## mark mccutcheon final theory

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- phillip1882
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### mark mccutcheon final theory

so i got his book and have read it several times.

basically he proposes all atoms increases in size, and that this is the underlying cause of gravity.

then he further uses that idea to explain magnetism, electricity, and light.

many of his ideas make a lot of sense. in particular the structure of the atom he proposes.

he assumes that all energy is fundamentally electrons, in various structures.

the electrons of an atom are charge-less particles that bounce off the core at a rapid pace.

there are four ways to disprove it.

1) the gravity at the top of a sky scraper should be greater than at the bottom.

2) a large object going down a vacuum tube should take a shorter amount of time than a small one.

3) the earth should wobble in a 24 hour period. (this is his explanation for the tides)

4) the gravity on the far side of the moon is greater than the near side.

basically he proposes all atoms increases in size, and that this is the underlying cause of gravity.

then he further uses that idea to explain magnetism, electricity, and light.

many of his ideas make a lot of sense. in particular the structure of the atom he proposes.

he assumes that all energy is fundamentally electrons, in various structures.

the electrons of an atom are charge-less particles that bounce off the core at a rapid pace.

there are four ways to disprove it.

1) the gravity at the top of a sky scraper should be greater than at the bottom.

2) a large object going down a vacuum tube should take a shorter amount of time than a small one.

3) the earth should wobble in a 24 hour period. (this is his explanation for the tides)

4) the gravity on the far side of the moon is greater than the near side.

good luck have fun

- Eebster the Great
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### Re: mark mccutcheon final theory

phillip1882 wrote:so i got his book and have read it several times.

basically he proposes all atoms increases in size, and that this is the underlying cause of gravity.

then he further uses that idea to explain magnetism, electricity, and light.

many of his ideas make a lot of sense. in particular the structure of the atom he proposes.

he assumes that all energy is fundamentally electrons, in various structures.

the electrons of an atom are charge-less particles that bounce off the core at a rapid pace.

I haven't read the book, but it has a very strong anatine odor.

there are four ways to disprove it.

1) the gravity at the top of a sky scraper should be greater than at the bottom.

It won't be. The reason is that at the bottom of a skyscraper, the Earth is closer, and at the top, the Earth is farther away. The skyscraper does also exert a gravitational force of its own, but it is much smaller. However, if you generally mean measuring gravitational forces around objects on Earth, we can in fact measure it for natural features like mountains and produce gravitational maps of the Earth. And in precise experiments, we can measure the gravitational force exerted by a mass less than a gram.

2) a large object going down a vacuum tube should take a shorter amount of time than a small one.

Why?

3) the earth should wobble in a 24 hour period. (this is his explanation for the tides)

It does not. This would be observable in astronomical measurements that are taken thousands of times a day around the world.

4) the gravity on the far side of the moon is greater than the near side.

It isn't. Lunar orbiters stay in orbit, which would be impossible if the gravitational field were greatly asymmetrical.

So I guess it's disproved, though a better point may simply be that electrons are not neutral, since they are the moving particles responsible for creating the electric currents used by the screen you are looking at. Electrons and protons can be deflected by magnetic fields, while neutrons cannot.

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### Re: mark mccutcheon final theory

Yeah if you want to make up absolute nonsense, at least have the courtesy to hide it in quantum shenanigans. The tides are a done fuckin deal.

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- Eebster the Great
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### Re: mark mccutcheon final theory

Though at the level of details, tides are still cool. When topography and bathymetry and geodesy and stuff are taken into account, it gets incredibly complicated really fast. They are obviously very well understood by some people, but not by me. The Bay of Fundy has a maximum tidal range of 17 meters. That's like a four-story building.

EDIT: I just found a review of the book that says "Velikovsky and von Daniken move over!" That's pretty high praise.

EDIT: I just found a review of the book that says "Velikovsky and von Daniken move over!" That's pretty high praise.

- phillip1882
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### Re: mark mccutcheon final theory

1) the gravity at the top of a sky scraper should be greater than at the bottom.

It won't be. The reason is that at the bottom of a skyscraper, the Earth is closer, and at the top, the Earth is farther away. The skyscraper does also exert a gravitational force of its own, but it is much smaller. However, if you generally mean measuring gravitational forces around objects on Earth, we can in fact measure it for natural features like mountains and produce gravitational maps of the Earth. And in precise experiments, we can measure the gravitational force exerted by a mass less than a gram.

yes i was fairly sure that was the case. the reason that under his theory that would be true, is a bit hard to explain, to summarize, the size of an object determines its gravity, for example if you took a long metal rod, say over a mile, and put a counter weight on the bottom, the gravity at the top should be proportional to the distance in an increasing manner.

