[Trolltrolls]

We know that the value of g increases as we go from the equator towards the poles.

How do flat earthers explain this? As per a picture of their model (with Antarctica at the edge), g would have to be maximum at the center, decrease, and then increase, while on a flat plane. How? I need math, nothing else.

Also, g decreases with depth. This occurs due to application of a theorem (Shell Theorem) by dividing the Earth into a number of shells. How is this possible in a flat Earth? Again, I need math.

And, why would all the space agencies of the world have a conspiracy? Is there any basis for this? I think not. It's conjecture. It's like me believing that other people are robots. It's crazy, and might even be a symptom of a mental disorder (irrational thinking, paranoia).

And what is the explanation for day/night?

I've been reading some other threads and so far I've seen flat earthers:

1. Modify laws into their favour.
2. Act condescendingly and not give a straightforward answer (EVER)
3. Not reply to a particular comment which concerns the two.
4. Act condescendingly and not give a straightforward answer (EVER)

[Dr David Thork]

We know that the value of p increases as we go from the equator towards the poles.

Do you mean p as in capital P for pressure, or Greek rho for density? I thought you meant g for gravity but p and g are a long way from each other on a keyboard so it doesn't seem like a very likely typo.

Re-reading the rest, there isn't a change of g towards the equator ... but density and pressure do change at the poles due to Hadley cells. Unfortunately the equator has the lowest pressure and density on average so you're all over the place here.

g decreases with depth. Really? g is higher on Mount Everest than it is in Holland and much higher than in the Mariana Trench under the Atlantic. you could dig down into Everest and it would still be higher than in that trench. Apparently. I think you should learn more about what causes gravity on the round earth model before you come here criticising the flat earth one.

[Boots]

Fg at sea level = GM/r² = 6.674*10^-11*5.972*10²⁴/(6.371*10⁶)² = 9.819N/Kg

Fg at peak of Everest = GM/(r+h)² = 6.674*10^-11*5.972*10²⁴/(6.371*10⁶+8848)² = 9.792N/Kg

Fg at bottom on Mariana Trench = GM/(r-h)² = 6.674*10^-11*5.972*10²⁴/(6.371*10⁶-10994)² = 9.854N/Kg

There are a couple of other factors such as the Mariana Trench and Mt Everest being about 15 and 30 degrees (respectively) North of the equator where Fg measures slightly different due to earth's rotation. Also, I think M would technically need to be adjusted for the Mariana trench calculation and probably an opposing force accounted for caused by the mass above the trench. I don't know how to calculate these, but differences caused by rotation are minuscule and at 1/600th of the way to earths center I think the shift in Mass would also be small.

[douglips]

We know that the value of p increases as we go from the equator towards the poles.

Re-reading the rest, there isn't a change of g towards the equator ... but density and pressure do change at the poles due to Hadley cells. Unfortunately the equator has the lowest pressure and density on average so you're all over the place here.

You are mistaken - g does change with latitude.

https://www.pharmamanufacturing.com/articles/2016/mass-measurement-precision-small-objects-pharmaceutical-production/

See especially this image:


These people are just trying to make pharmaceuticals, they don't care about round or flat earth, so this isn't some conspiracy, it's a real effect that can be explained by centrifugal force in round earth theory, and as far as I know has no explanation in flat earth theory.

g decreases with depth. Really? g is higher on Mount Everest than it is in Holland and much higher than in the Mariana Trench under the Atlantic. you could dig down into Everest and it would still be higher than in that trench. Apparently. I think you should learn more about what causes gravity on the round earth model before you come here criticising the flat earth one.

Here Baby Thork is correct - the Shell idea you are talking about only works with a sphere of uniform density, but the density of the earth varies with depth so the shell theory is too simple to explain g here.

[Trolltrolls]

We know that the value of p increases as we go from the equator towards the poles.

Do you mean p as in capital P for pressure, or Greek rho for density? I thought you meant g for gravity but p and g are a long way from each other on a keyboard so it doesn't seem like a very likely typo.

Re-reading the rest, there isn't a change of g towards the equator ... but density and pressure do change at the poles due to Hadley cells. Unfortunately the equator has the lowest pressure and density on average so you're all over the place here.

g decreases with depth. Really? g is higher on Mount Everest than it is in Holland and much higher than in the Mariana Trench under the Atlantic. you could dig down into Everest and it would still be higher than in that trench. Apparently. I think you should learn more about what causes gravity on the round earth model before you come here criticising the flat earth one.

Oops, I meant g. Sorry!

g does change with latitude. g'= g(9. - rw²cos²Q (theta)
Since cos0= 1, g is minimum at the equator. I can give you the derivation if your want.
I suggest you read point 3 and 4 of my original post.
g'=GM/R² (at x>=R) If we put in R+8848 metres, it will obviously come out to be lesser than sea level. You can even use binomial theorem.
g'=g(1+h/R)^-2
g'=g(1-2h/R)
Do your math.

We know that the value of p increases as we go from the equator towards the poles.

Re-reading the rest, there isn't a change of g towards the equator ... but density and pressure do change at the poles due to Hadley cells. Unfortunately the equator has the lowest pressure and density on average so you're all over the place here.

