Does gravity change as density increases?

[Raider5678]

Ok, while reading about the curvature about space time, black holes, and gravity, I came across a formula that tells you how small something has to be to become a black hole. Since a black hole has a huge amount of gravity at the event horizon, I came to the conclusion that perhaps gravity changes as density increases. The formula is as follows:

R=(2GM)/(c^2)

This means that you can calculate how small something has to be to become a black hole by "doubling the object's mass, multiplying it by the universal gravitational constant, and dividing the entire thing by the speed of light squared"

This is a quote from the website http://io9.gizmodo.com/5974372/at-what-point-would-the-earth-become-a-black-hole

I don't know how reliable this website is but the formula seems pretty sound and other people go with it so I trust it enough. Unless it's not a trustworthy site. This formula may also seem pretty easy to figure out, but the universal gravitational constant is this:

6.67408 × 10^-11 m³ kg^-1 s^-2

^{10 to the -11 power is a really tiny number, negative 1 trillion if I did my math correct. But then again I might not have because I'm only in 7th grade. Either way m3 I have no idea what it stands for, nor kg, or s. Mass, kilograms, and something else is my best guess. If someone could please explain if density does modify gravity, and if you could explain the formula that would be great!}

^Thanks.

[swansont]

Ok, while reading about the curvature about space time, black holes, and gravity, I came across a formula that tells you how small something has to be to become a black hole. Since a black hole has a huge amount of gravity at the event horizon, I came to the conclusion that perhaps gravity changes as density increases. The formula is as follows:

R=(2GM)/(c^2)

This means that you can calculate how small something has to be to become a black hole by "doubling the object's mass, multiplying it by the universal gravitational constant, and dividing the entire thing by the speed of light squared"

This is a quote from the website http://io9.gizmodo.com/5974372/at-what-point-would-the-earth-become-a-black-hole

I don't know how reliable this website is but the formula seems pretty sound and other people go with it so I trust it enough. Unless it's not a trustworthy site. This formula may also seem pretty easy to figure out, but the universal gravitational constant is this:

6.67408 × 10^-11 m³ kg^-1 s^-2

^{10 to the -11 power is a really tiny number, negative 1 trillion if I did my math correct. But then again I might not have because I'm only in 7th grade. Either way m3 I have no idea what it stands for, nor kg, or s. Mass, kilograms, and something else is my best guess. If someone could please explain if density does modify gravity, and if you could explain the formula that would be great!}

^Thanks.

 

 

No. Gravity depends on the mass (in Newtonian physics; in GR it's more generalized to energy) and how far away you are from the mass. So if you increase the density but reduce the volume to keep the mass the same, and your distance from the center of that mass is the same, the gravitational pull on you would be the same.

 

You can write the mass as density*volume, so you can have an equation that has density in it, but the important dependence is on the product density*volume

[ajb]

The formula you give is okay, it is based on the event horizon of a spherically symmetric non-rotating black hole. So, as a hand-waving argument this is perfectly okay.

 

Your question about Newton's constant, the m there is meters.

 

Your statement about a black hole having a huge amount of gravity is very slack. The usual thing one could mean by this is the surface gravity, which for a non-rotating black hole is 1/4M where M is the mass (in appropriate units).

[Strange]

 

R=(2GM)/(c^2)

 

Yes, this formula is correct. It is called the Schwarzschild radius: https://en.wikipedia.org/wiki/Schwarzschild_radius

 

the universal gravitational constant is this:

6.67408 × 10^-11 m³ kg^-1 s^-2

^{10 to the -11 power is a really tiny number, negative 1 trillion if I did my math correct. But then again I might not have because I'm only in 7th grade. Either way m3 I have no idea what it stands for, nor kg, or s.}

 

m³ is metres cubed (i.e. volume)

kg is kilograms (and kg^-1 means 1/kg or "per kilogram")

s is seconds, so s^-2 means 1/s² (or "per second squared")

Edited March 10, 2016 by Strange

[MigL]

For galactic center sized black holes ( million or billion solar mass ) the density needed for collapse can be less than that of water.

While for a star a couple of times bigger than our sun, even the density of neutronium will not allow for collapse.

[Raider5678]

 

 

No. Gravity depends on the mass (in Newtonian physics; in GR it's more generalized to energy) and how far away you are from the mass. So if you increase the density but reduce the volume to keep the mass the same, and your distance from the center of that mass is the same, the gravitational pull on you would be the same.

 

You can write the mass as density*volume, so you can have an equation that has density in it, but the important dependence is on the product density*volume

Ok thanks, that helps a lot.

 

The formula you give is okay, it is based on the event horizon of a spherically symmetric non-rotating black hole. So, as a hand-waving argument this is perfectly okay.

 

Your question about Newton's constant, the m there is meters.

 

Your statement about a black hole having a huge amount of gravity is very slack. The usual thing one could mean by this is the surface gravity, which for a non-rotating black hole is 1/4M where M is the mass (in appropriate units).

Right, hand waving argument. Not quite sure what it is but I'm guessing I did something wrong here. Either way the statement I made earlier was pretty slack, but the 1/4M equation I can't figure out. A black Holes mass, divided by 4, is a black hole's gravity if it's a non-rotating black hole. The mass would be measured in the appropriate units such as kilograms and grams? So a 1000 kilogram black hole, would have 250 gravity. I'm not sure what the appropriate unit for the 250 gravity would be, but if you could explain the equation a little better that would help. Thanks for the meters part though!

