# Gravitational force of a neutron star

1. Nov 26, 2015

### JohnnyGui

Hello all,

I could be looking at this the wrong way but here it goes:

From what I understand, if a star collapses into itself to form a neutron star, it would become more compact, denser and heavier with a larger gravitational attraction.

Now, if I understand correctly, the size of gravitational attraction depends on the total amount of mass that a star has.
If a star collapses into itself to form a neutron star, it will still have the same total amount of mass (let's assume that the star doesn't shed off any of its outer layers before collapsing) but in a more denser and smaller volume.
However, if it still has the same total amount of mass, shouldn't a neutron star have the same gravitational attraction as before the star collapsed? I know that the mass has become denser, but the volume also shrinked into a smaller size which "compensates" for the increase in density and thus keep the total amount of mass constant. How then can it be that a neutron star has a larger gravitational attraction than before?

I have read explanations about one being closer to the center of the star if it turns into a neutron star and thus experience a larger gravitational attraction, but I'd still think that my argument stands:
If one is standing on a star before collapsing, he will have the same amount of total amount of mass beneath him as when he's standing on the star's neutron version; it is denser but it's also smaller in volume.

Am I missing something here?

2. Nov 26, 2015

### DrStupid

Outside the volume of the original star: yes
Inside the volume of the original star: no

3. Nov 26, 2015

### rootone

Standing on the neutron star surface, you are closer to the center of mass.
Gravity is always higher when closer to a massive object than it is when further away from it.
Mercury experiences stronger gravity of the Sun than Earth does, yet it's the same Sun.

4. Nov 26, 2015

### JohnnyGui

Thank you guys for the replies.

Based on your answers, I'm starting to conclude that it's not the total amount of mass that determines the size gravitational attraction but the density. Someone who's floating at a distance from a star (analogue to your Mercury and Earth example) can be considered as someone having a lower density of mass beneath him (the star mass + some vacuum) and thus undergoing a lower gravitational attraction.

A question to verify my understanding:
Assume there's a star that has homogenous subdensities in its massive body and it sheds off its outer layers instead of collapsing onto itself. Can I say that if a person stands on that star after it has shed off its outer layers, he will experience the same graviational attraction as when he stands on the star before it sheds off its outer layers? The reason I'm thinking this is because the total density beneath the person's feet would stay the same; after the star sheds its outer layers there's less mass but there's also a smaller volume and thus the density would stay the same. Is this correct?

5. Nov 26, 2015

### Ken G

It's not the density, consider Newton's formula. The gravitational force at the surface is the mass divided by the square of the radius. That depends on the density times the radius if you think about it. So this is why Earth has a larger surface gravity than Mars, even though their density is not much different-- the difference is the radius. Thus, if a star sheds mass, but keeps fixed density (it wouldn't, but we can imagine), then its surface gravity would drop, because it would be proportional to the radius of the star.

6. Nov 27, 2015

### JohnnyGui

Can't I say that the gravitational force of the earth is greater than mercury's despite their similar densities simply because the earth has more total mass with the same density? The only outcome to have more mass but with a similiar density is to make it bigger in volume; thus having a larger radius. That's what I meant in my opening post; that the total amount of mass beneath you determines the gravitational attraction.
Having said that, if a star sheds mass while keeping its density, it must end with a lower amount of mass and thus a lower gravitational attraction, like you said indeed.
From those 2 examples I conclude that the total amount of mass is determinable for the gravitational attraction.
However, this conclusion leads to my problem; if a star collapses, it will still have the same amount of mass and yet, if you stand on it, you'll undergo a greater gravitational attraction compared to if you stand on it before it collapses. This scenario implies that the total amount of mass is not the only factor that determines the gravitational force of a body. You might say it's the distance to the mass object but in both cases (a star and a neutron star) the distance to it is 0 since you're standing on both of them.

What's wrong in this approach?

7. Nov 27, 2015

### rootone

What is wrong is saying that distance to the center of gravity is where your feet are, so if you are standing on the surface the distance is zero.
What is right is that center of gravity is where the center of mass is.
You experience high gravity on a neutron star because it has a small radius and surface is not far from the center.
On a normal star the surface is more distant from the center, so less gravity is felt.

8. Nov 27, 2015

### JohnnyGui

Thank you, I think I'm now at the root of my problem here. Is it possible to answer why exactly the center of gravity determines the force instead of the total amount of mass beneath you without saying that it's just the way how nature works? I think I'm trying to understand the roots of how gravity works here.

9. Nov 27, 2015

### Ken G

You tried two approaches:
1) the total mass is what matters for the gravity
2) the density is what matter for the gravity.
But both proved wrong. This is because what actually matters is the total mass divided by the square of the radius. It is very important in physics to use the actual formula. It's good to seek more intuitive understanding, like mass and density, but not at the expense of ignoring the formula, that's just not going to work.

Yes, the conclusion is wrong. Either #1 or #2 above are wrong conclusions, the formula is what is correct.
It is not respecting the formula for gravitational force. That should be your starting point. If you want to get some intuition into that formula, imagine that the mass is a kind of "source" of gravity, but the gravity from that source "spreads out radially", like bullets fired in all directions from a machine gun. At any given distance from the source (the center of the star), the source (or "bullets") are distributed over the surface area of a sphere with radius equal to that distance.

