Gravity...

[Janus58]

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

The gravity would be zero, and the air pressure extremely high. There are two competing factors at work. As you increase the depth you are adding to the column of air above you. Since the weight of the upper air rests on the lower air and compresses it, the deeper you go the more mass cm^3 the air will have at that depth. ( we get the same effect with our normal atmosphere. 99.99 percent of it is under an altitude of 100 km, but it would only have to be 8.5 km keep if it were at a uniform density.) Moderating that some will be the fact that local gravity will go down at the same time, so that same cc of air will weigh less than air of the same density would at the center. The end result would be a very high pressure at the bottom of the shaft where the air is supporting all the weight of the air above it.

 

[Quantum Quack]

So hypothetically one could be in a weightless environment yet exist in an atmosphere ( abet high pressure) at the bottom of the shaft... fascinating conceptually...
Reminds me of the floating islands in the Sci fi movie AVATAR

hee hee -fun video

This leads to the notion that gravity is an inverse "*force" that being the closer one gets to the source the weaker it gets.
As distinct from magnetic "force" which could be considered as a converse.

 

[Daecon]

I think it's more a case of instead of gravity pulling you "down", it pulls you in every direction simultaneously, therefore gravity in each direction is cancelled out by the exact same amount of gravity in the opposite direction. If that makes sense?

 

[Quantum Quack]

I think it's more a case of instead of gravity pulling you "down", it pulls you in every direction simultaneously, therefore gravity in each direction is cancelled out by the exact same amount of gravity in the opposite direction. If that makes sense?

If one subscribes to conventional understanding of gravity yes I would agree...

 

[BdS]

ignoring real stuff like heat etc

well now that you got me thinking about heat...

Is the constantly applied forced pressure what creates the heat inside the Earth and maybe even the sun? Gravitational induced pressure is maybe not like contained pressure in a cylinder. Cylinder pressure is forced once and heat generated during pressurising and then contained and no more heat is generated, where as gravitational pressure could be a constantly applied/forced pressure that is constantly generating heat.

 

[origin]

well now that you got me thinking about heat...

Is the constantly applied forced pressure what creates the heat inside the Earth and maybe even the sun? Gravitational induced pressure is maybe not like contained pressure in a cylinder. Cylinder pressure is forced once and heat generated during pressurising and then contained and no more heat is generated, where as gravitational pressure could be a constantly applied/forced pressure that is constantly generating heat.

When something is compressed it will generate heat, such as the air in a cylinder as you said. Once the material is compressed it will no longer generate heat. So if you compress air in a cylinder it will heat up, if you maintain the pressure no further heat will be produced and the air in the cylinder will cool to the ambient temperature.

That being said the pressure from gravity is vital to the mechanics of the sun. The heat of fusion at the core of the sun 'tries' to push the star apart but the pressure from gravity prevents the that. Look up star formation for more information.

 

[Billy T]

... This leads to the notion that gravity is an inverse "*force" that being the closer one gets to the source the weaker it gets. As distinct from magnetic "force" which could be considered as a converse.

no. It is a balanced to total of zero force. Same as on a cable between two identical locomotives pulling on the opposite ends of the cable equally.

 

[origin]

This leads to the notion that gravity is an inverse "*force" that being the closer one gets to the source the weaker it gets.

The same sort of thing occurs with an electric field. In the center of a charged spherical insulator the electric field will be zero.

 

[BdS]

no. It is a balanced to total of zero force. Same as on a cable between two identical locomotives pulling on the opposite ends of the cable equally.

To the whole system it is balance to zero force, but not to the cable. The cable is still experiencing a force and the force is being consumed by the bound atoms holding themselves together in the cable. If the locomotives pull harder the cable will snap and then it becomes clear where the force was being consumed.

 

[Billy T]

To the whole system it is balance to zero force, but not to the cable. The cable is still experiencing a force and the force is being consumed by the bound atoms holding themselves together in the cable. If the locomotives pull harder the cable will snap and then it becomes clear where the force was being consumed.

One does not consume force, one applies it. If a finite size object's mass center is at the Barycenter it, like the cable between two locomotives, will experience forces trying to pull it into parts.

In fact, if it is not ripped apart, a finite size, stationry object will "fall towards" one of the two masses making the Barycenter, with the greater mass as its gradient is stronger there. I'm not certain but think a relatively small mass can not even orbit a large number of orbits about the Barycenter - the larger one will "steal it" eventually I think.

 

[Daecon]

One does not consume force, one applies it. If a finite size object's mass center is at the Barycenter it, like the cable between two locomotives, will experience forces trying to pull it into parts.

In fact a finite size, stationry object will "fall towards" one of the two masses making the Barycenter, with the greater mass as its gradient is stronger there. I'm not certain but think a relatively small mass can not even orbit a large number of orbits about the Barycenter - the larger one will "steal it" eventually I think.

I think the larger object will still fall towards the smaller of the two, just not as much as the smaller falls towards the larger.

The barycenter between any two objects isn't going to be in the exact middle of the larger object, but ever-so-slightly off to the side that is closer to the smaller object, while still inside the mass of the larger. So both objects will still fall towards the barycenter of the two.

At least, that's how I understand it. If I'm wrong I'm sure someone can correct me.

 

[Billy T]

I think the larger object will still fall towards the smaller of the two, just not as much as the smaller falls towards the larger.

The barycenter between any two objects isn't going to be in the exact middle of the larger object, but ever-so-slightly off to the side that is closer to the smaller object, while still inside the mass of the larger. So both objects will still fall towards the barycenter of the two.

At least, that's how I understand it. If I'm wrong I'm sure someone can correct me.

No. Both objects making the barycenter are orbiting about it. Neither is falling in the common sense of that term towards the other. Both are accelerating towards the other. For simplicity, I'll assume both have circular orbits about the Barycenter.
Unless one is much larger mass than the other, the Barycenter is inside neither. It is by definition that point along the line between their mass centers where the gravitational pulls are equal and opposite. I. e. M1/ (R1)^2 = M2/(R2)^2 = B (f0r gravity at Barycenter.) As the gradient falls off as R^-3 , so if R1 > R2 we see:
B/R1 < B/R2 so the gradient from the small mass (R1 from the barycenter) will be less as I stated earlier.

 

[Daecon]

Sorry, of course you're right.

I was thinking of stationary objects like in post #50, not orbiting bodies.

 

[BdS]

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.

I drew a quick 2d example.


A - BCDE (blue lines) 90 deg cone from your point inside the shell creating 4 quadrants inside shell mass Z
A - FHIG (Grey lines) direction the gravitational force will attract you to that point (FGHI) in the respective quadrants. Its not center of mass for that quadrant, its the balance point in that quadrant for the shell mass and average distance from quadrants mass that you will be attracted to. right?

While I was drawing Dig. B and was trying to find the point you will be attracted to in quadrant ACD, I was trying to set the force line AH to intercept halfway between CD. I quickly realised that was incorrect. If you have a 90 deg cone you will always be attracted to the 45 deg point in that quadrant. The 45 deg force line automatically finds the balance point for you in that quadrant, is that correct?

 

[Billy T]

... If you have a 90 deg cone you will always be attracted to the 45 deg point in that quadrant. The 45 deg force line automatically finds the balance point for you in that quadrant, is that correct?

I think so. The mass on the arc I to B (and that on arc C to H) contribute to the force in the direction of F and by symmetry cancel each other out along the I,H line. Likewise by same argument the other two arc parts of those quadrants only make gravity for pulling your point G towards F.