You are mistaken - g does change with latitude.

https://www.pharmamanufacturing.com/articles/2016/mass-measurement-precision-small-objects-pharmaceutical-production/

See especially this image:


These people are just trying to make pharmaceuticals, they don't care about round or flat earth, so this isn't some conspiracy, it's a real effect that can be explained by centrifugal force in round earth theory, and as far as I know has no explanation in flat earth theory.

g decreases with depth. Really? g is higher on Mount Everest than it is in Holland and much higher than in the Mariana Trench under the Atlantic. you could dig down into Everest and it would still be higher than in that trench. Apparently. I think you should learn more about what causes gravity on the round earth model before you come here criticising the flat earth one.

Here Baby Thork is correct - the Shell idea you are talking about only works with a sphere of uniform density, but the density of the earth varies with depth so the shell theory is too simple to explain g here.

https://www.newscientist.com/article/dn24068-gravity-map-reveals-earths-extremes/
The calculated g' is still lower than the g at sea level. Read para 2.
Also, please provide a source that the shell theorem is only applicable to a sphere of uniform density. I cannot find it.

Fg at sea level = GM/r² = 6.674*10^-11*5.972*10²⁴/(6.371*10⁶)² = 9.819N/Kg

Fg at peak of Everest = GM/(r+h)² = 6.674*10^-11*5.972*10²⁴/(6.371*10⁶+8848)² = 9.792N/Kg

Fg at bottom on Mariana Trench = GM/(r-h)² = 6.674*10^-11*5.972*10²⁴/(6.371*10⁶-10994)² = 9.854N/Kg

There are a couple of other factors such as the Mariana Trench and Mt Everest being about 15 and 30 degrees (respectively) North of the equator where Fg measures slightly different due to earth's rotation. Also, I think M would technically need to be adjusted for the Mariana trench calculation and probably an opposing force accounted for caused by the mass above the trench. I don't know how to calculate these, but differences caused by rotation are minuscule and at 1/600th of the way to earths center I think the shift in Mass would also be small.

The g calculation of E at bottom of Mariana trench is wrong.
The correct formula is obtained from the shell theorem:
g' GM(R-d)/R³ = g(1-d/R)

[Boots]

Fg at sea level = GM/r² = 6.674*10^-11*5.972*10²⁴/(6.371*10⁶)² = 9.819N/Kg

Fg at peak of Everest = GM/(r+h)² = 6.674*10^-11*5.972*10²⁴/(6.371*10⁶+8848)² = 9.792N/Kg

Fg at bottom on Mariana Trench = GM/(r-h)² = 6.674*10^-11*5.972*10²⁴/(6.371*10⁶-10994)² = 9.854N/Kg

There are a couple of other factors such as the Mariana Trench and Mt Everest being about 15 and 30 degrees (respectively) North of the equator where Fg measures slightly different due to earth's rotation. Also, I think M would technically need to be adjusted for the Mariana trench calculation and probably an opposing force accounted for caused by the mass above the trench. I don't know how to calculate these, but differences caused by rotation are minuscule and at 1/600th of the way to earths center I think the shift in Mass would also be small.

The g calculation of E at bottom of Mariana trench is wrong.
The correct formula is obtained from the shell theorem:
g' GM(R-d)/R³ = g(1-d/R)

Thanks. I was wondering about that as I mentioned. I'll look into it more.

[douglips]

We know that the value of p increases as we go from the equator towards the poles.

Re-reading the rest, there isn't a change of g towards the equator ... but density and pressure do change at the poles due to Hadley cells. Unfortunately the equator has the lowest pressure and density on average so you're all over the place here.

You are mistaken - g does change with latitude.

https://www.pharmamanufacturing.com/articles/2016/mass-measurement-precision-small-objects-pharmaceutical-production/

See especially this image:


These people are just trying to make pharmaceuticals, they don't care about round or flat earth, so this isn't some conspiracy, it's a real effect that can be explained by centrifugal force in round earth theory, and as far as I know has no explanation in flat earth theory.

g decreases with depth. Really? g is higher on Mount Everest than it is in Holland and much higher than in the Mariana Trench under the Atlantic. you could dig down into Everest and it would still be higher than in that trench. Apparently. I think you should learn more about what causes gravity on the round earth model before you come here criticising the flat earth one.

Here Baby Thork is correct - the Shell idea you are talking about only works with a sphere of uniform density, but the density of the earth varies with depth so the shell theory is too simple to explain g here.

https://www.newscientist.com/article/dn24068-gravity-map-reveals-earths-extremes/
The calculated g' is still lower than the g at sea level. Read para 2.
Also, please provide a source that the shell theorem is only applicable to a sphere of uniform density. I cannot find it.

How hard did you look?
https://en.wikipedia.org/wiki/Shell_theorem
Excerpt, emphasis added:
"A corollary is that inside a solid sphere of constant density, the gravitational force varies linearly with distance from the centre, becoming zero by symmetry at the centre of mass."

This is pretty much obvious - imagine a sphere of radius 2, composed of one inner shell of radius 1 with a mass of 10^20 kg, and one outer shell with a mass of 10^10kg.

