Gravity...

[Billy T]

River said: "There are many that say the Earths gravity could not possibly capture the Moon" You replied:

Who are the many?

All who know a little a classical mechanics. In a two body interaction, (bodies E & M, for example) if M is initially not bound to E then it never will be. "Not bound" means M had wrt E, more than the escape velocity, or net positive energy which will be conserved.

If E has an atmosphere, which M passes thru, then the "net positive energy" can be reduced by the drag forces acting on M. If this reduction is enough to make the net energy negative, then E has captured M into a bound orbit.*

If there is a third body, T, then their dynamic gravitational interaction can transfer some of the M's net positive energy to the other two, usually mainly to T if E is much more massive that T. I.e. M can be captured in a three body interaction. T may or may not be still bound to E.

* in the case of earth, there is considerable energy in the earth's rotation and it is decreasing with the moon gaining what earth has lost. Thus an initially bound moon could conceptually become free - unbound, but in our case the moon is just moving farther away - climbing up a gravitational hill. The energy transfer rate is slowing down, with the net result that the moon will not escape, unless solar wind or interaction with Jupiter etc. helps it.

BTW, I think it is unlikely that there was some T bound to M that passed by E with a three body interaction that left M bound to E while T is now orbiting the sun. I. e. I think some object hit the earth and basted many things (a lot of them self bound liquids initially) many of which did collect into the moon with a few, perhaps, escaping to be now, near-earth orbit asteroids.

 

[Janus58]

Here on Earth you are always attracted to the center of the Earth, but most of the Earth's mass is not located at the center.

That is because the vast bulk of the Earth's mass is closer to the center than you are. If, however, you were to to dig down to few hundred miles below the surface, the pull would not be as strong. At the center, it would be zero. The gravity you feel on the surface is the sum of all the parts of the earth pulling on you. Some of it is pulling on you sideways, but since all these sideways pulls are more or less equal, they cancel out. You only feel the net "downward" components of all those pulls.

Or let's take another example. Let's say that the Earth has taken the from of a hollow shell with an equal mass to it present mass. To someone some distance from the Earth or standing on the surface, it would make no difference. they would still feel the same pull as before towards the center of the Earth. But someone in the interior of the shell would feel no gravity at all no matter where he was. At all points inside the shell the individual pulls from all the parts of the shell cancel each other out. Newton proved this with what is now called his Shell theorem. (The above assumes a uniform density of our shell).

So as far as your question about the black hole at the center of the galaxy goes, no, it is not, nor could it be due to the sum of the mass of the galaxy surrounding it.

 

[Billy T]

30 minute edit time limit stopped me from adding to pot 21:

I also think it is even less probable that some space object we now call the moon just skimmed thru the earth's atmosphere and was captured.

AS always, Janus58 is correct, but proof of Shell theorem is so easy I'll indicate how it goes: Consider a thin shell (one of many each of which is by this one shown to have same external or internal effect).
First pick any interior point and from it make a cone (which is two of what most think a cone is that share a common apex, that I'll call C & D. as cone C's apex is Closer to the shell than cone D's is). Then note there is more of the uniform density shell's mass intercepted by cone D than by Cone C but it is farther away from the point you pick - the R^2 of the greater mass is exactly cancelled by the R^-2 of the separation, so no net force on any interior point.

Now for any exterior point, with the cone's axis passing thru the center of the shell, the one of the two cones with common apex cuts thru the shell in two places. The more distant intercept has exactly the extra mass to compensate for its greater distance - I.e. both contribute the same gravitational force to the external point chosen. Only a little algebra to show that twice the intercepted mass if at the center of the shell makes the identical external attraction.

 

[Billy T]

Again the 30 minute limit stopped me. It is total intercepted mass not "twice" I needed to say.

Also note real mass distributions are not uniform in density. Thus an object in orbit not far from the surface will not be elliptical, but with "bumps." That is how we know that California has been getting about half the water it uses from the wells and why low moon orbit will crash into the surface (Bump amplitude greater than orbit altitude, certainly does that, but the accumulated effects of mass concentration can crash even higher altitude orbit objects.)

