# Density and Gravity?

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Gravity is higher on a object with a bigger mass and attracts smaller objects to it. But I have a theory. I will use this picture as an asset.

What if m2 has a bigger density but is much smaller than m1, which is larger and is not extremely dense as m2. Would the gravitational pull be higher on m2 or would m1 still have more gravitational pull than m2?

IAstroVIz

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The Newtonian force of gravity is given by the following formula:

F = G {m_1 m_2}/{r^2} See Wikipedia

Assuming your two bodies are attracted to a third body that is placed at the same distance from your two bodies, the more massive body will attract with more force. Density is not a factor.

IIRC an object with mass distributed (e.g., galaxy) is the same as the mass of a point object of equal mass at the center of the distributed mass, for things outside the volume of distribution (e.g., things outside the galaxy).

Edited by EdEarl
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Someone standing on the surface of m2 will feel more "gravitational force" than someone standing on the surface of m1. Basically, their "r" would be smaller, thanks to the higher density giving a smaller volume from the same mass.

But as above, anything would feel an equal "gravitational force" at an equal distance from m2 and m1.

e.g. if we squeezed the Earth, making it's density higher, and volume smaller, but with the same mass, then people standing on the surface of Earth would feel stronger gravity; but the Moon would continue to orbit pretty much * the same as it does now, as it wouldn't notice any difference.

(* besides effects from Earth not actually being completely smooth, and not having even density throughout.)

Edited by pzkpfw
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So to simplify it, if you would condense the earth to an apple, it would have the same gravitational force like it did before?

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So to simplify it, if you would condense the earth to an apple, it would have the same gravitational force like it did before?

Assuming that you are located at the same distance from the center of the Earth that the Earth's surface currently is, yes.

Force of gravity is dependent on mass and distance, so just saying that the force will be "the same" is tricky. At the same distance from the center, the force will be the same if the mass doesn't chance. You can get closer to smaller objects, however, and as you get closer, the strength of gravity will increase.

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So to simplify it, if you would condense the earth to an apple, it would have the same gravitational force like it did before?

If you could squeeze Earth hard enough, that it became a black hole, then anything trying to stand on it would certainly learn about its greater density.

But the Moon wouldn't notice.

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If you could squeeze Earth hard enough, that it became a black hole, then anything trying to stand on it would certainly learn about its greater density.

That would be about the size of a grape, rather than an apple.

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Strange

That would be about the size of a grape, rather than an apple

So that's how the camel passed through the eye of the needle.

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To correct you, the earth need to be as big as a peanut...

So density affects gravity! Or is it more complicated?

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That would be about the size of a grape, rather than an apple.

To correct you, the earth need to be as big as a peanut...

...

$r_s = \frac{2GM}{c^2}$

http://www.wolframalpha.com/input/?i=%282*6.67384e-11*5.972e24%29%2F%28299792458%5E2%29+

ie grape sized

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To correct you, the earth need to be as big as a peanut...

So density affects gravity! Or is it more complicated?

It is misleading to say density affects gravity. It is better to say that as two masses get closer to each other, the force of gravity between them gets stronger, and it is possible to get closer to a dense object than one that is not dense.

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So density affects gravity! Or is it more complicated?

If the sun turned into a black hole (about 3km across) it would not make any difference to the Earth's orbit. (But it would get very cold.)

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So density affects gravity! Or is it more complicated?

There is no direct relationship to density. One can extract a density relationship under certain conditions/assumptions (e.g. surface gravity, uniform composition), but it is not useful outside of those special cases, i.e. there is no generally true, useful relationship that involves density.

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There is no direct relationship to density. One can extract a density relationship under certain conditions/assumptions (e.g. surface gravity, uniform composition), but it is not useful outside of those special cases, i.e. there is no generally true, useful relationship that involves density.

I don't know if this is a sensible question. but would the gravitational gradient be steeper around a denser object relative to another object of the same mass but lower density?

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I don't know if this is a sensible question. but would the gravitational gradient be steeper around a denser object relative to another object of the same mass but lower density?

