Evidence for a Young Earth (Not Billions of Years Old)

[Subduction Zone]

No because the mass causing the potential decreases as r squared.



And the mass towards the center decreases, therefore the potential is still highest at the surface where there is greatest mass.

-2 and -4 are both negatives. It is only you that thinks -2 instead of -4 would mean it is a positive and must point away from the center.....




It is zero. That’s why an object will feel no force, because it has no potential to move in any direction.....

You can’t tell me an object moves towards earth at an increasing velocity because the potential becomes more negative, then tell me it won’t move anywhere because it is more negative. That would imply an infinite negativity....



Ok, fine, we will go with fantasies..... the force still decreases as mass decreases. You do understand you experience less force on the moon than earth because the moon has less mass. So which has the greatest potential. The object with more mass or less?



No, there is only needed an increasing force as you go upwards to the maximum at the surface. But you need not start with the same force at the center as you do at the surface. Just increase it with increasing mass below you.....



No, the net force at the center is zero and only increases with distance from the center. We were not discussing how much fuel you would need to expend due to the increasing distance and force. But the amount of force needed to keep moving and not fall back. Don’t try to change the subject into total energy expended versus the force required to keep moving....



And since the gravitational field is weaker below the surface due to less mass, less thrust and velocity is required initially, only needed to increase to the maximum at the surface.

As I said above. Don’t try to change it into Tatar energy required due to increasing distance.....



It’s not irrelevant. It’s science that escape velocity need not be obtained with sufficient fuel and propulsion. You just don’t want to hear it....




Since we can arbitrarily set the zero point anywhere, then the result is just a mere coordinates choice, as is the answer......

Kind of relative.....



SR on all three....





We typically don’t ever need it except for things in the solar system under unique conditions....



I’m not the one that jumped from a discussion on time dilation (age of earth) to talk of orbital mechanics.

I’ve been trying to keep it on track despite all of your diversionary attempts to sidetrack the issue.....

Okay, so math is a concept that we need to work on a bit. You do realize that you just said that material should come up freely from a well, don't you? In fact by your "logic" an object within a well will "fall" up.

 

[Polymath257]

No because the mass causing the potential decreases as r squared.

Actually, r cubed: volume times density.

And the mass towards the center decreases, therefore the potential is still highest at the surface where there is greatest mass.

And higher still once you get off the surface. It continues to increase *up to zero* as we move farther away from the center.

-2 and -4 are both negatives. It is only you that thinks -2 instead of -4 would mean it is a positive and must point away from the center.....

Let me put it like this. If the potential at 'infinity' (just means a distant observer) is 0, then the potential at the surface is, say, -4. Then the potential at the center will be something like -6. The potential is never positive.

It is zero. That’s why an object will feel no force, because it has no potential to move in any direction.....

Once again, you are confusing force with potential and both with escape velocity.

You can’t tell me an object moves towards earth at an increasing velocity because the potential becomes more negative, then tell me it won’t move anywhere because it is more negative. That would imply an infinite negativity....

If something falls from a great distance, it's velocity at any point will be the escape velocity of that point (this is almost the definition). Now, it accelerates down to the surface. And then, from the surface, it would accelerate more, so the escape velocity at the center is MORE than that at the surface. It is certainly *not* zero, although the force is zero there.

Ok, fine, we will go with fantasies..... the force still decreases as mass decreases. You do understand you experience less force on the moon than earth because the moon has less mass. So which has the greatest potential. The object with more mass or less?

Why change to the moon? yes, the force is smaller from an object with less mass. Now, what does that say about the force/potential/escape velocity inside a spherical body?

No, there is only needed an increasing force as you go upwards to the maximum at the surface. But you need not start with the same force at the center as you do at the surface. Just increase it with increasing mass below you.....

This only shows you don't understand what the escape velocity is. Wow.

No, the net force at the center is zero and only increases with distance from the center.

Up until the boundary with the core. Then, according to the graph you gave, the force decreases. it decreases faster once we get away from the surface.

We were not discussing how much fuel you would need to expend due to the increasing distance and force. But the amount of force needed to keep moving and not fall back. Don’t try to change the subject into total energy expended versus the force required to keep moving....

No. We were discussing escape velocity, the gravitational potential, the gravitational force, and the curvature.

Here's a couple of quick questions:

1. How does the potential relate to the escape velocity? (Easy answer, although perhaps unexpected)

2. How does the force relate to the potential. (Again, an easy answer. Assuming you know calculus and basic mechanics).

3. In a spherically symmetric body, how does the amount of mass closer to the center relate to the force at a point? (Yet another easy answer)

And since the gravitational field is weaker below the surface due to less mass, less thrust and velocity is required initially, only needed to increase to the maximum at the surface.

Irrelevant for the escape velocity. You are determining how much force is required to counteract gravity. That is very different than the escape velocity.

As I said above. Don’t try to change it into Tatar energy required due to increasing distance.....

Hmm.....do you know that the escape velocity is directly related to that total energy?

It’s not irrelevant. It’s science that escape velocity need not be obtained with sufficient fuel and propulsion. You just don’t want to hear it....

Right. The escape velocity need not ever be reached if a force is always provided that is just more than the force of gravity at each point.

But we were asking for the escape velocity, not that force. We were also asking about the gravitational potential, which is yet another thing.

