke story. If that's true then hahahahaha I was laughing so you wouldn't say I don't understand jokes.
So bye, dear Michael
Make a puzzle for the weekend
I'll solve it, you say I didn't solve it so we have something to deal with in at least a week.
Good Day
Sabdarmish Yehuda
Yehuda honey
For several days now I have been following the riddle of the ants that you and Michael are dealing with. But at this stage my understanding, which is limited due to other priorities at this time, is hindered from understanding how solving this issue will benefit humanity. Living and breathing in many and other fateful works of ants, I did not consider it appropriate at this time to bother you for your honor and for the honor of verifying the winning proof you put forward for the solution of the specific "problem"/riddle that perhaps serves to preserve the refinement of my brain... ….who knows?).
In any case, greetings from all the parallels/triangles/and all the numbers... in the global distances.
By the way, have you noticed that recently our Michael has developed a special sense of humor? Don't you think he deserves a Kinneret triangle for that?
Hugin
Yehuda:
Just like that, so that you understand what is called a proof, I will take your answer and point out its flaws:
"After one second everyone who is in the first cm is on their way out"
1. Set "out"
2. Proven!
"After two seconds everyone who is in the second is on their way out"
proven!
"And so on until the fiftieth cm."
proven!
"The same thing is similar from the other side as well"
does not require further proof.
"so that the last maximum will move 50 cm without collision,"
Maximum of what?
"That means 50 more seconds on top of the first fifty."
That is, a maximum of 100 seconds.
Yehuda:
There is no doubt that you really excel in chattering.
You didn't have to quote me your "answer" because I was the one who referred you to it after you didn't remember where it was found.
This is not a solution, but your capacity for self-criticism does not exist and this is expressed in all your words - starting with the illusory "solutions" and ending with baseless slanders.
Any amount of sand you try to pour in the eyes of the readers will not turn a lie into the truth. The solution to the circle problem with the X will remain right on your nose and your anger (as your quarrel with him will remain a disgrace) your solution to the first ant problem will remain unsolved and your solution to the second ant problem (the algorithm) will remain non-existent.
I have no doubt that you won't solve the problem you call contemporary either, but I'm sure you'll invent some additional nonsense to confuse the readers.
First of all, Mr. Michael, I really answered the question as I understood it, in a XNUMXth grade class about fifty years ago, they talked about a minimal common denominator, if you force me, I will go interview the well-remembered teacher Shmuel, although he has already moved to a parallel universe, but I'm sure Hogin, who likes To me she will be ready to use connections within her powers, she will go there, with me or without, and return with a sympathetic answer.
So please don't slander.
Now I have to calm down again
Good night
Sabdarmish Yehuda
For Michael and Geva and others
I gave the answer there in response number 18:-
(a quote)
To Michael
The answer is this:-
After one second everyone in the first cm is on their way out
After two seconds everyone who is in the second is on their way out and so on until the 50th cm. The same is similar from the other side so that the latter will move a maximum of XNUMX cm without a collision,
That means 50 seconds more than the first fifty.
That is, a maximum of 100 seconds.
That's according to my logic.
What now? Does it meet the requirements?
Holidays for joy and happiness
Sabdarmish Yehuda
End of quote.
But Michael will confuse his head that this is not the answer, and he will come up with a million and one reasons and make eights just to not admit to Yehuda's wonderful ability to solve his riddles - in any case the ones that have a solution, not to circle with the X's.
Of course, the fact that Michael's explanation matched infinite numbers is just an interesting curiosity and nothing more, but Michael's answer about the infinite situation is completely incorrect! Because everyone knows that the riddle talks about a finite number of ants. And surely everyone knows that a pole with an infinite number of ants would turn into a black hole and therefore no ant would be able to get out of the pole at any time, not in a hundred seconds, not in a million seconds. So please respect the wisdom expressed here.
After me, I will rest from the slanders for not being wronged. I will approach Mr. Michael's current puzzle.
Good night for now
Sabdarmish Yehuda
Hill:
The "power" of Yehuda's calculations is evident in the "solution" he gave in response 18 in the following link:
https://www.hayadan.org.il/hounory-doctors-weizmann-0511076
As I said - this is an open-ended solution that no one except Yehuda was convinced of, and Yehuda himself did not know if it would work for an infinite number of ants (and denials will not help - although they will surely be justified - because Yehuda himself admitted this in response 74).