2) a large object going down a vacuum tube should take a shorter amount of time than a small one.

Why?

because the size of the object determines its gravity, i can post his formula if you're interested.

3) the earth should wobble in a 24 hour period. (this is his explanation for the tides)

It does not. This would be observable in astronomical measurements that are taken thousands of times a day around the world.

yes i looked up the wobble of the earth on wikipedia, and found no mention of such a rapid wobble.

4) the gravity on the far side of the moon is greater than the near side.

It isn't. Lunar orbiters stay in orbit, which would be impossible if the gravitational field were greatly asymmetrical.

this i wasn't aware of. yes then his theory is definitely bunk.

the reason why to me it seemed plausible, is that everything to do with gravity can be expressed in terms of formulas that don't require force.

in particular, falling objects obey the formula distance fell = 1/2 *acceleration*time^2,

and orbiting objects obey the formula velocity^2 *distance to orbiting body = k where k is a constant.

good luck have fun

- Eebster the Great
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### Re: mark mccutcheon final theory

The first formula is actually not correct. It's an approximation that applies in a uniform gravitational field. For instance, if I fall a few feet to the ground, the gravitational field is about the same throughout my fall, so you can use y =

The formula v² r = k, as you put it, can't be right. For instance, if I jump, while I'm traveling up, v > 0 and d > 0, but at the peak of my jump, v = 0. You might mean v²/r is constant, but that's also wrong. v²/r is the centripetal acceleration of any object in circular motion and is not related specifically to gravity. Again, it comes from the definitions of acceleration and velocity. In uniform circular motion, both velocity and radius are constant, so centripetal acceleration is also constant. However, gravitational orbits are elliptical, not circular, and when the Earth (for instance) is at the part of its orbit where it's closer to the Sun, it actually moves faster, and when farther away, slower. So as r goes up, v goes down, and v²/r goes way down. As r goes down, v goes up, and v²/r goes way up. It is not constant.

It is true that when dealing with gravity, the mass of the small body often cancels out. In Newtonian gravity, this is because the force of gravity is proportional to the mass (F = GmM/r²), and acceleration (due to any force) is always inversely proportional to the mass (F = ma). So ma = GmM/r², the m's cancel, and we get a = GM/r². The acceleration depends only on the mass of the planet, the distance from the planet, and the gravitational constant G. This is why bodies of different masses fall at the same rate (neglecting air resistance).

In Einsteinian gravity, this is actually because gravity is not a force. The apparent ("fictitious") force results from observing inertial (freefalling) objects from non-inertial reference frames, like watching a straight pitch while standing on a rotating platform. It will appear to be pushed toward the wall by a mysterious "centrifugal force", but it's really moving straight and you are the one accelerating. The idea that someone in freefall can be inertial and someone standing still on the surface seems strange, and it has to do with the way mass curves spacetime in his theory. It's too much to go into here, but it's interesting stuff.

^{1}/_{2}g t², where g is the constant acceleration of gravity during that fall. Note that any object initially at rest with constant acceleration a travels a distance^{1}/_{2}a t² in a time t. This actually has nothing to do with gravity and is an equation of motion that comes directly from the definition of acceleration. In reality, gravitational fields are not uniform. As you fall toward the surface of the Earth from very far away, the force of gravity will increase, and the acceleration will therefore also increase in direct proportion. The gravitational force is strongest at the surface of the Earth.The formula v² r = k, as you put it, can't be right. For instance, if I jump, while I'm traveling up, v > 0 and d > 0, but at the peak of my jump, v = 0. You might mean v²/r is constant, but that's also wrong. v²/r is the centripetal acceleration of any object in circular motion and is not related specifically to gravity. Again, it comes from the definitions of acceleration and velocity. In uniform circular motion, both velocity and radius are constant, so centripetal acceleration is also constant. However, gravitational orbits are elliptical, not circular, and when the Earth (for instance) is at the part of its orbit where it's closer to the Sun, it actually moves faster, and when farther away, slower. So as r goes up, v goes down, and v²/r goes way down. As r goes down, v goes up, and v²/r goes way up. It is not constant.

It is true that when dealing with gravity, the mass of the small body often cancels out. In Newtonian gravity, this is because the force of gravity is proportional to the mass (F = GmM/r²), and acceleration (due to any force) is always inversely proportional to the mass (F = ma). So ma = GmM/r², the m's cancel, and we get a = GM/r². The acceleration depends only on the mass of the planet, the distance from the planet, and the gravitational constant G. This is why bodies of different masses fall at the same rate (neglecting air resistance).

