iwpoe wrote:

Timocrates wrote:

Even in this case we are granting that things would be moved and forces/pressures would exist.

Again, in my mind, all we are doing here is negating the alleged near perfect vacuum of space. Indeed, I believe that given the demonstrable and measurable strength of the earth's gravity, I do not believe it is even plausible to conclude that the earth ever would have attracted or kept an atmosphere at all, because at some point you had to have a situation where the earth was like the moon is supposed to be (devoid of atmosphere), but then the pressure differential would have been absolute between any atmospheric molecules, which then would have been in direct contact with the void/vacuum of space. But we know gravity is not a strong enough force to resist such a vacuum. So how did we acquire an atmosphere in the first place?

Again, we have to entertain "spaceship earth" for this theory to work. At best, this necessitates creation and a Creator.

Well this is an interesting part of natural history, but it is far different from the idea you previously articulated of gravity being unable to retain the atmosphere.

Vaccums don't work like you think they do and you are confused- probably because you model them on Dysons and not conceptually.

Well, I believe my disagreements with modern cosmological doctrine actually arises in part from an understanding of the consequences of a vacuum. I'm not even sure which doctrine we are supposed to believe: is there not by natural necessity a positive force acting on things to fill a vacuum or not? Sometimes it seems like I am being asked to believe that outer space is effectively a vacuumless void; other times, that it really is a vacuum but gravity somehow manages to negate its consequences. I believe I have raised real, demonstrable problems with either conception.

Last edited by Timocrates (7/19/2016 7:00 pm)

One small answer before going to bed :D

Check Quora. Doing some mathematics, you can reach to the same formuals given by the person in Quora. I do understand that gravity becomes weaker. But at the same time the velocity of gas molecules also decrease. Remember, that it is force that is reducing the velocity. So even if you decrease the force but have some finite force you are still reducing the velocity. If you look at it this way it would make sense. And one last note is that if the gas molecules do not have velocity they cannot go to vacuum. 

EDIT: Check Quora please, I think the guy made a nice explanation. Here I am giving general picture. There is actually some percentage of the gas molecules leaving the earth but it is very insignificant. I will try to read your post more carefully and completely so I can address all the issues but sorry for my very small and limited answers. Now that I am thinking, I believe you are right in some sense but you must also consider that how much gas is leaving the atmosphere. With qualitative language we cannot arrive at the answer. But according the graph that you put the molecules at high heights still have velocity (because of temperauter) but you must also consider that gas is very diluted there and that height gravity is almost the same at sea level. (90 KM is nothing compared to 6400 km which is the radius of earth)

Last edited by nojoum (7/19/2016 7:27 pm)

"The earth spinning (at the equator) at 1,030mph, though this only causes someone to be 0.3% lighter at the equator as opposed to the poles. Complete nonsense and total fiction, in my mind. There is just no way a 1,000+ mph spin velocity only reduces a person's effective weight by 0.3%. It's actually quite ridiculous to even suggest this."
Why is it ridiculous?    Here are the calculations.  https://davidrielle.files.wordpress.com/2013/03/do-i-weigh-less-at-the-equator-than-at-the-north-pole.pdf  Please read it with an open mind.  Please don't say "Well, he simplified it here, therefore it's wrong."  He simplified it for a popular audience.  Suffice it to say other physicists have done the refined calculations and do them every day.  GPS wouldn't work if they hadn't.  The refined calculations will not change the 0.3% change in apparent weight to some bigger number that will satisfy you; it will change it to something that still rounds to 0.3%.
 

ArmandoAlvarez wrote:

"The earth spinning (at the equator) at 1,030mph, though this only causes someone to be 0.3% lighter at the equator as opposed to the poles. Complete nonsense and total fiction, in my mind. There is just no way a 1,000+ mph spin velocity only reduces a person's effective weight by 0.3%. It's actually quite ridiculous to even suggest this."
Why is it ridiculous?    Here are the calculations.  https://davidrielle.files.wordpress.com/2013/03/do-i-weigh-less-at-the-equator-than-at-the-north-pole.pdf  Please read it with an open mind.  Please don't say "Well, he simplified it here, therefore it's wrong."  He simplified it for a popular audience.  Suffice it to say other physicists have done the refined calculations and do them every day.  GPS wouldn't work if they hadn't.  The refined calculations will not change the 0.3% change in apparent weight to some bigger number that will satisfy you; it will change it to something that still rounds to 0.3%.
 

Well, this is the first time you've presented something explaining how a 1,030mph spin velocity just disappears.  A 1,030mph spin velocity will produce tremendous force and pressure - that is the same velocity a jet engine's fan spins at. It's the actual velocity - not the RPMs - that matter. The RPMs will be constant on any spinning disc or sphere: they make one revolution in the same time at every point. For this theory to work, gravity has to be much stronger around the equator for everything to balance out, but there's no reason why gravity would be stronger around the equator. The force in questions looks like this:

Earth's axis = Vertical line
Distance from axis of rotation = vertical line = increasing force

| (Pole = about 0mph)
|_
|__ (45 N Latitude, about 515mph)
|___
|____ (Equator = 1,030mph)

The RPMs as we move from either pole to the equator will remain perfectly constant at one rotation per day/24hrs. It's the actual necessary velocity that is the relevant force, because points on the equator have to travel more distance in the same time. This is important for mechanical engineering because it well tell us what sort of stresses and pressures we can expect to be exerted on the material in question, as things could easily explode as a consequence of centrifugal forces.

