Newbie
Then how does the moon know that the earth has an iron core and not a marshmallow core? That is, if gravity is all about distorting space, why does a more dense object with radius X exert more gravitational pull than a less dense object with the same radius. I'm thinking Einstein here and not quantum physics.
Established Member
An object's gravitational pull is proportional to its mass, not its volume. Why that is, I'm not sure anyone knows.
Banned
Hmm?Originally Posted by nauthiz
Inverse Square (third time!)
If the Earth were to turn into a Black Hole, and you stood (stood? Whatever. Bear with me) at the line of where the Surface of the Earth was before it turned into a black hole- You would feel 1 G.
The Moon would continue to feel the Same Pull of gravity from the Earth as it did before too.
However, if you move toward the New Surface of the Earth, the gravity you feel will increase the closer you get to the new surface.
As far as the OP- A dense object contains more matter- more mass- then a not dense object.
Established Member
OK, fine, if you want to be pedantic about it: The magnitude of a massive object's gravitational pull on a second, distant massive object is proportional to the product of the masses of the two objects and inversely proportional to the square of the distance between the two objects.Hmm?
Inverse Square (third time!)
Banned
I don't study Tick Feet.
Order of Kilopi
Because each individual massive particle that Earth is composed of has its own gravitational field, and the Earth's field is simply the sum of all of those tiny fields. A marshmallow core of the same size would have fewer, lighter massive particles and the result would be a much smaller gravitational field. Density only controls how spread out those masses are, and close to the center you can get before running into the planet/moon/star. Density affects surface gravity...but only because the distance of the surface from the center changes.
Banned
Oh by the way- I wasn't clear. When I quoted you- the only response to you was the [ Hmmm?]
The rest was response to the OP.
Cjameshuff- I gotta learn the pretty words like you. You said pretty much the same thing I did- only much better
Established Member
If gravity were a geometric property of the Higgs (or some similar) field as opposed to space-time, how could we tell?
Established Member
BONG! A new winner of the occasionally granted worst pun of the week award. Proposed award, which has received no support, is an honorary banning for an infintisimally small time interval.I don't study Tick Feet.
OP, the answers so far have been good. It is mass, not density or volume, that determines gravitational attraction. And the mass can be considered to be located as a point at the center of the object.
Established Member
The Higgs boson doesn't offer any explanation of gravity, just mass. Since we don't have a working model of quantum gravity, all Higgs really offers with respect to gravity is another way to work out how much mass to plug into relativity's equations.
That said, in a case where there are two competing theories of gravity the only way we'd be able to tell which one is better is by running experiments to see which one more closely matches what actually happens in nature.
Newbie
My background is in literature so I'm no genius. However, it seems that you geniuses are doing nothing more than stating the alternative theory (the "quantum" theory) of how gravity works, reciting definitions, etc. I must have phrased my question in too simplistic a manner, which I figure is how the universe wants me to phrase questions and answers. Really, a bowling-ball-sized balloon should bend space in the same way a bowling ball does. And in fact these two objects pull themselves to the earth at exactly the same speed/velocity/whatever.
Order of Kilopi
Can I be a genius with you guys?
The reason the balloon and the bowling ball are pulled at similar rates is because the earth's influence overwhelms their own. Otherwise, the bowling ball "bends space" a hundred times greater than the balloon.
Order of Kilopi
Acceleration. The acceleration of an object toward another due to gravity depends only on the mass of that other object. This is because:
F = G*m1*m2/r^2
And:
A = F/m
So:
A1 = G*m1*m2/r^2/m1 = G*m2/r^2
and:
A2 = G*m1*m2/r^2/m2 = G*m1/r^2
Since both the balloon and bowling ball are so small, the Earth's acceleration toward them is unnoticeable, leaving their acceleration toward Earth...which is identical. Their gravitational fields are not, however...the bowling ball has a much stronger field, proportional to its mass. If you put the balloon in a vacuum chamber and reduced the pressure in the chamber so the balloon swelled to double the size, its gravitational field would remain exactly the same...the balloon would have the same number of particles with the same total mass, and their individual gravitational fields would sum up to the same overall field. They would simply be spread out over more space.
Moderator
"If gravity is an artifact of the geometry of space"
It's not the geometry of space, but of space-time. The two are very different concepts, and I'm unable to accurately explain the latter. But that is where I feel your misunderstanding comes from.