 

 

Yes, this formula is correct. It is called the Schwarzschild radius: https://en.wikipedia.org/wiki/Schwarzschild_radius

 

 

m³ is metres cubed (i.e. volume)

kg is kilograms (and kg^-1 means 1/kg or "per kilogram")

s is seconds, so s^-2 means 1/s² (or "per second squared")

Thanks, the universal gravitational constant is a lot easier if you know what the variables are.

For galactic center sized black holes ( million or billion solar mass ) the density needed for collapse can be less than that of water.

While for a star a couple of times bigger than our sun, even the density of neutronium will not allow for collapse.

So the bigger something is, the less density is required to collapse in on itself if I am understanding this correctly. Is that correct?

 

Ok now, why does a black hole have more gravity than its previous counter part that collapsed in on itself?

Edited March 10, 2016 by Raider5678

[Strange]

Ok now, why does a black hole have more gravity than its previous counter part that collapsed in on itself?

 

It doesn't. So, for example, if the Sun suddenly turned into a black hole nothing would change; we would continue to orbit it in the same way. (Well, it would get very dark and cold ...)

 

The difference is that you can get much closer to a black hole than you can an "ordinary" object of the same mass.

[Raider5678]

 

It doesn't. So, for example, if the Sun suddenly turned into a black hole nothing would change; we would continue to orbit it in the same way. (Well, it would get very dark and cold ...)

 

The difference is that you can get much closer to a black hole than you can an "ordinary" object of the same mass.

Ok, that makes sense. But why does light get caught in a black hole?

[dimreepr]

Read this.

[Strange]

Ok, that makes sense. But why does light get caught in a black hole?

 

It is because gravity is caused by the curvature of space-time. At the "surface" (event horizon) space is so curved that there is no path that leads out of the black hole.

 

Note that the "escape velocity" description is quite popular. But it doesn't quite work. For example, you can throw a rock from the surface of the Earth at less than escape velocity and it will go some way and then fall down again. So that suggest that light could temporarily escape a black hole before being pulled back. It can't.

[Raider5678]

Read this.

Ok I get it, the closer you are to the center of the mass the higher the gravity is, since the density of black holes is as dense as it can get, the object is really small, allowing things to get way closer to the center of the gravity. The event horizon is the distance from the center that the gravity is so high, any light that hits the event horizon enters a gravitational pull way too powerful to escape, hence the black hole is the distance from the center that light can't escape from, therefore making it black. Ok, this coupled with the other posts completely work together and answer my question. Thanks!

[ajb]

Not quite sure what it is but I'm guessing I did something wrong here.

You did nothing wrong. It is just that the conditions for an object to become a black hole are not fully understood. However, we know that under some general and physically reasonable assumptions that they do form. So the formula you have is okay as a general argument for the length scales involved. That is all I meant.

 

... but the 1/4M equation I can't figure out.

You can look up surface gravity on wikipedia. I am not sure if you will follow the derivation, given your age.

 

A black Holes mass, divided by 4, is a black hole's gravity if it's a non-rotating black hole. The mass would be measured in the appropriate units such as kilograms and grams?

I suspect there should be some factors of the speed of light. But we can pick natural units so that any constants are numerically equal to 1. (I will have to check carefully)

 

 

So a 1000 kilogram black hole, would have 250 gravity. I'm not sure what the appropriate unit for the 250 gravity would be, but if you could explain the equation a little better that would help. Thanks for the meters part though!

The surface gravity of a black hole is basically the acceleration of a test body at the event horizon. In general this is not well defined, but for 'nice' black holes it is. (They need a particular kind of symmetry)

 

It is like the g ~ 10 m/s^2 at the surface of the Earth.

 

 

 

Ok now, why does a black hole have more gravity than its previous counter part that collapsed in on itself?

We are confused by the meaning of 'more gravity'.

[Mordred]

I believe he's defining force of gravity at a radius. The problem is he didn't know the mass doesn't change but the radius does.

 

For example if your 1 unit of measure from the pre-existing star. The gravity is weaker than being 1 unit of measure from the BH as your closer to the center of gravity for that object.

Though the previous posts already explained this aspect and the OP understands it

Edited March 11, 2016 by Mordred

[Raider5678]

You did nothing wrong. It is just that the conditions for an object to become a black hole are not fully understood. However, we know that under some general and physically reasonable assumptions that they do form. So the formula you have is okay as a general argument for the length scales involved. That is all I meant.

 

Ok.

 

You can look up surface gravity on wikipedia. I am not sure if you will follow the derivation, given your age.

I read it, and while I didn't understand all of it, I looked up gravity of earth and read that, which made slightly more sense. Then I went back and read that page again and it made much more sense, although I don't fully understand 100% of the formulas

 

I suspect there should be some factors of the speed of light. But we can pick natural units so that any constants are numerically equal to 1. (I will have to check carefully)

 

Ok.

 

The surface gravity of a black hole is basically the acceleration of a test body at the event horizon. In general this is not well defined, but for 'nice' black holes it is. (They need a particular kind of symmetry)

 

It is like the g ~ 10 m/s^2 at the surface of the Earth.

 

The Gravity of Earth gave me slightly more understanding of what your talking about. As for a black hole wouldn't its maximum fall speed be the speed of light?

We are confused by the meaning of 'more gravity'.

 

Stronger gravitational pull than the previous object is what I meant, but I now understand that it's the distance from the center.

P.S. How fast would earth have to be rotating to completely counteract the gravity at the equator?