Since we can arbitrarily set the zero point anywhere, then the result is just a mere coordinates choice, as is the answer......

Kind of relative.....

And the default is to set the potential to be zero at infinity. Yes, we can declare it to be zero at the center, but then the potential will always be positive and still increasing as we move away from the center. You can't choose both the center and infinity to have zero potential.

SR on all three....

Wrong. Time dilation starts with SR. Universal expansion and universal acceleration are both GR effects.

We typically don’t ever need it except for things in the solar system under unique conditions....

Right. But it was *you* who was claiming we have to worry about acceleration due to universal expansion. Now you are claiming we don't. if you do, then you are forced into GR.

I’m not the one that jumped from a discussion on time dilation (age of earth) to talk of orbital mechanics.

I’ve been trying to keep it on track despite all of your diversionary attempts to sidetrack the issue.....

We were talking about time dilation. That meant considering curvature and thereby the potential, which lead into classical mechanics.

Here's a fun one:

Q: As observed by a distant observer, clocks are slowed as you get closer to a gravitating object. That slowing corresponds to the time dilation for a certain velocity. Which velocity? @Subduction Zone, you might like this one.

 

[Justatruthseeker]

The graph I commented on pretty clearly graphs acceleration (force divided by mass) with radius. This was the graph *you* originally gave. it is NOT a graph of the potential.

Now, the graph @Subduction Zone gave *is* a graph of the potential, with a minimum at the center.

/E: In fact, based on *your* graph of the acceleration, the inflection for the potential (which happens at the maximum force) is about at the boundary of the outer core.

Since I first posted that graph before the other......

 

[Justatruthseeker]

Okay, so math is a concept that we need to work on a bit. You do realize that you just said that material should come up freely from a well, don't you? In fact by your "logic" an object within a well will "fall" up.

And hence your all’s avoidance of discussing density.

I asked you all if you wanted to start with seismic reflection interpretation, earth and space rocks, or experiment.

You all avoided.....

 

[Polymath257]

Since I first posted that graph before the other......

I'm not following you. Since you posted it before the other......what? You should have got the identification correct? You should know it was the acceleration and not the potential?

 

[Polymath257]

And hence your all’s avoidance of discussing density.

I've been pretty clear about density and how it relates to this material. It is density that determines the potential. It is density that appears in Einstein's equations for gravity.

I asked you all if you wanted to start with seismic reflection interpretation, earth and space rocks, or experiment.

You all avoided.....

The question is whether the potential is zero at the center of the Earth (it isn't under standard definitions). Another question is whether it is a maximum at the surface (it isn't--it continues to increase above the surface). Yet another question is whether the curvature is zero when the force is zero. The answer is no. The relation between force and curvature is more complicated than that.

 

[Justatruthseeker]

Actually, r cubed: volume times density.

Correct, was thinking of the force.

And higher still once you get off the surface. It continues to increase *up to zero* as we move farther away from the center.

Exactly, so correct your comrade in arms contention that it increases towards the center..... I’m not the one confused on this point....

Let me put it like this. If the potential at 'infinity' (just means a distant observer) is 0, then the potential at the surface is, say, -4. Then the potential at the center will be something like -6. The potential is never positive.

And -6 is less than -4?????? Of course, it is why you recognize it increases to zero at infinity.....

Once again, you are confusing force with potential and both with escape velocity.

I’m not confusing anything. If there is no force there is absolutely no potential..... the ball has no potential to go anywhere......

If something falls from a great distance, it's velocity at any point will be the escape velocity of that point (this is almost the definition).

which will be less increasing to the maximum at the surface....

Now, it accelerates down to the surface. And then, from the surface, it would accelerate more, so the escape velocity at the center is MORE than that at the surface. It is certainly *not* zero, although the force is zero there.

No. At the surface it’s acceleration rate would decrease in proportion as the mass decreases. If it was accelerating at a value of 10 at the surface it would then continue to accelerate (minus air drag) at 9 then 8, 7, 6, etc until its acceleration was zero at the center. You are confusing continued increasing velocity with continued increasing acceleration. Velocity can continue to increase as acceleration decreases.

Just as in space I can accelerate at 10g, reduce my acceleration to 9G, 8G 7G etc and my velocity will continue to increase even as my acceleration decreases.....

Then as it passed the center it would begin to feel the force of more and more mass beneath it until it reached the maximum once again at the surface. Except this time it would feel a increasing decelerating force.

Why change to the moon? yes, the force is smaller from an object with less mass. Now, what does that say about the force/potential/escape velocity inside a spherical body?

The mass becomes less as you near the center. What part didn’t you understand?????

This only shows you don't understand what the escape velocity is. Wow.

It shows only that you don’t understand, even if you understand the moons escape velocity/force is less because of less mass beneath you. Now apply what you understand as the mass beneath you decreases....

Up until the boundary with the core. Then, according to the graph you gave, the force decreases. it decreases faster once we get away from the surface.

And down past the core boundary it decreases to zero, while outward from the surface it only reaches zero at a speculative infinity...... so which really decreases faster? One that stays non zero to infinity or one that reaches zero in approximately 3,958.8 miles?????

But you forgot I stated based upon their flawed density profiles. I asked which you wanted to discuss first. Seismic reflection interpretation, earth and space rocks or experiment. You avoided...

No. We were discussing escape velocity, the gravitational potential, the gravitational force, and the curvature.