In Einsteinian gravity, this is actually because gravity is not a force. The apparent ("fictitious") force results from observing inertial (freefalling) objects from non-inertial reference frames, like watching a straight pitch while standing on a rotating platform. It will appear to be pushed toward the wall by a mysterious "centrifugal force", but it's really moving straight and you are the one accelerating. The idea that someone in freefall can be inertial and someone standing still on the surface seems strange, and it has to do with the way mass curves spacetime in his theory. It's too much to go into here, but it's interesting stuff.

- phillip1882
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### Re: mark mccutcheon final theory

the equation v^2 *r = k

is essentially a rewrite of Kepler's third law, which states( distance^3 /time^2) = k

and is only meant to explain orbits, not free falls.

is essentially a rewrite of Kepler's third law, which states( distance^3 /time^2) = k

and is only meant to explain orbits, not free falls.

good luck have fun

- gmalivuk
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### Re: mark mccutcheon final theory

phillip1882 wrote:↶

the equation v^2 *r = k

is essentially a rewrite of Kepler's third law, which states( distance^3 /time^2) = k

and is only meant to explain orbits, not free falls.

Orbits are free falls, though.

Kepler's third is about the average distance and the time for the entire orbit. That equation doesn't hold for instantaneous velocity.

- Eebster the Great
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### Re: mark mccutcheon final theory

And that law does in fact depend on the inverse-square law of gravity, and it is not merely true by definition. (Nor does it rely solely on the fact that the gravitational force points directly at the Sun, like the first two laws.)

### Re: mark mccutcheon final theory

Right. Kepler derived his laws empirically, it was about seven decades later that Newton derived them from his laws of motion & gravitation.

a = GM/r² is true in the centre of mass frame, but when the second body has considerable mass relative to the first, then we also need to consider the acceleration of the first body towards the second. See Don't heavier objects actually fall faster because they exert their own gravity?

a = GM/r² is true in the centre of mass frame, but when the second body has considerable mass relative to the first, then we also need to consider the acceleration of the first body towards the second. See Don't heavier objects actually fall faster because they exert their own gravity?

Last edited by PM 2Ring on Tue May 28, 2019 7:33 pm UTC, edited 1 time in total.

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### Re: mark mccutcheon final theory

Eebster the Great wrote:And in precise experiments, we can measure the gravitational force exerted by a mass less than a gram.

It isn't. Lunar orbiters stay in orbit, which would be impossible if the gravitational field were greatly asymmetrical.

Planet. Weighing scale. Half-gram object. Simples!

Maybe interesting Scott Manley video about lumpy Lunar gravity: https://www.youtube.com/watch?v=EadClM4Y45A

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### Re: mark mccutcheon final theory

Going on a tangent here:

I don't see that answer section talking about (general) relativity. How does relativity affect the acceleration/time to impact? Do heavier objects fall faster (towards an even heavier object) on all mass scales, including big black holes? (What if the objects don't start at zero speed but in (circular) orbit around each other and they have to radiate their energy via gravity waves?)

PM 2Ring wrote:a = GM/r² is true in the centre of mass frame, but when the second body has considerable mass relative to the first, then we also need to consider the acceleration of the first body towards the second. See Don't heavier objects actually fall faster because they exert their own gravity?

I don't see that answer section talking about (general) relativity. How does relativity affect the acceleration/time to impact? Do heavier objects fall faster (towards an even heavier object) on all mass scales, including big black holes? (What if the objects don't start at zero speed but in (circular) orbit around each other and they have to radiate their energy via gravity waves?)

- Eebster the Great
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### Re: mark mccutcheon final theory

Sableagle wrote:Eebster the Great wrote:And in precise experiments, we can measure the gravitational force exerted by a mass less than a gram.

It isn't. Lunar orbiters stay in orbit, which would be impossible if the gravitational field were greatly asymmetrical.

Planet. Weighing scale. Half-gram object. Simples!

You are measuring the gravitational force of the entire Earth pulling on that half gram object. I'm talking about measuring the gravitational force of an object weighing less than a gram on an even tinier object.

### Re: mark mccutcheon final theory

phillip1882 wrote:1) the gravity at the top of a sky scraper should be greater than at the bottom.