On the surface, the gravity force from the inner shell is 10^10 times stronger than the force from the outer shell.
If you dig down to the interface between the shells, you lose the relatively tiny gravity from the outer shell, but now you're half the distance to the inner shell and your gravitational force goes up by the r^2 factor.

If you want me to actually plug numbers into the formulae you provided I'm happy to do so, I'm just thinking you'll realize how obvious this is.

In round earth theory, the density of the earth's core is much greater than that of the crust or ocean:
https://web.archive.org/web/20150602060826/http://geophysics.ou.edu/solid_earth/prem.html
Accordingly, depending on how deep a hole you dig and where, your local observation of g may increase as you dig.

[Dr David Thork]

Fg at sea level = GM/r² = 6.674*10^-11*5.972*10²⁴/(6.371*10⁶)² = 9.819N/Kg

Fg at peak of Everest = GM/(r+h)² = 6.674*10^-11*5.972*10²⁴/(6.371*10⁶+8848)² = 9.792N/Kg

Fg at bottom on Mariana Trench = GM/(r-h)² = 6.674*10^-11*5.972*10²⁴/(6.371*10⁶-10994)² = 9.854N/Kg

There are a couple of other factors such as the Mariana Trench and Mt Everest being about 15 and 30 degrees (respectively) North of the equator where Fg measures slightly different due to earth's rotation. Also, I think M would technically need to be adjusted for the Mariana trench calculation and probably an opposing force accounted for caused by the mass above the trench. I don't know how to calculate these, but differences caused by rotation are minuscule and at 1/600th of the way to earths center I think the shift in Mass would also be small.

Well thanks for the maths, but you haven't included all of the variables. You know you haven't or else NASA wouldn't provide potato diagrams of gravity maps. For a start, mountains have a lot of rock in them, and that localised mass adds gravity. The M that you set as a constant but is actually a complex variable. If you look at a gravity map, the Himalayas has the highest gravity on earth (you calculated the lowest) ... so your calculations are bunk.


Although the entire NASA concept is just junk ... but that's another story.

[nickrulercreator]

Fg at sea level = GM/r² = 6.674*10^-11*5.972*10²⁴/(6.371*10⁶)² = 9.819N/Kg

Fg at peak of Everest = GM/(r+h)² = 6.674*10^-11*5.972*10²⁴/(6.371*10⁶+8848)² = 9.792N/Kg

Fg at bottom on Mariana Trench = GM/(r-h)² = 6.674*10^-11*5.972*10²⁴/(6.371*10⁶-10994)² = 9.854N/Kg

There are a couple of other factors such as the Mariana Trench and Mt Everest being about 15 and 30 degrees (respectively) North of the equator where Fg measures slightly different due to earth's rotation. Also, I think M would technically need to be adjusted for the Mariana trench calculation and probably an opposing force accounted for caused by the mass above the trench. I don't know how to calculate these, but differences caused by rotation are minuscule and at 1/600th of the way to earths center I think the shift in Mass would also be small.

Well thanks for the maths, but you haven't included all of the variables. You know you haven't or else NASA wouldn't provide potato diagrams of gravity maps. For a start, mountains have a lot of rock in them, and that localised mass adds gravity. The M that you set as a constant but is actually a complex variable. If you look at a gravity map, the Himalayas has the highest gravity on earth (you calculated the lowest) ... so your calculations are bunk.


Although the entire NASA concept is just junk ... but that's another story.

His calculations are not bunk, they're quite accurate. They're predictable AND have been repeated and confirmed.

The localized gravity map simply adds onto the already-known differences in gravity based on latitude.

As for your last paragraph, what evidence do you have to prove that NASA's data is bunk?

[ShowmetheProof]

His calculations are not bunk, they're quite accurate. They're predictable AND have been repeated and confirmed.

The localized gravity map simply adds onto the already-known differences in gravity based on latitude.

As for your last paragraph, what evidence do you have to prove that NASA's data is bunk?-nickrulercreator

They don't.  The fact remains that NASA hasn't done much lying to us, and doesn't usually.

[JohnAdams1145]

Quick note:
Tom Bishop is correct in one of his other posts about there being more centripetal acceleration as you get closer to the equator. If any of you wish to measure the actual value of g and compare it to the theoretical, you need to take this into account by adding it in to whatever you measure on a scale.

[Trolltrolls]

Quick note:
Tom Bishop is correct in one of his other posts about there being more centripetal acceleration as you get closer to the equator. If any of you wish to measure the actual value of g and compare it to the theoretical, you need to take this into account by adding it in to whatever you measure on a scale.

I think I included it somewhere, because that is the reason g varies with latitude.
g'=g + mrw²cos²Q

[Boots]

Can you confirm what g' stands for? I assume it stands for the actual force felt per Kg at any given location as opposed to g which only accounts for earth's mass and radius. (ie. GM/r²)

[JohnAdams1145]

Boots,
That's correct, although that quantity is also sometimes just taken as g, as g is an experimentally observed acceleration and varies a lot depending on where you are. GM/r^2 assumes that Earth is a uniform density sphere, which it slightly deviates from.