Here's a couple of quick questions:

1. How does the potential relate to the escape velocity? (Easy answer, although perhaps unexpected)

The potential goes up as the escape velocity requirements go down... or if you prefer the potential goes down as the escape velocity goes up.

But potential starts at infinity inwards. Escape velocity starts at the surface outwards.

2. How does the force relate to the potential. (Again, an easy answer. Assuming you know calculus and basic mechanics).

The potential goes down as the force goes up..... or if you prefer the potential goes up as the force goes down....

But potential starts at infinity, force starts at the center. In direct opposition to one another.

3. In a spherically symmetric body, how does the amount of mass closer to the center relate to the force at a point? (Yet another easy answer)

The force decreases as the mass decreases.....
As the potential decreases as the escape velocity decreases.... your “surface” is now smaller in diameter....

You have two choices. Either ignore the mass above you and treat the situation as a completely new planet, in which the old calculations do not apply. Or give up your shell theorem so you can still pretend the mass above you is valid to consider before you move above it. Which essentially creates a new planet and a new calculation.

Comprehend. Once below the surface the old calculations no longer apply. That mass no longer adds into the equation. But you are trying to reduce the size of the mass while keeping the old mass calculations and results. Conflict without you even realizing it..... you can’t use the old value data points.... they no longer enter the calculations as the mass above you no longer enters the calculations...

Irrelevant for the escape velocity. You are determining how much force is required to counteract gravity. That is very different than the escape velocity.

The escape velocity can be 1 mph with sufficient force....

“A rocket moving out of a gravity well does not actually need to attain escape velocity to escape, but could achieve the same result (escape) at any speed with a suitable mode of propulsion and sufficient propellant to provide the accelerating force on the object to escape.”

All that needs be achieved is the sufficient accelerating force.

Hmm.....do you know that the escape velocity is directly related to that total energy?

Calculated from the surface. Once below the surface the mass needed to escape from is less..... it increases to the maximum at the surface. You only require more energy because you have more distance. I could walk from the center and my weight would slowly increase and the force I expended would slowly increase to a maximum at the surface.....

Right. The escape velocity need not ever be reached if a force is always provided that is just more than the force of gravity at each point.

And the force of gravity decreases below the surface to zero at the center. So only at each point upwards is the force needed to be more and the escape velocity increases to the maximum at the surface....

But we were asking for the escape velocity, not that force. We were also asking about the gravitational potential, which is yet another thing.

You answered your own question. The force needed just needs to be more than the force of gravity at each point, which increases from the center to the maximum at the surface.....

And the default is to set the potential to be zero at infinity. Yes, we can declare it to be zero at the center, but then the potential will always be positive and still increasing as we move away from the center. You can't choose both the center and infinity to have zero potential.

And then it would decrease to zero at infinity, once it reached it’s maximum at the surface. It would then increase in proportion to force and decrease in proportion to force. Just as force remains positive but increases from zero at the center, reaches it’s maximum at the surface, then decreases again to zero at infinity.....

So what was your objection against two zero points again, we already use it with force.... are you saying you object to force having two zero points?????

Wrong. Time dilation starts with SR. Universal expansion and universal acceleration are both GR effects.

No, accelerated expansion occurs because gravity is too weak to stop it and so can be neglected. We once again are in the realm of SR with high velocities where the force of gravity can be neglected.....

Don’t kid yourself....

Right. But it was *you* who was claiming we have to worry about acceleration due to universal expansion. Now you are claiming we don't. if you do, then you are forced into GR.

No I’m not. Gravity doesn’t apply to expanding space. Gravity is what prevents space from expanding so we can’t measure it in the lab.... it can be ignored when discussing expansion.....

We were talking about time dilation. That meant considering curvature and thereby the potential, which lead into classical mechanics.

We were not discussing gravitational time dilation, but dilation due to velocity at high speeds where gravity can be neglected. SR..... you all keep erroneously bringing the others into the conversation....

Here's a fun one:

Q: As observed by a distant observer, clocks are slowed as you get closer to a gravitating object. That slowing corresponds to the time dilation for a certain velocity. Which velocity? @Subduction Zone, you might like this one.

The energy that the gravitating object imparts....

What the distant observer sees is irrelevant.

If the object had to slow down then his clocks sped up more than the gravity slowed them. It’s why GPS clocks run a tiny bit faster from distance from a gravitational force and slower still from its velocity.

So if you brought a GPS clock back to earth it would run slightly slower due to gravity, but speed up as it slowed down to surface speed. They run slower, they would speed up more to match earth clocks.

I understand you think the GPS clocks see earth clocks slower and so the reality probably confuses you. But once again. Earth clocks don’t slow because GPS clocks are launched into orbit. What they think is irrelevant to the reality....

 

[Subduction Zone]

And hence your all’s avoidance of discussing density.

I asked you all if you wanted to start with seismic reflection interpretation, earth and space rocks, or experiment.

You all avoided.....

I have no problem discussing density. But since you do not even understand basic Newtonian mechanics what is the point? What do you want to know about density?

 

[Subduction Zone]

Correct, was thinking of the force.



Exactly, so correct your comrade in arms contention that it increases towards the center..... I’m not the one confused on this point....



And -6 is less than -4?????? Of course, it is why you recognize it increases to zero at infinity.....