Not only this is refuted by direct measurements (as Eebster the Great stated), but we can even measure the relativistic time dilation caused by the difference of gravity between these two points. A clock at the top of a skyscraper really does ticks faster, because there's less gravitational time dilation there.

Speaking of gravitational time dilation: This effect must be taken into account for GPS navigation to work. Every time you use a map on your cellphone, you're confirming general relativity to an outstanding precision. So any alternative model that gives a wildly different prediction to either the strength of gravity or its effect on time, can safely be declared wrong.

### Re: mark mccutcheon final theory

Flumble wrote:Going on a tangent here:PM 2Ring wrote:a = GM/r² is true in the centre of mass frame, but when the second body has considerable mass relative to the first, then we also need to consider the acceleration of the first body towards the second. See Don't heavier objects actually fall faster because they exert their own gravity?

I don't see that answer section talking about (general) relativity. How does relativity affect the acceleration/time to impact? Do heavier objects fall faster (towards an even heavier object) on all mass scales, including big black holes? (What if the objects don't start at zero speed but in (circular) orbit around each other and they have to radiate their energy via gravity waves?)

It gets messy, but you can see a GR version of the

F = GMm/r² formula here.

In Newtonian mechanics, any orbits involving 2 bodies can be solved analytically, but for 3 or more bodies there's no general solution. There are special symmetrical cases, but in general we have to resort to numerical methods (IOW, doing step by step simulations on a computer) to plot the trajectories, and to decide if a given system is stable.

In GR, there's no general solution for 2 bodies. And there's the added complications that different observers measure times & distances differently, depending on their location & state of motion.

When NASA need to do very accurate orbit calculations they use Newtonian gravity with some GR adjustments bolted on, aka post-Newtonian expansion.

- Eebster the Great
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### Re: mark mccutcheon final theory

PsiCubed2 wrote:Speaking of gravitational time dilation: This effect must be taken into account for GPS navigation to work. Every time you use a map on your cellphone, you're confirming general relativity to an outstanding precision.

Indeed, the correction is measured in microseconds per day and is less than one part in a billion. That miniscule difference still would produce a cumulative error in distance of about 10 km/day. And that's still tens of thousands of times larger than the drift between atomic clocks on the ground.

- gmalivuk
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### Re: mark mccutcheon final theory

PM 2Ring wrote:It gets messy, but you can see a GR version of the

F = GMm/r² formula here.

Deriving it is messy, but the end result is actually pretty clean, once you notice that it's just the Newtonian formula with a correction factor of 1/sqrt(1-s/r), where s is the Schwarzschild radius.

This same factor shows up in gravitational redshift and time dilation, and is thus pretty similar to how gamma=1/sqrt(1-v^2/c^2) shows up in special relativity.

(In fact, iirc, the two factors are equal if v is the escape velocity from distance r from the center of a "Schwarzschild mass", which I think is a handy shorthand for the neutral static spherically symmetric mass alone in the universe that the Schwarzschild metric applies to.)

### Re: mark mccutcheon final theory

Eebster the Great wrote:4) the gravity on the far side of the moon is greater than the near side.

It isn't. Lunar orbiters stay in orbit, which would be impossible if the gravitational field were greatly asymmetrical.

I hate to say it, but actually probes do have trouble staying in lunar orbit for very long due to gravitational asymmetry.

And gravity is, generally, slightly stronger on one side but he got the side wrong, the field is actually slightly stronger on the near (left, in below diagram) side:

Of course this has no bearing on the rest of his garbage.

- Eebster the Great
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### Re: mark mccutcheon final theory

I mean, it varies at extremes by 1.5% and on average from one side or the other by much less than 1%. That's kind of like the Earth. It is, for all intents and purposes, symmetrical. And lunar orbiters can potentially stay up for decades without station-keeping depending on the altitude.

I wasn't trying to suggest the Moon was actually a perfectly uniform sphere.

I wasn't trying to suggest the Moon was actually a perfectly uniform sphere.

- gmalivuk
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### Re: mark mccutcheon final theory

Also, surface gravity varies with elevation no matter how close to spherically symmetric the whole body is. It's higher in craters and lower on mountaintops, and the difference could theoretically be made as large as you like just by building a sufficiently tall and thin "spike" on the surface. This wouldn't affect orbits beyond the mass that's in the spike itself (plus the danger of hitting the thing, of course).

I suspect the presence of Earth is the main thing that makes lunar orbits hard. Earth's gravity can differ by as much as 2% from one side of the Moon to the other, and that difference increases with altitude (while anomalies from the Moon itself get smoothed out).