When pilots are being trained to handle g-forces, there is a reason they are put at the outer edge of a spinning machine and not at its centre or halfway up the radius. He will experience virtually no g-forces if he were placed at the centre and the spinning would probably only make him nauseous and dizzy. The maximum force is at the outer edge, which can easily cause a grown, fit, healthy young man to simply pass out. The mid-point of the radius and the outer edge are both moving at the same RPMs: it is the greater velocity at the outer edge that is relevant. The earth is spinning on its axis, which means as we approach the equator the forces generated by the earth's spin will be constantly increasing; whereas, gravity comes from a fixed centre point and spreads its force equidistantly. 

Now who could seriously entertain the idea that the effect of the difference of force here would only amount to a 0.3% reduction in a person's weight?  The g-forces experienced in a centrifuge will increase almost exponentially as you move away from the axis of rotation; but apparently not on earth. Your article is dealing with the centripetal (inward) not the centrifugal (outward) forces that are generated, which latter force is the relevant force: this is why pilot g-force test machines are called centrifuges.

Now as the author himself admits, mass remains constant regardless of any forces present. We are not even concerned with apparent weight here; we are actually concerned with the force effectively generated by the spin and what effect this, in turn, will have on a mass in the relevant location (i.e. on the equator). Obviously lighter and looser objects will be move likely to be thrown off or away by this force whereas heavier, more strongly attached items will be less likely. The apparent weight is only relevant when decided whether or not the force will be sufficiently powerful to throw off that weight.

Edit: Added:
I am also having issue with the author of that article's statement here:

Do I Weigh Less at the Equator..? wrote:

For simplicity's sake I am assuming that the radius of the earth doesn'tchange between the North pole and the Equator...

It's actually the distance from the axis of rotation that matters here. While this may not affect the subsequent calculations much, I am nervous to the extent that this value is actually quite irrelevant. The North Pole should not experience much or any centrifugal force even though it, too, is approximately the same distance from the centre of the earth than the Equator is. But it's the distance from the axis of rotation that matters in these calculations: not the radius as such.

Also, I'm not seeing how his gravitational force math is working out. He says that for something having a mass of 80kg, when we factor in the acceleration due to earth's gravity, which he states is 9.8m/s, we get an effective weight of 176lbs. But this is just converting kilograms to pounds (80kg x 2.2 = 176lbs). So what was the reason for brining up acceleration due to gravity in the first place? It apparently has no bearing or consequence whatsoever: it just assumed already to effectively negate the consequences of things like centrifugal forces that might otherwise be expected to throw things off of it.

I'm concerned that this force is again by definition doing the work of largely just negating any consequence to the spin velocity of the earth right off the bat. Because what is exactly is at issue is the fact that we simply do not expect an 80kg person to actually weigh 176lbs at the equator; but once again, gravity just magically causes this to happen. Then, apparently, we have only the relative angle of rotation taken in to subtract some weight and, voila, you get a 0.3% reduction in weight - because we had assumed that someone's weight is effectively the same at the equator as at the poles to begin with; whereas, isn't this what is exactly supposed to be in question?

Last edited by Timocrates (7/20/2016 3:44 pm)

Timocrates wrote:

nojoum wrote:

ArmandoAlvarez wrote:

"The earth spinning (at the equator) at 1,030mph, though this only causes someone to be 0.3% lighter at the equator as opposed to the poles. Complete nonsense and total fiction, in my mind. There is just no way a 1,000+ mph spin velocity only reduces a person's effective weight by 0.3%. It's actually quite ridiculous to even suggest this."
Why is it ridiculous?    Here are the calculations.  https://davidrielle.files.wordpress.com/2013/03/do-i-weigh-less-at-the-equator-than-at-the-north-pole.pdf  Please read it with an open mind.  Please don't say "Well, he simplified it here, therefore it's wrong."  He simplified it for a popular audience.  Suffice it to say other physicists have done the refined calculations and do them every day.  GPS wouldn't work if they hadn't.  The refined calculations will not change the 0.3% change in apparent weight to some bigger number that will satisfy you; it will change it to something that still rounds to 0.3%.
 

What part of it do you find nonsense? It is a legitimate and sound demonstration.
Your common sense is betraying you. The speed of earth is for sure very high but conside the radius of movement. 6400 km is a lot. To understand it better, you think of making a very sharp turn with a car and making a turn with much higher radius. Or in fact if you increase the radius to infinity the force will become zero. Because you are moving in a line.
 

But you are still moving circularly at 1,030mph or 500+m/s. If we constructed a centrifuge with a radius the same as the earth's, and spun it at 1,030mph, the forces generated would be enormous, because it is not RPMs that matter here: if they did, then there would be no difference between the mid-point of the radius of a centrifuge and its outer edge.

This brings to mind other problems in modern mechanics. The actual velocities of things are treated effectively as irrelevant. Hence, though everything on earth is moving at something like 66,000+ mph, we can still bump into each other and not expect a violent catastrophe. Two cars colliding at 1 and 2 mph respectively will produce minimal force and damage (on each other); two cars moving at 100 and 101mph, much more. But these forces all just manage to disappear, just as the centrifugal force of the earth's spin does.

Dear Timocrates, with all the respect, I must say that it is better to study physics more deeply and also to avoid qualitative language. You can calculate the force, it will be small compared to the weight of the body( mg)
I understant that it is not nice to say this. But please, do study the fundamentals of physics by David Halliday.
I cannot simply start teaching here because I am not in the position to be a teacher and you to be a student of mine. But these things are like saying earth is flat. Or that if God does not need a cause why does the universe needs a cause? Or what caused God?

Last edited by nojoum (7/20/2016 3:26 pm)