I’m not confusing anything. If there is no force there is absolutely no potential..... the ball has no potential to go anywhere......

which will be less increasing to the maximum at the surface....



No. At the surface it’s acceleration rate would decrease in proportion as the mass decreases. If it was accelerating at a value of 10 at the surface it would then continue to accelerate (minus air drag) at 9 then 8, 7, 6, etc until its acceleration was zero at the center. You are confusing continued increasing velocity with continued increasing acceleration. Velocity can continue to increase as acceleration decreases.

Just as in space I can accelerate at 10g, reduce my acceleration to 9G, 8G 7G etc and my velocity will continue to increase even as my acceleration decreases.....

Then as it passed the center it would begin to feel the force of more and more mass beneath it until it reached the maximum once again at the surface. Except this time it would feel a increasing decelerating force.





The mass becomes less as you near the center. What part didn’t you understand?????



It shows only that you don’t understand, even if you understand the moons escape velocity/force is less because of less mass beneath you. Now apply what you understand as the mass beneath you decreases....


And down past the core boundary it decreases to zero, while outward from the surface it only reaches zero at a speculative infinity...... so which really decreases faster? One that stays non zero to infinity or one that reaches zero in approximately 3,958.8 miles?????

But you forgot I stated based upon their flawed density profiles. I asked which you wanted to discuss first. Seismic reflection interpretation, earth and space rocks or experiment. You avoided...



The potential goes up as the escape velocity requirements go down... or if you prefer the potential goes down as the escape velocity goes up.

But potential starts at infinity inwards. Escape velocity starts at the surface outwards.


The potential goes down as the force goes up..... or if you prefer the potential goes up as the force goes down....

But potential starts at infinity, force starts at the center. In direct opposition to one another.


The force decreases as the mass decreases.....
As the potential decreases as the escape velocity decreases.... your “surface” is now smaller in diameter....

You have two choices. Either ignore the mass above you and treat the situation as a completely new planet, in which the old calculations do not apply. Or give up your shell theorem so you can still pretend the mass above you is valid to consider before you move above it. Which essentially creates a new planet and a new calculation.

Comprehend. Once below the surface the old calculations no longer apply. That mass no longer adds into the equation. But you are trying to reduce the size of the mass while keeping the old mass calculations and results. Conflict without you even realizing it..... you can’t use the old value data points.... they no longer enter the calculations as the mass above you no longer enters the calculations...




The escape velocity can be 1 mph with sufficient force....

“A rocket moving out of a gravity well does not actually need to attain escape velocity to escape, but could achieve the same result (escape) at any speed with a suitable mode of propulsion and sufficient propellant to provide the accelerating force on the object to escape.”

All that needs be achieved is the sufficient accelerating force.



Calculated from the surface. Once below the surface the mass needed to escape from is less..... it increases to the maximum at the surface. You only require more energy because you have more distance. I could walk from the center and my weight would slowly increase and the force I expended would slowly increase to a maximum at the surface.....



And the force of gravity decreases below the surface to zero at the center. So only at each point upwards is the force needed to be more and the escape velocity increases to the maximum at the surface....


You answered your own question. The force needed just needs to be more than the force of gravity at each point, which increases from the center to the maximum at the surface.....



And then it would decrease to zero at infinity, once it reached it’s maximum at the surface. It would then increase in proportion to force and decrease in proportion to force. Just as force remains positive but increases from zero at the center, reaches it’s maximum at the surface, then decreases again to zero at infinity.....

So what was your objection against two zero points again, we already use it with force.... are you saying you object to force having two zero points?????



No, accelerated expansion occurs because gravity is too weak to stop it and so can be neglected. We once again are in the realm of SR with high velocities where the force of gravity can be neglected.....

Don’t kid yourself....




No I’m not. Gravity doesn’t apply to expanding space. Gravity is what prevents space from expanding so we can’t measure it in the lab.... it can be ignored when discussing expansion.....



We were not discussing gravitational time dilation, but dilation due to velocity at high speeds where gravity can be neglected. SR..... you all keep erroneously bring the others into the conversation....


The energy that the gravitating object imparts....

What the distant observer sees is irrelevant.

If the object had to slow down then his clocks sped up more than the gravity slowed them. It’s why GPS clocks run a tiny bit faster from a gravitational force and slower still from its velocity.

So if you brought a GPS clock back to earth it would run slightly slower due to gravity, but speed up as it slowed down to surface speed. They run slower, they would speed up more to match earth clocks.

I am not the confused one. We were discussing gravitational potential, not gravity itself.

You need to pay attention.

 

[Subduction Zone]

Since I first posted that graph before the other......

Yes, and I have posted that particular graph for years too.. So what?

 

[Subduction Zone]

I'm not following you. Since you posted it before the other......what? You should have got the identification correct? You should know it was the acceleration and not the potential?

He posted a graph showing the gravity profile of the Earth above and below the surface. He has conflated gravitational potential with the acceleration due to gravity or so it seems.

 

[Polymath257]

Correct, was thinking of the force.

Exactly, so correct your comrade in arms contention that it increases towards the center..... I’m not the one confused on this point....

And -6 is less than -4?????? Of course, it is why you recognize it increases to zero at infinity.....

The *magnitude* is larger for -6 than for -4.

I’m not confusing anything. If there is no force there is absolutely no potential..... the ball has no potential to go anywhere......