I suspect the presence of Earth is the main thing that makes lunar orbits hard. Earth's gravity can differ by as much as 2% from one side of the Moon to the other, and that difference increases with altitude (while anomalies from the Moon itself get smoothed out).

### Re: mark mccutcheon final theory

Eebster the Great wrote:I mean, it varies at extremes by 1.5% and on average from one side or the other by much less than 1%. That's kind of like the Earth. It is, for all intents and purposes, symmetrical. And lunar orbiters can potentially stay up for decades without station-keeping depending on the altitude.

I wasn't trying to suggest the Moon was actually a perfectly uniform sphere.

No I know, but orbiting a probe around the moon for significant periods *is* a bit more of an issue than around the Earth, there are orbits where long lifetimes can be had, but you have to plan it quite well (https://www.lpi.usra.edu/lunar/document ... P_3394.pdf).

### Re: mark mccutcheon final theory

Eebster the Great wrote:I mean, it varies at extremes by 1.5% and on average from one side or the other by much less than 1%. That's kind of like the Earth. It is, for all intents and purposes, symmetrical. And lunar orbiters can potentially stay up for decades without station-keeping depending on the altitude.

Actually, the Moon's gravity is far more asymmetrical than the Earth's. It is actually asymmetrical enough to render low orbits unstable, which was a very real problem for the early lunar missions (before NASA learned to take these variations into account).

There's no analogue problem for low earth orbits. As long as a satellite is above the atmosphere, it will follow a Keplerian orbit for a very very long time. In fact, the maximum deviation of the Earth's gravity from that of a perfect ellipsoid is around 0.01%.

For more information, see here:

Lunar mascons

The earth's gravity field (scroll down a bit for the relevant diagram from which I took the 0.01% figure from)

EDIT TO ADD:

Sorry for partially repeating what p1t1o said. He posted his comment while I was writing mine.

- gmalivuk
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### Re: mark mccutcheon final theory

Still, the *surface* gravity is not the most useful illustration of that.

### Re: mark mccutcheon final theory

To be clear, mccutcheon is still a nutbar.

- gmalivuk
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### Re: mark mccutcheon final theory

phillip1882 wrote:the equation v^2 *r = k

is essentially a rewrite of Kepler's third law, which states( distance^3 /time^2) = k

The correct equation for this is known as the vis-viva equation:

v^2 * r = GM(2 - r/a) where 'a' is the semi-major axis.

Clearly, if r changes, then 2-r/a changes and therefore v^2 * r changes.

(If r doesn't change, then yes, you have v^2 * r = GM, the relationship between orbital speed and radius for perfectly circular orbits. But that ignores all real-life orbits, which are elliptical and which thus have an r that changes.)

- Eebster the Great
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### Re: mark mccutcheon final theory

I have to admit that I did not expect the deviations to be that much. After subtracting the free air anomaly, differences in extremes can still be upwards of 1.5 cm/s

But it's not, like, actually a lot. That's still pretty small.

^{2}, or 0.9%. Even on average from one side to another, the difference is a few tenths of a percent, which is a lot.But it's not, like, actually a lot. That's still pretty small.

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### Re: mark mccutcheon final theory

PsiCubed2 wrote:The earth's gravity field (scroll down a bit for the relevant diagram from which I took the 0.01% figure from)

That's not the same information, though. The map of the Moon is about gravity at the surface, and the image at that link is about gravity at sea level. Furthermore, it's about the difference between actual gravity and what would be expected if Earth were more symmetrically dense, but even that expectation varies between the poles and the equator due to rotation and the equatorial bulge.

The strongest gravity on Earth's surface is 0.715% more than the weakest. Sure, that's less than the 1.5% on the Moon, but almost two orders of magnitude more than the +/- 0.01% anomaly at sea level.

### Re: mark mccutcheon final theory

True. Surface gravity maps are quite misleading (especially when you realize that the difference of gravity due to altitude alone from the highest to lowest point on the moon would be around 2%).

Another thing to consider, is that a raw difference in gravity does not necessarily translate into an "asymmetry". The gravity field of a gently oblated ellipsoid (a good approximation of the earth) is quite symmetric. The lunar mascons are distributed more randomly, hence their noticable effect on the stability of lunar orbits.

Another thing to consider, is that a raw difference in gravity does not necessarily translate into an "asymmetry". The gravity field of a gently oblated ellipsoid (a good approximation of the earth) is quite symmetric. The lunar mascons are distributed more randomly, hence their noticable effect on the stability of lunar orbits.

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