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Order of Kilopi
How do you reach this conclusion?
Everyone is entitled to his own opinion, but not his own facts.
Banned
Good question.
In other words he said, "Don't confuse me with facts, math or definitions, I've invented my own Common Sense Idea."
Established Member
In the same way, yes. To the same extent, no.
Think of the rubber sheet analogy: If you put a bowling ball or a balloon on a sheet of rubber, both will make the sheet sag down a little bit. But the bowling ball will make it sag a lot more than the balloon.
Banned
The way he said it implied that he thinks it should warp it by the same amount.
This is really when the Rubber Sheet analogy fails spectacularly and needs to be abandoned.
Because the warping of spacetime occurs at all levels- not just at a surface like the 2 D sheet.
Established Member
Maybe, but for an analogy that appeals to everyday experience, you're not going to get much better than the rubber sheet. A two-dimensional surface deforming into a third dimension is a lot easier to visualize than a four-dimensional manifold.
The rubber sheet at least gets the essential concepts: The space in question isn't flat, more massive objects warp it more than less massive objects, and objects in motion will follow geodesics across the sheet's surface rather than moving in straight lines.
Order of Kilopi
It works alright if you restrict yourself to point sized masses, and only use it as an illustration of how a distortion of spacetime leads to objects moving together. Unfortunately, it very often seems misunderstood and stated as gravity pulling the objects down, and then pulling other objects "downhill" on the sheet, when the point is that distances along the sheet surface are distorted.
It also works for a visualization of the gravitational field, as long as you keep in mind that it is just that, not an actual elastic sheet with objects sitting on it, and that the field of each of those objects is the sum of the fields of its parts. It needs to be computed by machine to be useful for this, though...human ideas about the tension of a sheet will lead you astray. For example, a ring placed on a real rubber sheet will stretch it flat across the bottom, when a real ring does not have a flat internal gravitational field...you need a hollow spherical shell for that.
Zayfart: Given a spherical mass, the gravitational field outside the sphere does not depend on its size, only its mass...increasing the size without changing the mass will bring some parts closer to a given test point, but move other parts away, and the total influence at that point remains the same.
Banned
I understand When it works- I was saying that this is not one of those times.
Established Member
On the other hand, wouldn't the 2-dimensional equivalent of a hollow spherical shell would be a ring?
Order of Kilopi
And the 2D equivalent of a gravitational field wouldn't be 1/r^2, it would be 1/r. (while sphere area is proportional to the square of the radius, circle circumference is proportional to radius)
It's far more useful to consider a 2D slice of a 3D system. The dynamics are then what you would observe between bodies moving in a plane, which is roughly the case for most situations (and always the case for 2-body interactions).
Banned
Why are you of that opinion? We know from thousands upon thousands of experiments that it's mass that bends space, not size.
No. Not only do these objects not pull themselves to the Earth (they're pulled by the Earth), but the "speed/velocity/whatever" for which you're searching is called "acceleration," and since that's proportional to the product of the two masses, that acceleration is different, not the same.And in fact these two objects pull themselves to the earth at exactly the same speed/velocity/whatever.
Established Member
The force is proportional to the product of the two masses, so they would be different. But since inertial mass is identical to gravitational mass, both objects experience the same acceleration. If we were ignoring other forces such as air resistance, that is.
cjameshuff ran through the equations to show it up above.
Banned
No. Actually, yes, but only if both smaller masses (the feather and the hammer) are simultaneously present. If it's just one or the other, since the force between the feather and the Earth and that between the hammer and the Earth, are different.
Proportional to their masses, you say?
Not quite! It's proportional to the product of each mass with the Earth, which yield different acceleration than would be yielded if the feather and the hammer were dropped together.
So, no, the acceleration is not the same. It is, however, so close to being the same that, for all intents and purposes, it might as well be the same.
Order of Kilopi
As I wrote in that post, the acceleration is proportional to the product of their mass with that of the Earth, divided by their mass...and thus dependent only on the mass of the Earth. The gravitational acceleration of one body toward another depends only on the mass and distance of that other object.
The Earth will be accelerated in the other direction by an amount proportional to the mass of the object being dropped, and so the Earth and a massive object will meet more quickly than the Earth and a low-mass object, but the falling objects still accelerate at the same rate. The difference is entirely in the acceleration of the Earth in the opposite direction.
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