The force being zero at a point does NOT mean the potential is zero.

which will be less increasing to the maximum at the surface....

And increasing further below the surface.

No. At the surface it’s acceleration rate would decrease in proportion as the mass decreases. If it was accelerating at a value of 10 at the surface it would then continue to accelerate (minus air drag) at 9 then 8, 7, 6, etc until its acceleration was zero at the center. You are confusing continued increasing velocity with continued increasing acceleration. Velocity can continue to increase as acceleration decreases.

Sorry if what I said was ambiguous. That the acceleration continues at all means the velocity would be larger at the center.

Just as in space I can accelerate at 10g, reduce my acceleration to 9G, 8G 7G etc and my velocity will continue to increase even as my acceleration decreases.....

Sort of my point. The escape velocity continues to increase below the surface.

Then as it passed the center it would begin to feel the force of more and more mass beneath it until it reached the maximum once again at the surface. Except this time it would feel a increasing decelerating force.

The mass becomes less as you near the center. What part didn’t you understand?????

Which mass? The mass inside that radius. But that isn't relevant for computing the escape velocity because as we go up, there is more mass.

It shows only that you don’t understand, even if you understand the moons escape velocity/force is less because of less mass beneath you. Now apply what you understand as the mass beneath you decreases....

No, it is a combined effect of both the mass and the radius. For example, if the mass of the moon was concentrated in a sphere whose radius was 1/10 of the actual value, the escape velocity would be a bit more than 3 times as much.

In the case of the center of the earth, as you pointed out, the force continues to increase and you have to do work against that force, which means an increased escape velocity at the center verses the surface.

And down past the core boundary it decreases to zero, while outward from the surface it only reaches zero at a speculative infinity...... so which really decreases faster? One that stays non zero to infinity or one that reaches zero in approximately 3,958.8 miles?????

And, to get back to the point, what does this say about the potential?

But you forgot I stated based upon their flawed density profiles. I asked which you wanted to discuss first. Seismic reflection interpretation, earth and space rocks or experiment. You avoided...

*sigh*

The potential goes up as the escape velocity requirements go down... or if you prefer the potential goes down as the escape velocity goes up.

But potential starts at infinity inwards. Escape velocity starts at the surface outwards.

The potential is *defined* as zero at infinity. The escape velocity is *defined* to be the velocity required to get to infinity with no additional thrust. And no, it doesn't 'start' at the surface. If anything, it starts at infinity just like the potential: that the the reference 'point'.

The potential goes down as the force goes up..... or if you prefer the potential goes up as the force goes down....

But potential starts at infinity, force starts at the center. In direct opposition to one another.

Wrong. And this is a major issue. The force is the rate of change of the potential, directed to fastest decrease of potential (also known as the negative gradient). For spherical symmetry, if the force is directed inward, then the potential is increasing with increasing radius. If the force is directed outward, the potential is decreasing with increasing radius.

The force decreases as the mass decreases.....
As the potential decreases as the escape velocity decreases.... your “surface” is now smaller in diameter....

Nope. As the potential decreases (gets more negative), the escape velocity *increases*. In fact, if v is the escape velocity and V is the potential, then v=sqrt(-2V).

You have two choices. Either ignore the mass above you and treat the situation as a completely new planet, in which the old calculations do not apply. Or give up your shell theorem so you can still pretend the mass above you is valid to consider before you move above it. Which essentially creates a new planet and a new calculation.

Nope. To compute the *force*, the shell theorem shows only the mass closer in produces the net force. But that isn't the case for the *potential* which is the (negative of) energy per mass required to lift to infinity.

Comprehend. Once below the surface the old calculations no longer apply. That mass no longer adds into the equation. But you are trying to reduce the size of the mass while keeping the old mass calculations and results. Conflict without you even realizing it..... you can’t use the old value data points.... they no longer enter the calculations as the mass above you no longer enters the calculations...

The mass above no longer affects the *force*. But it is quite relevant to find the potential because to lift to infinity you have to go past all the rest of that mass.

[QUOThe escape velocity can be 1 mph with sufficient force....[/QUOTE]

LMAO. This shows you really don't understand what the term 'escape velocity' means. The escape velocity *means* the velocity required to escape to infinity if no additional thrust is provided.

“A rocket moving out of a gravity well does not actually need to attain escape velocity to escape, but could achieve the same result (escape) at any speed with a suitable mode of propulsion and sufficient propellant to provide the accelerating force on the object to escape.”

Yes, the velocity required of a rocket need not be the escape velocity if it continuously provides thrust. But that doens't mean the escape velocity of the Earth can be 1 mph.

All that needs be achieved is the sufficient accelerating force.

Which is why the *definition* of escape velocity requires no such accelerating force.

Calculated from the surface. Once below the surface the mass needed to escape from is less..... it increases to the maximum at the surface. You only require more energy because you have more distance. I could walk from the center and my weight would slowly increase and the force I expended would slowly increase to a maximum at the surface.....

And if you started with a velocity upwards at the center, you would slow down and not be going as fast when you reach the surface. To 'escape', you still need the escape velocity once you reach that surface, so the escape velocity from the center is MORE than that at the surface.

This is trivial if you comprehend what the term 'escape velocity' means.

And the force of gravity decreases below the surface to zero at the center. So only at each point upwards is the force needed to be more and the escape velocity increases to the maximum at the surface....