No - you can use a point mass instead of a sphere for all gravity calculations I believe (excepting tidal influences on the central mass). But you can get much closer to the centre of a high density mass. Force scales with 1/r^2 so the change in one meter is higher at say r/2 than at r, 2*r or 3*r - and in the situation of an earth mass black hole you can get to r_earth and r_earth/2 without hitting the ground

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No - you can use a point mass instead of a sphere for all gravity calculations I believe (excepting tidal influences on the central mass). But you can get much closer to the centre of a high density mass. Force scales with 1/r^2 so the change in one meter is higher at say r/2 than at r, 2*r or 3*r - and in the situation of an earth mass black hole you can get to r_earth and r_earth/2 without hitting the ground

So, you always use the centre, and not the outer boundary, for the starting calculations?

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What if m2 has a bigger density but is much smaller than m1, which is larger and is not extremely dense as m2. Would the gravitational pull be higher on m2 or would m1 still have more gravitational pull than m2?

Yes but if I recall pressure=force/area so the gravitation pull of m2 would exert a greater downward pressure.

Edited by fiveworlds
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Changing the density requires changing one of two factors - the mass or the volume.

For an object at a given fixed distance away from the center point of the planet, such as a moon, one of two things can happen:

1. If the mass increases, then the gravitational attraction of the object also increases.
2. If the volume decreases, then the gravitational affect on the distant object remains unchanged.

In both cases, the surface gravity of the planet would increase, as given by

$g = \frac{4\pi}{3}G \rho r$

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So, you always use the centre, and not the outer boundary, for the starting calculations?

Yeah - if you think about it we would have a divide by zero problem here on the earth's surface otherwise. r in these circumstances is distance from centre of mass

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Yeah - if you think about it we would have a divide by zero problem here on the earth's surface otherwise. r in these circumstances is distance from centre of mass

Right, OK. I wondered if it was the surface because G is at minima in the centre, isn't it?

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Right, OK. I wondered if it was the surface because G is at minima in the centre, isn't it?

G is a constant. g varies

Yes attraction drops a 1/r^2 till you reach the surface - and then assuming a sphere with uniform density drops at 1/r till the centre

edit

this is confusing - will re-write

At a great distance r is large and r^2 is huge - the gravitational attraction grows as r decreases. The highest gravity you will experience is on the surface. After you pass r = r_earth - ie the surface - then the volume of earth, that is the mass beneath you starts to decrease as well.

As you progress through the earth the attraction will decrease - the falling distance to the centre of mass means that 1/r^2 gets bigger (ie closer to one) BUT the amount of mass (ie the volume) varies with r^3. So as r decreases the r^2 in the denominator changes - but not as quickly as the mass ie r^3 in the numerator. So gravitational attraction falls linearly with distance below the surface

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Right, OK. I wondered if it was the surface because G is at minima in the centre, isn't it?

And this is where density comes in. Because the relationship between distance from the center and the strength of gravity that we use above the surface breaks once you go past it. As you go down, more and more of the mass is "above" you, no longer contributing to the gravity you feel underfoot, and in fact counteracting it, until you reach the center where everything balances out.

With denser objects, you can simply get closer to the center before breaking the surface.

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G is a constant. g varies

Yes attraction drops a 1/r^2 till you reach the surface - and then assuming a sphere with uniform density drops at 1/r till the centre

edit

this is confusing - will re-write

At a great distance r is large and r^2 is huge - the gravitational attraction grows as r decreases. The highest gravity you will experience is on the surface. After you pass r = r_earth - ie the surface - then the volume of earth, that is the mass beneath you starts to decrease as well.

As you progress through the earth the attraction will decrease - the falling distance to the centre of mass means that 1/r^2 gets bigger (ie closer to one) BUT the amount of mass (ie the volume) varies with r^3. So as r decreases the r^2 in the denominator changes - but not as quickly as the mass ie r^3 in the numerator. So gravitational attraction falls linearly with distance below the surface

Right, thanks.

And this is where density comes in. Because the relationship between distance from the center and the strength of gravity that we use above the surface breaks once you go past it. As you go down, more and more of the mass is "above" you, no longer contributing to the gravity you feel underfoot, and in fact counteracting it, until you reach the center where everything balances out.

With denser objects, you can simply get closer to the center before breaking the surface.

Yep. Got it. Ta. MTW Gravitation is on my to-do-before-I-die list. Just got to beef up my maths... a touch.

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• 1 month later...

The Eye of the Needle was a low, narrow gate to the city. The servant and service entrance. Camels had to bow their heads and have packs removed to enter. Most camels were left outside. Most of the packs are taken in by the humble ass.

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