It should be clear to anyone with an understanding that someone who cannot know if his "proof" is correct in any situation, then it doesn't matter how much politics he wraps things in - he himself knows that even he was not able to convince himself of his "proof" for this situation.
I explained what a least common multiple is.
It is a term that cannot be interpreted except in its correct meaning.
Yehuda comes and instead of admitting that he didn't understand a simple Hebrew phrase - inserts a wrong Hebrew phrase that is used to express the same meaning and asks me why I didn't use this phrase.
My answer is that there is no need to speak politely if you arrive and there is certainly no need to resort to phrases that those who missed which class (such as the one dealing with adding fractions) missed.
The phrase I used is simple and Yehuda understood it but he accuses me of not understanding it immediately. This expression is not based on any prior knowledge other than understanding the words "double" "shared" and "minimal".
Yehuda tries to claim that he does not understand these words, but by using the incorrect phrase "common denominator" he shows that he actually understands them.
Therefore, the minimum common multiple of 30 and 18 is really 90
Nice that I said in general - 18 has multiples which are the numbers
18, 36, 54, 72, 90, 108, 126, 144, 162, 180, 198 etc.
30 has multiples which are the numbers
30, 60, 90, 120, 150, 180, 210 etc.
The common multiples are 90, 180, etc.
The minimum common multiple is 90
It's just a matter of reading comprehension. You can understand this even if you miss many classes.
You can't get along with it only if you don't understand Hebrew.
It can be argued that you don't get along with it only if you don't understand Hebrew or if you are a Jew.
Dear Mr. Geva
Those who have a memory in their head will also remember Yehuda Sabdarmish's instructive solution about crossing and colliding ants. I no longer remember where the puzzle was, but that's where the strength of my calculations was visible.
By the term "least common multiple" do you mean the "lowest common denominator"?, why don't you also explain your questions to people who missed a skin or two?,
For example:- What is the minimum common multiple of 30 and 18.
Do you mean 90?
Allow me, please, to fill a hole in my education (one of the few I have) and then you will be blessed both in the upper and the lower.
Good and blessed week
Sabdarmish Yehuda
Hill:
He did not solve.
He guessed a correct answer (in retrospect, I'm sorry I didn't ask about 72 ants crossing the stick in 45 minutes because then he wouldn't have guessed correctly either) and then tried to give an unfounded argumentation for the answer he guessed (the argumentation he gave can be formulated correctly and proven but he did none of the things These) but I assume that you have known Yehuda long enough to understand that such trifles will not deter him from trying to deceive the public.
By the way, have you seen the algorithm?
What do you think?
Yehuda
I don't remember you solving the ant puzzle.
Yehuda:
The fact that you want to tell me that your explanations are no less good than mine and even that you say this does not make your words correct.
Your solution to the original puzzle of the ants was bad and when the correct solution was presented you even admitted that it gave you an answer to a question you didn't know the solution to before.
All of this, of course, refers to the first question since for the second you didn't give any answer - nothing - not a word - you just said that I don't have a solution either.
I have already mentioned that I will present the solution - there is no doubt that in the future you will appropriate it for yourself, but even I did not imagine that the future would come in such close proximity that allows anyone to see that you are lying.
Regarding the least common multiple - probably the one that lacks a lesson in Hebrew because it is a Hebrew expression that says exactly what it says. You know that different numbers have multiples, some of the multiples are common - among the common multiples there is one that is the smallest - that is - minimal.
Besides, even if you refer to the product of the numbers then it is clear that, contrary to what you say, the set of numbers can be (like nothing) the set {2} or the set {2, 507, 555} or the set {3, 500, 700, 800}.
I didn't see you admit when I'm right.
The most that can be said is that you admit when you realize that I am right.
What disgraces you are the false accusations you throw at me time and time again because anyone can make a mistake.
Regarding the circle with the X, you are still wrong, but since I explained things in every possible way except building a XNUMXD model for you - I will not make any more effort on the matter. The question was given in the context of imagination and what I am aware of is that on the subject of spatial imagination...
First of all I want to tell you that in the ant puzzle my stories are no less good than yours and the fact is that I have reached the correct solution, the problem is that you have a psychological mental failure to praise people who reach achievements, and this is due to the abnormal evolutionary development of the genes for praise which in your case remained in the state of primitive monkeys who, as we know, did not praise each other . Not bad.