You answered your own question. The force needed just needs to be more than the force of gravity at each point, which increases from the center to the maximum at the surface.....

Go back and learn what the term 'escape velocity' actually means. What you said here makes it clear you don't understand the concept.

And then it would decrease to zero at infinity, once it reached it’s maximum at the surface. It would then increase in proportion to force and decrease in proportion to force. Just as force remains positive but increases from zero at the center, reaches it’s maximum at the surface, then decreases again to zero at infinity.....

So what was your objection against to zero points again, we already use it with force....

No, the potential is NOT proportional to the force. You need to learn what the term 'potential' means also.

No, accelerated expansion occurs because gravity is too weak to stop it and so can be neglected. We once again ate in the realm of SR with high velocities where the force of gravity can be neglected.....

Don’t kid yourself....

Nope. SR cannot handle cosmology, which is where the accelerated expansion arises.

No I’m not. Gravity doesn’t apply to expanding space. Gravity is what prevents space from expanding so we can’t measure it in the lab.... it can be ignored when discussing expansion.....

Not even close. It is the term called the 'cosmological constant' be Einstein that, when put into the equations for *gravity* produces an outward pressure leading to accelerated expansion. This is very much a GR and a gravitational effect.

We were not discussing gravitational time dilation, but dilation due to velocity at high speeds where gravity can be neglected. SR..... you all keep erroneously bring the others into the conversation....

If you are talking about universal expansion and curvature, you are not doing SR.

The energy that the gravitating object imparts....

What the distant observer sees is irrelevant.

If the object had to slow down then his clocks sped up more than the gravity slowed them. It’s why GPS clocks run a tiny bit faster from a gravitational force and slower still from its velocity.

So if you brought a GPS clock back to earth it would run slightly slower due to gravity, but speed up as it slowed down to surface speed. They run slower, they would speed up more to match earth clocks.

Ignored as non-responsive to the question.

 

[Justatruthseeker]

The *magnitude* is larger for -6 than for -4.

Your contradicting yourself since you just agreed earlier it “increased” until it reached zero. You are not even self consistent in your arguments. Your reasoning can be rejected....

The force being zero at a point does NOT mean the potential is zero.

Lol. If there is no force there is no potential to move in any direction. If there is a potential to move towards an object, there is a force.

“Potential energy is equal (in magnitude, but negative) to the work done by the gravitational field moving a body to its given position in space from infinity.”

If there is no force no work can be done by the gravitational field because the gravitational field is zero.

You clearly understand nothing.....

All your claims can be rejected....

And increasing further below the surface.

No. There is less mass. Less gravitational field, less work being done, less potential energy....

“Potential energy is equal (in magnitude, but negative) to the work done by the gravitational field moving a body to its given position in space from infinity.”

Sorry if what I said was ambiguous. That the acceleration continues at all means the velocity would be larger at the center.

So what. Has nothing to do with the escape velocity except that it would already have exceeded it....

Sort of my point. The escape velocity continues to increase below the surface.

No. The escape velocity is greatest at the surface where the mass is greatest.

Which mass? The mass inside that radius. But that isn't relevant for computing the escape velocity because as we go up, there is more mass.

Until it reaches maximum at the surface. Below the surface you can not count the mass above you until you reach it. Escape velocity increases from the center to the maximum at the surface. It does not increase below the surface.

No, it is a combined effect of both the mass and the radius. For example, if the mass of the moon was concentrated in a sphere whose radius was 1/10 of the actual value, the escape velocity would be a bit more than 3 times as much.

No it wouldn’t. You can condense a 1 lb ball of bubble gum to 1/10 of its size and it will still weigh 1 lb. it has no more mass than it originally had. That is pure theory and has never been validated in any experiment. EVER.

“Mass is the measure of matter in an object. The mass of an object doesn't change if that object is heated, bent, stretched, squeezed or compressed, or transported from one place to another on earth or even to a position out in space.

According to the above definition, if we compress or expand an object, the mass of the object remains same. What is the experimental evidence for this property of the matter?

An object was never compressed or expanded as part of an experiment to check the validity of the definition of mass.”

An object on the moon has the same mass as an object on the earth. The same mass it has in space....

The mass won’t change unless you add more matter. With one exception. The energy added in compressing it will add a small amount of mass.

You need that fiction to support your invalid belief in point masses.....

In the case of the center of the earth, as you pointed out, the force continues to increase and you have to do work against that force, which means an increased escape velocity at the center verses the surface.

I pointed out no such thing.... the force decreases inwards from the surface. Which means initially less force is required increasing as you near the surface to its maximum at the surface. The force continues to increase from zero, not from the surface value.....

Imagine it takes 1 unit of force to move against 1 unit of mass (for simplicity). The surface is 10 units of mass 10 meters in diameter the mass decreasing by 1 unit per meter.

The center is zero. 1 meter above the center is 1 unit of mass requiring 1 unit of force. 2 meters above is two units of mass requiring two units of force. On and on until we reach the surface which requires 10 units of force. Each unit of force requires 1 mph to escape. 10 mph is the maximum velocity required to escape. You need not start at 10 then increase it by 10 as you want to be the case. Your thinking is flawed....

The potential is *defined* as zero at infinity. The escape velocity is *defined* to be the velocity required to get to infinity with no additional thrust. And no, it doesn't 'start' at the surface. If anything, it starts at infinity just like the potential: that the the reference 'point'.