 

[krash661]

About the Moon :

There are many that say the Earths gravity could not possibly capture the Moon . So using the Moon as an example of gravity is mistaken .

explain how it is there then. it's that simple.

 

[krash661]

And there are many who believe that summer is caused by Earth being closer to the Sun. Doesn't mean it's true - it just means that ignorance is popular.

It is an excellent example, since the Earth's gravity affects the Moon (tidal lock, defines its orbit) and the Moon's gravity affects the Earth (tides.)

top surface of earth is in a tilt proportional to the sun.[ top is closer ]. hence, summer when the portions lineup, hence winter when they don't.

 

[krash661]

I am trying to figure out if a blackhole could be created by all the surrounding masses gravity, it would be the point where all the surrounding masses gravity converge. Like a hybrid point where things would be attracted to the common center of all the mass. Example, all the mass in our galaxy is creating a blackhole in the center of our galaxy by the convergence of all the gravity fields from the surrounding mass.

I suspect the idea is flawed, because the hybrid point would actually contain no gravity, because you would be pulled in all directions at once and the gross effect would be canceled out. I've read people say that at the center of the Earth you would feel no gravity, because you would be pulled in all directions at once.

black holes can be looked at as a rip in a dimensional plane to move energy to create equilibrium within a system[also tho, a system is so chaotic that it's not the same in every moment]. just a thought tho. it may be perceived as a singularity due to transformation from one system to another. energy is being built up and condensed on one side as it can not push through at a fast enough rate[a delay due to phasing and such], while being stretched into another end of another system. in this system it will appear as a singularity[from a threshold point]. some believe this is the birth of the big bang, only on the other side of this concept.

 

[krash661]

The point where all the fields converge would be dense in gravitational fields and is the point where you would be attracted to.


I am aware of the normally explanation. Its just an idea I am curious about.

fields are nothing more than a fluctuation state.

 

[krash661]

Here on Earth you are always attracted to the center of the Earth, but most of the Earth's mass is not located at the center.

correct, it's derived from the center.

 

[Quantum Quack]

At the center, it would be zero. The gravity you feel on the surface is the sum of all the parts of the earth pulling on you.

and the sum of all parts of all those individual particles would also be zero gravity and so on until you find by applying reductio ad absurdum, the source of gravity is actually zero (nothing)!! ( chuckle)

 

[Russ_Watters]

Here on Earth you are always attracted to the center of the Earth, but most of the Earth's mass is not located at the center.

In some cases it can be easier to think of the fact that you are attracted to all particles in the universe, separately. But when you add together all the forces from all the particles that make-up the earth, the sum of all of those vectors points toward the center. That should be obvious, given that when you face straight down, there are equal amounts of "earth" to your left and right.

Applying that at greater distances, it is easy to see then that when you are between the earth and moon, they pull you in both directions and the net force could be toward either, depending on your distance. Much further away, you are still toward both, but the net is toward the center of mass of the system.

 

[James R]

river:

About the Moon :

There are many that say the Earths gravity could not possibly capture the Moon . So using the Moon as an example of gravity is mistaken .

Please provide a link or citation to one of the "many" who say this. You have been asked several times.

 

[BdS]

Let's say that the Earth has taken the from of a hollow shell with an equal mass to it present mass.

That’s a nice example. When we say you would feel no gravity/force in the center it does not mean that there is no force on you. Your mass is still experiencing a force, but is being pulled in all directions to make it feel like there is no gravity or force moving you in a particular direction.

What will the effects of that be to me (my mass) though, since im being pulled in all directions?
A. Will it cause my mass to compress (pressurise)
B. Will it be the opposite and cause my mass to inflate (negative pressure)
C. Nothing will happen to my mass
D. or?

At all points inside the shell the individual pulls from all the parts of the shell cancel each other out.
But someone in the interior of the shell would feel no gravity at all no matter where he was.

If I was inside the shell and I was not in the middle, I am located more to the right of the interior, will I

A. be attracted to the right side of the shell?
B. be attracted back to the middle of the shell?
C. just stay where I am?
D. or?

By what you saying I will answer C, because Im closer to the center than all the other mass? Would I be correct?

 

[BdS]

If we had a uniform density shell where ever I moved to on the inside of the shell I will be moving the center of mass where things on the outside would be attracted to. Would I be forcing things to move to the new center with only my masses amount of gravity or my mass plus the whole shells mass?