And about your new puzzle
I quote your words:-
"The least common multiple of any two of them is greater than a thousand." End quote.
So it is true that 2 times 501 is greater than a thousand, and also 3 times 400 is greater than a thousand, but 2 times 3 is less than a thousand. Therefore, since that is the case, they cannot be in the group of members of the puzzle which says that the minimum common multiple of any two of them is greater than a thousand.
From here I understood that the members must be over 31.
And it's true that you differentiate between two definitions "...that you refer to the product of numbers instead of their least common multiple", well I'm sorry to tell you that I just missed this lesson in the XNUMXth grade, so I'll have to check what the difference is, or in your wisdom, please clarify the difference for me .
Nevertheless, thanks for the formula you brought, and indeed I saw that it was not going to be 1.5 and as you showed it would be 3.5.
As soon as I know the difference I will try to understand what the poet meant by his riddle and continue to try to solve.
And besides, I still think the circle and X charges are wrong, and I hope you've already learned that if you were right, I'll admit it wholeheartedly despite the disgrace and humiliation and depression it will put on me.
So have a good day
And a good week to all
Sabdarmish Yehuda
I did everything so that the order of the English letters wouldn't go wrong and it still didn't work.
In the line with the strange combination of brackets and English letters I meant Ln of N
To the cool commenter:
I do not intend to present the solution to this riddle.
Certainly not before Yehuda says his last word about her, but probably not at all.
I know that this will make Yehuda repeat the sentence in which he usually comforts himself and say that I have no solution, but I hope to resist the temptation.
As for the hint - I am ready to give one (and only one).
The claim is true for any number - not just a thousand.
I will use this response to expand something on the formula (which does not belong to the matter) that Yehuda tried to talk about.
In fact - if we talk about some N instead of 1000 then the sum of the inverses of all the numbers is approximately
(N) Ln
In role 32 or 33 the root of N will then play (according to the wrong consideration presented by Yehuda).
The sum of all the inverses from the root onwards is the sum of the inverses up to N minus the sum of the inverses up to the root of N and since the Ln of the root of N is half of the Ln of N we get that the sum of the inverses from the root onwards is approximately half of the Ln of N
In the case of 1000 that comes out more than 3.45.
In general - this of course aspires to infinity together with N and therefore the consideration presented by Yehuda - in addition to being wrong, also does not give the desired result.
To the cool commenter:
It was a serious miss 🙂
Not only did you not get the expected block but you only referenced two numbers and not any number of numbers.
Yehuda:
I am not a liar and therefore I will not admit your lie.
The circle with the X has a solution and I even said what it is.
You, for some reason, fail even to understand the solution given and therefore, in your own way, you try to make a mess.
Since in your response you admit that you do not know the answer to the algorithm question and since I said in advance that I am ready to waste this question on you, then here is the answer in front of you (so that in the future you can tell us that you also solved it):
As I have already mentioned - the rod looks at every stage like a rod with the ants placed on it passing through each other.
This includes every aspect of the appearance of the rod and in particular we can say that ants came down from the rod in the problem scenario from its left side exactly at the same moments when ants came down from the rod from the left side in the scenario where the ants pass each other.
The same goes for ants coming down from the right side.
In particular, this also includes the number of ants that descended from each side.
If at the beginning of the process there are, for example, 35 ants going to the right and 55 ants going to the left, then it is clear that in the scenario of the ants passing each other this way (I will call it the suit scenario) there will be 35 ants on the right side and 55 ants on the left side.
Therefore, it is clear that even in the scenario where the ants return following a collision as opposed to when they came (I'll call it the collision scenario), 35 ants will descend from the right side and 55 from the left side (and as mentioned - they will descend exactly at the same times when the ants descend in the suit scenario)
So what is different in the two scenarios?
What differs between them is the identity of the ants descending at each moment.
How do we know if ants are coming down from each side in the collision scenario?
It's simple, if we understand that in this script the order of the ants on the straight can't change and if 35 ants come down from the right, these should be exactly the 35 ants on the right on the pole - with the first one coming down from the right, the first from the right, followed by the second from the right, and so on.
When will every ant come down?
As mentioned - the first ant to come down from the right will do so at the exact time when the first ant was coming down from the right in the suit script.