So flawed it’s not even funny....

Once earths escape velocity is reached no more acceleration or force is required to move an object to infinity.....

“The escape velocity from Earth is about 11.186 km/s (6.951 mi/s; 40,270 km/h; 36,700 ft/s; 25,020 mph; 21,744 kn)[1] at the surface.....

.....Once escape velocity is achieved, no further impulse need be applied for it to continue in its escape. In other words, if given escape velocity, the object will move away from the other body, continually slowing, and will asymptotically approach zero speed as the object's distance approaches infinity, never to come back.”

Go ahead, calculate it from infinity....

Wrong. And this is a major issue. The force is the rate of change of the potential, directed to fastest decrease of potential (also known as the negative gradient). For spherical symmetry, if the force is directed inward, then the potential is increasing with increasing radius. If the force is directed outward, the potential is decreasing with increasing radius.

Potential energy is not a force.... potential energy is the rate of change against the force....

“Climbing stairs and lifting objects is work in both the scientific and everyday sense—it is work done against the gravitational force.”

Not with it.... I haven’t lost you yet have I?

“If the object is lifted straight up at constant speed, then the force needed to lift it is equal to its weight mg. The work done on the mass is then W = Fd = mgh. We define this to be the gravitational potential energy (PEg) put into (or gained by) the object-Earth system. This energy is associated with the state of separation between two objects that attract each other by the gravitational force.”

It is gained by separating two objects of mass, not bringing them together.....

Nope. As the potential decreases (gets more negative), the escape velocity *increases*. In fact, if v is the escape velocity and V is the potential, then v=sqrt(-2V).

The potential increases as you separate objects of mass, not bring them together....

Nope. To compute the *force*, the shell theorem shows only the mass closer in produces the net force. But that isn't the case for the *potential* which is the (negative of) energy per mass required to lift to infinity.

And the energy per mass required to lift it to infinity is now less because less work needs done against less gravitational force...... it increases away from the center, not increases towards the center..... it increases as objects of mass are separated, not brought together.....

The mass above no longer affects the *force*. But it is quite relevant to find the potential because to lift to infinity you have to go past all the rest of that mass.

And hence that potential increases as you go upwards...... you are lifting against a smaller force “until” your radius increases and the mass increases and the force increases and the potential increases.....

LMAO. This shows you really don't understand what the term 'escape velocity' means. The escape velocity *means* the velocity required to escape to infinity if no additional thrust is provided.

So you keep saying and so the science keeps disagreeing with you....

Escape velocity - Wikipedia

“A rocket moving out of a gravity well does not actually need to attain escape velocity to escape, but could achieve the same result (escape) at any speed with a suitable mode of propulsion and sufficient propellant to provide the accelerating force on the object to escape.”

At a constant 1 mph I would eventually reach the point where 1 mph was enough to be escape velocity. You don’t need to jump at 11.186 km/s to leave the surface. You fall back only because the force is not constant....

Yes, the velocity required of a rocket need not be the escape velocity if it continuously provides thrust. But that doens't mean the escape velocity of the Earth can be 1 mph.

Yes it does. I can leave the earths surface at 1 mph by jumping. The initial force I expend just stops instead of continues. If that same force were to continue I would escape the earth.

“A rocket moving out of a gravity well does not actually need to attain escape velocity to escape, but could achieve the same result (escape) at any speed with a suitable mode of propulsion and sufficient propellant to provide the accelerating force on the object to escape.”

What part of “at any speed” didn’t sink in????

Which is why the *definition* of escape velocity requires no such accelerating force.

because no specific force is required, just a constant force. Our limited technology simply uses high force to achieve high velocity quickly to overcome fuel expenditure problems.....

And if you started with a velocity upwards at the center, you would slow down and not be going as fast when you reach the surface. To 'escape', you still need the escape velocity once you reach that surface, so the escape velocity from the center is MORE than that at the surface.

Wrong. If I can escape the surface at any velocity, I can escape from the center at any velocity as long as my thrust is constant..... the force required at the center is less, increasing to the maximum at the surface....

No, the energy expenditure is more from the center, the escape velocity will never require more force than that needed to escape the entire gravitational influence of the earth....

This is trivial if you comprehend what the term 'escape velocity' means.

Which you don’t. It is the since you like velocity instead of force..... the velocity required to escape the entire gravitational influence of the earth. The earths gravitational influence decreases below the surface not increases..... it is it’s maximum at the surface.....

Go back and learn what the term 'escape velocity' actually means. What you said here makes it clear you don't understand the concept.

You need to go back and learn what it means. It’s clear you are the one that doesn’t comprehend even the basics.....

It is the velocity (force) required to overcome the entire gravitational influence of the earth. Said influence decreasing as one nears the center. You need never apply more force than that required at the surface.

No, the potential is NOT proportional to the force. You need to learn what the term 'potential' means also.

We have already seen your concepts are flawed. Potential increases as gravitational objects are separated, not brought together.... it’s why the negative confuses you......

 

[Justatruthseeker]

Continued

Nope. SR cannot handle cosmology, which is where the accelerated expansion arises.