 

[James R]

BdS:

That’s a nice example. When we say you would feel no gravity/force in the center it does not mean that there is no force on you. Your mass is still experiencing a force, but is being pulled in all directions to make it feel like there is no gravity or force moving you in a particular direction.

There's no net force.

What will the effects of that be to me (my mass) though, since im being pulled in all directions?
A. Will it cause my mass to compress (pressurise)
B. Will it be the opposite and cause my mass to inflate (negative pressure)
C. Nothing will happen to my mass
D. or?

Nothing will happen, because there's no net force on any of your mass.

If I was inside the shell and I was not in the middle, I am located more to the right of the interior, will I

A. be attracted to the right side of the shell?
B. be attracted back to the middle of the shell?
C. just stay where I am?

You'll stay where you are, because there's no net force on any mass inside the shell (assuming it is spherical).

If we had a uniform density shell where ever I moved to on the inside of the shell I will be moving the center of mass where things on the outside would be attracted to. Would I be forcing things to move to the new center with only my masses amount of gravity or my mass plus the whole shells mass?

Things outside the shell feel a force that is the linear combination of the force from the shell's entire mass (acting as if from a single point at the shell's centre) and the force from any other mass that is inside the shell (acting from the location of that mass).

 

[Billy T]

... There's no net force. Nothing will happen, because there's no net force on any of your mass.
You'll stay where you are, because there's no net force on any mass inside the shell (assuming it is spherical). ...

Perhaps I'm wrong but lets consider the mass inside is a small piece of a neutron star with mass 10^6Kg and it is only 1 cm from point "a" on the inside of the uniform shell, which has a mass total of only 1Kg and kilometer radius.

I think the part of the shell near the huge small mass would deform and touch the big mass. Even if that is "cheating" as shell is assumed to be "infinitely rigid" are you sure the much greater local "warping of space" by the huge mass does not make net attraction of point a to the huge mass? I.e. the inverse square law is only an approximation of the gravity field.

If point a does move into contact, then so does the "big mass" but only by about 10^(-6) cm.

 

[BdS]

Nothing will happen, because there's no net force on any of your mass.

I asked that question because at the center of the Earth there is 0 gravity but the pressure is greatest at the center. On the inside of the Earth you have matter that is applying the pressure directly on you by physical contact pressing on your mass. I was thinking is that the case though, that the pressure on your mass has to be caused by matter in contact with your mass, or could it cause that pressure on you with just field interaction. The most logical is that physical contact would be required to cause pressure. You should be able to prove this by going in a mine shaft and the pressure will be greater, but how much greater? is the increased pressure proportional to just the atmospheres weight above you or the atmosphere and the solid mass located at a radius above yours? If I take a guess, just the malleable mass above you which in this case is the air/atmosphere.

It is interesting that gravity and pressure scale opposite each other inside the Earth. The lower the gravitation potential the higher the pressure.

 

[Quantum Quack]

Interesting!
So if you sunk a hypothetical shaft all the way to the core's center, I wonder what air pressure and what gravity you would experience?
ignoring real stuff like heat etc

 

[JJM]

#37, so as pressure approaches 1, then gravity potential approaches 0. are they corollary to each other? the potential is translated into higher pressure allow ability. at the point of equal density then gravity and pressure equals one. ergo both exhibit the potential.......since compressed force as F*2*2 is compressed compression pressure density as gravity density then it is also its' own potential.

 

[Billy T]

Interesting!
So if you sunk a hypothetical shaft all the way to the core's center, I wonder what air pressure and what gravity you would experience?
ignoring real stuff like heat etc

Assuming that air flowed down to drilling surface as hole was made and that convection (or at least conduction) still happens to warmer air so that the heat of air compression gets up to the surface, not much trouble to calculate the air density, Provided it does not begin to liquefy any components, like argon.

It is interesting to note that central part of the core is solid due to the high pressure, despite the high temperature. I doubt that either O2 or N2 would become solid, but perhaps become liquids, especially if they "keep cool" = 300K via convective heat transfer, as you suggested we could ignore the local heat outside the tube to center.