In the suit scenario - the first ant to descend from the right is the rightmost ant among those going right, so we know that in the collision scenario the first ant from the right will fall from the pole at the exact moment when the rightmost ant among the ants going right falls from it.
Since each ant walks at a constant speed until it falls, we can also conclude that the distance traveled by the first ant from the right until it falls is the same as the distance of the rightmost ant walking to the right from the right end.
In the same way, the second ant from the right will travel a distance identical to the distance of the second ant from the right among the ants going to the right from the right end and so on.
All this - up to the number of ants going to the right which is the number of ants going down to the right.
In the same way, the distance that the seventh ant from the left will travel is the same as the distance of the seventh ant from the left among the ants going to the left, from the left end.
In a more general formulation, if we have in the opening data M ants going to the right and N ants going to the left, then for every J that is smaller than M, the J ant on the right will travel a distance equal to its distance from the right side of the J ant on the right among the ants going to the right.
Symmetrically, for a small J equal to N, the Jth ant on the left will go a distance equal to its distance from the left side of the Jth ant on the left among the ants going to the left.
so that's it:
Your desire to accuse me of conspiracies is insatiable even though it is presented in all its ridiculousness every time.
It happened this time too but I've grown up and I understand that it will continue as long as you have access to a keyboard.
Regarding the need for the numbers to be different - there was no need to say this at all because the minimum common multiple of X with X is X and if the numbers are less than a thousand and the common multiple of any two of them is greater than a thousand it goes without saying that there are no numbers that repeat more than once.
The truth is, it's quite a shame you didn't see it.
It seems to me, based on the rest of your words, that you are referring to the product of numbers instead of their least common multiple.
Besides, you tell some strange story about 32 as a minimum number when it is clear that even 2 can be taken as a minimum number (alone, or together with all the odd numbers greater than 500).
In short - shame upon shame.
To Michael
A. I have solved your riddles whether you like it or not. The fact that you want an algorithm or ask questions that have no solution (circle with an X), is not my concern. Don't be petty and thank you for solving your riddles.
B. I also want to circumnavigate the Kinneret and even want to do it in a fast run, I'm just waiting a year or two for it to reach the diameter of a bathtub.
third. And for your old puzzle, a new one:
First of all you didn't say if the pairs of natural numbers must be different.
Assuming yes. So the lowest number must be 32 because 32*33= is the first pair whose product is above a thousand. The biggest number is 999 because you said they are all less than a thousand.
It remains to prove that the sum of the inverses from 1/32 to 1/999 is less than 1.5.
It seems to me that there is a formula that will sum up the sum of the aforementioned inverses.
Now I want to rest a bit.
good evening
Sabdarmish Yehuda
Wow this site really stinks when it comes to English
x*y>0
1/x + 1/y= x+y/x*y
x<1000, y1000
1/x + 1/y= x+y/x*y
The number is the largest, when its numerator is the largest possible, and the denominator is the smallest possible, therefore we will be strict:
1998/1001 = 1.996..
Um.. Michael? Apparently I failed once again.. I was able to prove that the sum of their inverses is less than 1.996 and not less than 1.5 as you requested.
Any clue?
Yehuda:
Regarding the black hole - I answered you because you directed your words to "Mr. Michael" and not to whom you want to warn.
I explained to you that the phenomenon you describe is a private case of tidal forces.
Now I will stop responding for today even though I haven't read the articles that came out tonight - because I'm going to the Kinneret.
I repeat - this is about the circumambulation of the Kinneret and not about unfriendliness.
Yehuda:
I have many and I explained to you why I don't waste them.
Why do you have to come up with conspiracy theories about it? Does my explanation not seem justified to you?
Here, I'll go back for you on another riddle I gave here (beyond the one with the ants that you're still dealing with and the one with the levies that you need to understand the answer I gave).
This, as mentioned, is a repetition of a riddle I already gave here:
There is a set of natural numbers that are all less than a thousand but the least common multiple of any two of them is greater than a thousand.
It was proved that the sum of their inverses is less than one and a half.
Clarification: The inverse of a number is one of the parts of the number.
Yes
Who are you?
After all, I wanted to warn the science commenters not to get close to a black hole. There is no reason for you to come down on me for that.
And regarding the puzzle that you don't want to add it's because of one of the two, either you don't have one, or you do and you have an unsociable personality.
for everyone:-
Anyone have a riddle for the weekend?