GR can’t handle it. It is 99.8% correct describing planetary motions where it has been tested to that extreme accuracy. Once outside the solar system it suddenly requires 96% ad-hoc theory to make it fit even a semblance of reality. Even if it needed none of it to be tested to that 99.8% accuracy. You just ignore that tested accuracy and what it is trying to tell you and keep sledgehammering it harder and harder to force it to fit.....

Not
even close. It is the term called the 'cosmological constant' be Einstein that, when put into the equations for *gravity* produces an outward pressure leading to accelerated expansion. This is very much a GR and a gravitational effect.

That same constant that was the “biggest mistake of his life”??????

Dark energy has nothing to do with GR. Was never predicted by GR, is nowhere in the equations for GR, and is one of those sledgehammer blows needed to force it to fit.....

And worse yet they had to add another inflationary unknown cause (not Dark Energy) to the early universe because one sledgehammer blow wasn’t enough..... bam, bam, bam..... by god we’ll make it fit yet..... if we keep slamming it hard enough and ignore that 99.8% accuracy under direct experimentation. Wouldn’t want cold hard facts to get in the way of theory.....

Cosmological constant - Wikipedia

“However, soon after Einstein developed his static theory, observations by Edwin Hubble indicated that the universe appears to be expanding; this was consistent with a cosmological solution to the original general relativity equations that had been found by the mathematician Friedmann, working on the Einstein equations of general relativity. Einstein reportedly referred to his failure to accept the validation of his equations—when they had predicted the expansion of the universe in theory, before it was demonstrated in observation of the cosmological red shift—as his "biggest blunder".“

Einstein added the cosmological constant to make a static case. But his original equations as shown by Friedmann already showed his original equations without the cosmological constant were consistent with observations.. And yet without adding back the cosmological constant you can’t make it fit observations. Because you have already added too many sledgehammer blows (dark matter, etc) and bent the original until it no longer fitted without more hammering. So because you ignored it’s tested accuracy you have bent it until it isn’t even GR anymore and require the laws of physics to work differently in different parts of the universe......

If you are talking about universal expansion and curvature, you are not doing SR.

I sure am since expansion is at 2.23c and continuing to ignore gravity as we speak by accelerating.....

Ignored as non-responsive to the question.

Ignoring the fact that GPS satellites would speed up when returned to earth. Directly responding to your question with cold hard empirical facts instead of fantasy......... ignoring cold hard facts that don’t fit your perceived beliefs is par for the course... now your ignoring the cold hard fact GPS clocks would need to be sped up if returned to earth (minus their speed adjustments already), just as all your life you have ignored the accuracy of GR and what it tried to tell you....

 

[Polymath257]

Your contradicting yourself since you just agreed earlier it “increased” until it reached zero. You are not even self consistent in your arguments. Your reasoning can be rejected....

Well, I evidently can't educate you about this.

My recommendation is to go and learn some math and then take a basic classical mechanics class at a local university.

Maybe someone else will have better luck showing you your mistakes.

 

[Justatruthseeker]

Well, I evidently can't educate you about this.

My recommendation is to go and learn some math and then take a basic classical mechanics class at a local university.

Maybe someone else will have better luck showing you your mistakes.

You haven’t educated yourself..... you have no hope of teaching your conflicting beliefs to those that are.

You want force to decrease, but think it requires a higher velocity to move against less force.

Ignore the experts telling you PGE increases as objects of mass are separated because all you can think about is negative force. Because you don’t understand the concept.

“Positive work must also be done against gravity (by an external force) to move the object away from the Earth. The object would gain potential energy if moved away from the Earth. ... AND THUS GRAVITATIONAL POTENTIAL ENERGY WILL BE NEGATIVE. ”

Work must be applied. GPE has no meaning to a single object. GPE can only be spoken of by the earth as the earth moves away from the sun..... gains it, and looses it as it approaches the sun.....

If the earth was the only mass in the universe it would possess no GPE because their would be no other force from another mass to move against.

An object does not possess GPE only because of its mass, but because of the force required to move it against another masses force.

You have just realized your error and this is your way of saving face without having to admit you were wrong.... it’s ok, I understand....

 

[Polymath257]

You have just realized your error and this is your way of saving face without having to admit you were wrong.... it’s ok, I understand....

Nope. It's clear you understand very little here. Once again, I suggest you actually take a university level course in this. You *might* just learn a few things.

 

[Justatruthseeker]

Nope. It's clear you understand very little here. Once again, I suggest you actually take a university level course in this. You *might* just learn a few things.

What, like how to ignore the tested accuracy of GR by insisting physics works differently in different parts of the universe?

Sorry, not interested in ignoring the tested accuracy of GR where physics has been found to work just fine and agrees with GR.....

Hit it with that sledgehammer a few more times....

 

[Polymath257]

What, like how to ignore the tested accuracy of GR by insisting physics works differently in different parts of the universe?

Sorry, not interested in ignoring the tested accuracy of GR where physics has been found to work just fine and agrees with GR.....

Hit it with that sledgehammer a few more times....

Once again, go take an actual physics class. Learn about the realities here and not just your misunderstandings. Then, after you have learned a bit, we can discuss this again.

 

[Justatruthseeker]

Once again, go take an actual physics class. Learn about the realities here and not just your misunderstandings. Then, after you have learned a bit, we can discuss this again.

I understand reality. Just as I understand the reality that GPS clocks would speed up if returned to earth, not slow down.....

It’s only you that is lost in theory and is oblivious to the reality....