10. Nov 27, 2015

### JohnnyGui

I think my problem is that I'm trying to understand the exact physical reasoning why the formula is formulated that way and trying out different approaches to see if it would fit the formula. So far, my approaches are indeed wrong but I can't find the exact reasoning why it's the center of the gravity that plays a role and not the density or total amount of mass. Sorry if I'm being so stubborn at trying to understand this.

11. Nov 27, 2015

### Ken G

Try the idea of a machine gun spitting out bullets in all directions, at a rate we can call "mass." Then ask how many bullets hit a shield of given area, as a function of distance from the machine gun. That's also how the strength of gravity works.

12. Nov 28, 2015

### JohnnyGui

That does help me understand it better, thank you.
I've got one question regarding you saying that the formule m / r2 actually being the same as the density × radius. Do you mean by this that the density is written as (m / r3)? Because that means that density × radius can be written as (m / r3) × r which is the same as m / r2.

I noticed with the formula (m / r3) × r that if a sphere has its radius reduced with a factor x while holding the same density ( (m / r3) = constant), that its mass would be reduced with a factor x3. Is this correct?

13. Nov 29, 2015

### Ken G

Correct.
Yes. So the combination of the big mass loss, and the slightly less increase in the force due to the smaller radius, together produce a net decrease in the force at the surface by the factor x in the case you describe. That's a lot like Earth vs. Mars, for example.

14. Nov 30, 2015

### Curiousandriven

A formula is only as good as its principals and principals have changed throughout history. I agree that we need to use the tools that have been given to us, but to say something is correct for the sole reasoning of it being created before us is counterproductive to us making progress. We are constantly learning knew things and things are changing. Please feel free to always challenge and understand any equation. It could lead to our next great discovery.

15. Nov 30, 2015

### Curiousandriven

Essentially they're saying that an object of size and density sheds some of its size but the density remains the same. Obviously if an object gets smaller but it's density stays constant it would have to shed mass. If you're playing with play dough you can squeeze it as hard as you want and it's density and overall mass will remain constant. The only way to make that play dough ball smaller is to take some away. It remains the same density but know with less play dough it has smaller size and less mass. Mass does effect gravity, but so does the distance you are away from the center of the object that you're experiencing the gravity from. Our sun stays the exact same mass but is compressed to half its size. You just reduced the distance to the center of its gravity and therefore greatly increased the gravity felt. Now if we're really getting down to things I believe that gravity is a constant and that mass is the only thing that effects the total amount of an object. The distance from center and density change the effect felt, but as an object its overall gravity remains constant to its mass.

16. Nov 30, 2015

### |Glitch|

The star does not have the same total amount of mass when it dies. If the star is approximately 4.8 solar masses or larger, the star sheds approximately 70% of its mass in a supernova when it dies. Just to be clear, a star "dies" when it is no longer able to fuse elements into new elements. It is that fusion process that keeps gravity from collapsing the star into a white dwarf, neutron star, or black hole. The nuclear reactions from this fusion process pushes the stellar material outward, while gravity attempts to collapse the star, thus achieving what is called an hydrostatic equilibrium.

In order to become a neutron star what is left after the supernova must be between 1.44 and 3.0 solar masses. If the remnant of the star is less than 1.44 solar masses it becomes a white dwarf. If the remnant of the star is greater than 3.0 solar masses it becomes a black hole. Therefore, the original mass of the star before going supernova must be approximately between 4.8 and 10 solar masses in order to become a neutron star.

17. Nov 30, 2015

### Jim Hasty

The radius distance between two objects is calculated from their 'center of mass'. So it makes no difference if the surface of either object 'shrinks' - if the distance r between them is constant - then the gravitational attraction remains constant. At least for Newton's equation F=Gm1m2/r^2. [However, according to Einstein's GR: you must also take into account energy which can make some difference. So if for example a star is giving up energy and shrinks to become a neutron star - as in your example. Then the loss of energy effects the gravity - i.e., the gravity effect is less. But if all mass and energy is held constant, and the radius r is constant - then Newton's equation is correct.]

18. Nov 30, 2015

### Ken G

This is a minor point, but I think it is worth making, because what you are describing is a pretty common misconception. Fusion actually plays almost no role in the hydrostatic equilibrium of a star, in the sense of creating an "outward push" that is due to fusion. Hydrostatic equilibrium involves a balance of pressure and gravity, and sets up on timescales characteristic of the free-fall time, typically just minutes. During that time, the amount of fusion that occurs is unimportant. Fusion only shows up on timescales like millions of years, the time it would take the star to change if there were no fusion going on. And if there were no fusion going on, what would happen to that "outward push" is that it would gradually increase as the star contracted, it just wouldn't increase as fast as gravity would. So when fusion ends, you never notice a reduction in the outward push of pressure.