Sabdarmish Yehuda
Good Day
Sabdarmish Yehuda
To Mr. Yehuda:
Why did you think you had to tell me such a trivial thing?
I was talking about the laws of nature. not about anything else.
Although what you described does not contradict this claim, I also wrote about tidal forces because they are a result of the laws of nature that can be harmful.
It turns out that you are not familiar with the term (although I have explained it many times even on this site) and therefore you chose to point out to me that these powers exist.
Regarding puzzles:
As I told you before - I have already posted enough puzzles here. After you solve the ones I posted I will consider whether to post more.
To Mr. Michael
The reality near a black hole is different from near a star of the same mass because the differences in the gravitational forces near the black hole between the head and the legs can cause the body falling into the black hole to be torn apart.
It will mostly manifest as a little black guy.
No puzzle for the weekend for the free comments section?
good evening
Sabdarmish Yehuda
If your fart particles have the right acceleration they can light a match.
To the commenter to the cool commenter:
I think you made the creator.
Dr. Dahan:
Good thing you wrote in parentheses "if I'm not wrong" because you are wrong.
The laws of time are no different near a black hole and in fact all the laws of nature remain the same.
The difference is in the perception of time by an observer from a distance the time of the one who is close to the black hole but the one who is close to the black hole does not feel anything about this.
The only thing that can harm us are the strong tidal forces, but these forces become significant only when you are very close to the black hole and in fact - again - are no different from the tidal forces exerted by any other star of the same mass located at the same distance.
Just so you know that a match lights up quickly and easily.
The chance that the LHS particle accelerator will create a stable black hole is about the same as the chance that I will be able to light a match with a fart...
Michael, you have a mistake...
Let's say you are right, and yes, the Earth will succeed in orbiting the sun, we will not get to see the black hole because the laws of nature will automatically change because (if I am not mistaken) the laws of time are different near a black hole and we humans will not survive, and the Earth will turn into chaos, but what luck The chance that a medium-sized black hole will pass by the solar system is currently minimal, huh?
And one more thing, tell me what is happening with the LHS particle accelerator, which in a certain situation may create an artificial black hole that will suck in all the cosmic rays?
The radius (Schwarzchild radius) of a black hole whose mass is the mass of the Sun is three km and as the mass increases, the radius increases proportionally to it. 10 solar masses - 30 km, 100 solar masses - 300 km.
Outside this radius, the gravity of a black hole is just like the gravity of a normal star with the same mass.
If someone approached the black hole to a distance less than its radius it was already guaranteed that they would never leave.
If it is at a greater distance - its fate depends on its speed relative to the hole and its distance from the hole - again - just like with normal stars.
In other words - it is not correct to say that the Schwarzschild radius is the only danger zone. It is true that this is the most dangerous area because the relative speed is no longer important, but also in other areas there is a danger that depends, as mentioned, on the relative speed from the hole.
So are we in danger?
As far as I understand - not really.
The speed of movement of these holes exceeds 4000 km per second.
Although this is the speed relative to the origin cluster, but since it far exceeds the speed of the movement of the stars in the galaxy, it can be seen as the speed of their movement relative to us as well.
I made a rough calculation and if I wasn't too wrong, then if such a black hole passes at a distance that is greater than the radius of the solar system, then our relative speed will be an escape speed, that is, one where we will not even be trapped in orbit around it.
If it passes closer, things become more complicated, but even then it is very likely that we will enter orbit around it, and it is even likely that we will continue to revolve around the sun that will surround it (this is important, of course, because without the sun our fate is decided).
If we take into account the fact that only a few hundred such black holes are even moving around in the galaxy - the chance that something bad will happen to us is minimal.
Answer to Dr. Dahan:
If there are black holes close to the solar system, theoretically they could indeed swallow the solar system.
But in practical terms, black holes have a gravitational effect on the bodies around it. And if there was a black hole near the solar system, it would likely have a great impact on us - and the small galactic region we are in would look completely different.
Wait, I don't understand something, if there are black holes that are close to the solar system, then theoretically they can swallow the solar system and end life on Earth, right?
I wonder what will happen if a black hole reaches us. 🙂
This is also what is written in the original Vanderbilt University article.
Sabdarmish Yehuda
Their risk zone - Schwarzschild radius, extremely tiny, only a few hundred kilometers.
Only hundreds of kilometers risk zone of a medium black hole?!