What is true is that when fusion stops, the star loses its equilibrium, and begins to very gradually change on timescales like a million years, but hydrostatic equilibrium is still maintained during that period, with or without fusion, because the evolutionary time is always way longer than the free-fall time (except in an actual supernova). So if the Sun ceased fusing for a year or two, there would be essentially zero effect on its pressure, even though the hydrostatic equilibrium must be maintained over timescales like minutes. Put differently, the pressure in a star does not come from fusion, it comes from internal heat/energy. Fusion is only a (very slow) term in the evolution of that internal heat/energy.

19. Dec 2, 2015

### JohnnyGui

Wow, thank you all for the great explanations.

That was indeed the conclusion I wanted to hear to understand why a shedding star loses gravitational attraction. Hence me asking about the formula :). I really tried to fully understand your explanation about the hydrostatic equilibrium and fusion in your post #18 but I have a bit of a difficulty with what you're trying to say regarding the different time scales. Is it possible to reformulate this?

That's what I was trying to understand. However, the thought that gravity and mass always should remain constant to each other led me conclude wrong that regardless of how dense an object is, as long as it has the same mass it should have the same gravity and thus have the same gravitational attraction without taking the radius from the center into account.

@|Glitch| : I'm actually surprised that, although a star loses that much of mass from its original state, it still somehow manages to have a larger gravitational force. I guess the factor by which the radius decreases is much larger than the factor by which the mass does?

20. Dec 2, 2015

### zylon

The total gravitation of the neutron star is the same as anything else of the same mass. But the force is concentrated on the smaller surface area. So if the neutron star has 10^-5 the diameter of the sun with the same mass as the sun, its surface area would be 10^-10 of the sun, and its surface gravity would be concentrated to 10^10 times that of the sun. (10^10 is one with ten zeros, i.e. ten billion!)

21. Dec 2, 2015

### Ken G

The issue is, what is the role of fusion in the hydrostatic equilibrium of a star. This is a somewhat dicey issue, because no stars are actually in hydrostatic equilibrium, that is always an approximation. One way to characterize the accuracy of the approximation is to compare the timescale for the star to change, to the free-fall timescale (which is the timescale for change if there was not an approximate hydrostatic equilibrium). For a star undergoing fusion, the timescale for change is the fusion timescale, which is usually very long-- something like ten billions years for the Sun. But the free-fall time is some number of minutes-- way way shorter. So that is the basis of using the hydrostatic equilibrium approximation, and expecting it to be very accurate.

So what is the role of fusion in that approximation? Not much-- without fusion, the timescale for change would still be way way longer than the free-fall time. However, it would be millions of years, instead of billions, so we can still recognize fusion as having some small role in making the equilibrium approximation accurate. But really what it does is, slow the evolution of the star, moreso than impose an equilibrium that would have been there anyway.

22. Dec 3, 2015

### nikkkom

This can't possibly be right, since the most massive stars go through their entire evolution in just a few million years.

I don't understand why you are agonizing over fusion role. It's quite clear: it supplies energy to heat the star.

23. Dec 3, 2015

### Ken G

The numbers were not tailored for the most massive stars, but the overall conclusions are the same there. Any main-sequence star that stops fusing for a free-fall time will remain in very good hydrostatic equilibrium throughout that time, so fusion is not responsible for hydrostatic equilibrium, other factors are (pressure is, which is essentially the internal kinetic energy content). Fusion is responsible for greatly extending the already long evolutionary timescale, which is quite a bit different than producing hydrostatic equilibrium. Perhaps the best way to say all this is that almost all stars would be in very good hydrostatic equilibrium, and would look pretty similar to how they look now, even if there was no such thing as fusion. What would be different mostly is an accelerated timescale for evolution, but still nowhere close to the free-fall time, and also what happens in late stages like red giants (where fusion alters the internal structure dramatically).
Actually, it is fairer to say that fusion replaces the heat the star is losing, which can certainly be counted as "heating the star" if we are clear on the meaning there, but this is not really what we care about here. My objection was not in saying that fusion appears as an energy source, it was the claim that it produces some kind of "outward force" essential to the hydrostatic equilibrium. This very often leads to the misconception that hydrostatic equilibrium is lost when fusion ends, which is quite untrue. To say this more formally, let me point out that hydrostatic equilibrium is a statement about the smallness of the time dependent term in the momentum equation. But fusion does not appear at all in the momentum equation, it appears only in the energy equation, which has a way longer characteristic timescale. What this means is, the effects of fusion on hydrostatic equilibrium are only felt indirectly, and over timescales that are way way longer than those involved in setting up hydrostatic equilibrium.

Last edited: Dec 3, 2015
24. Dec 3, 2015

### |Glitch|

It does not have a larger gravitational force. A neutron star has a much smaller gravitational force than it had before it became a neutron star because it is now ≈70% less massive. The neutron star is as small and as dense as it is because it has lost its ability to fuse new elements and generate heat. So there is no more outward pressure counteracting the effects of gravity.

25. Dec 4, 2015

### Curiousandriven

But if what's been being said is true then fusion plays a very small, and over an immense amount of time, role in the outward force of a star. So the fact that fusion is lost would have an extremely gradual affect on the hydrostatic equilibrium? From what I understand of the previous posts on the thread.