How to determine Gravity

How can I determine the gravity of Earth and moon using MM's math?

How exactly does the E/M field relate to magnetism and / or electricity?  Could I increase the E/M field density on a sphere using magnetism and / or electricity?

What experiments can I do to test MM's theories?

If everything is constantly expanding, would this explain "Pangea"?

parkham, Hello sir/ma'am.

I'm happy to oblige. I'm not him, but I like his ideas.

I see by the view count you're probably waiting for a quick answer.
I'm an unpaid laborer doing chores at the moment... Give me 6-8 hrs or so. Feel free to clarify.

Welcome

Thanks, Airman. I was popping in periodically, sorry about that. I appreciate any help you're willing to offer.

parkham,

Thanks for the opportunity to reply, it's a good way for me to review Miles' ideas. Of course, his answers would be much better. If you e-mail him he'll probably answer, so maybe we clean up any small misunderstandings first.

parkham wrote:How can I determine the gravity of Earth and moon using MM's math?

Since your first and last questions concern gravity, you may appreciate, or be in the middle of The Third Wave series, here are two of them:

171. The Third Wave: a Redefinition of Gravity, Part I. http://milesmathis.com/third.html No curved space, no force at a distance, no force-carrying particle. Gravity is explained here simply in terms of straight-line motion. Mass, inertia and gravity are all shown to be equivalent, reducible to the same motions. 9pp.

177. The Third Wave: a Redefinition of Gravity, Part VII. http://milesmathis.com/third7.html The Moon as proof of Expansion Theory. Mass and weight using both the gravitational field and the E/M field. 15pp.

Miles has many papers relating to gravity.

Back to the question, The Book of MM’s Math, hasn’t been written yet. Miles sure makes it look easy, but I for one am barely poking along.

It’s safe to say we would begin the same way we begin today, with Newton’s famous g=G(m1m2)/r^2. Using relative masses we could calculate earth’s apparent gravity.

We now know that that apparent gravity is actually the summation of two opposing fields, solo gravity and the charge field. The problem becomes how to determine the charge field component. I'll leave a more detailed answer to a discussion.

parkham wrote:How exactly does the E/M field relate to magnetism and / or electricity?  Could I increase the E/M field density on a sphere using magnetism and / or electricity?

By definition, the E/M field is electricity and magnetism. Miles went further by assigning the cause of electric and magnetic fields to individual photons. The so-called pre-electric field corresponds to the forward momentum of the photon, and the pre-magnetic field results from the photon’s tangential spin. We cannot see individual photons displaying E/M characteristics but they do; our technology is based on moving electrons and their associated magnetic fields, what’s doing the real work is photons.

Absolutely, you can increase the E/M field density by increasing voltage potentials or frequencies. Just keep in mind that electric currents and associated magnetic currentsfields result from net charge flow movement. By increasing electrical potential, you’re actually increasing the number of charge photons which push the electrons along.

parkham wrote:What experiments can I do to test MM's theories?

That depends on you of course. For 300 years, mainstream has been forced into fabricating solutions without benefit of the charge field. Everything needs to be rewritten. The most important thing to keep in mind is the existence of the charge field - it’s a fact of matter - any and all matter. Charge and gravity are in opposition at all scales.

parkham wrote:If everything is constantly expanding, would this explain "Pangea"?

Gravitational expansion is the theory where the earth’s gravity can be described as an earth radius doubling every seventeen minutes. As the earth expands, the measuring sticks also expand, so one is unaware of the expansion.

Pangea is a separate topic. As far as I know, Miles hasn’t talked about earth history. I personally believe that 65 million years ago the earth’s radius was half what it is today. The earth’s gravity was smaller and charge field stronger. I also believe that over time, matter tends to increase or decrease according to relative long term abundance or scarcity of general matter, i.e. photons and higher matter in our galactic neighborhood.

Discussion?
.

Thanks for taking the time to reply. Once I figure out the gravity equation using MM's compound idea, I'll post it here. I'm still trying to wrap my head around it.

I sent a msg, maybe it doesn't work. I assume you're having a problem with mass, is that it?

From moon.html :

"I have just postulated that the E/M field is proportional to 1/r4"  -

Is r4 = r^4.  Is r the radius of the planetary body?  If so, then for earth (radius = 6371 km), it would be (1/(6371^4)) ?

But, he also wrote to me that:  "The foundational e/m field strength or charge of a body is determined by its mass and density, simply multiplying the two together, see my papers on axial tilt and Bode's law for the simple math"

If I understand correctly, we need only the following things to determine the E/M field and gravity on given planetary body with a know constant (our Sun):

Radius
Density
Mass
Rotational speed (if any)


I'm getting referred to papers but I just can't figure it out.  I'm getting lost.  MM seems to have discovered something great, but I need some hand holding to get a formula for trying it out on other bodies.  

Specifically, would someone give an example, step by step, please?  

Example:  "Here is the formula, here is what we need to plug into the formula, here is how we get the numbers we need to plug into the formula."

I'm buried with papers on the why, I just don't know the how.

parkham wrote. From moon.html :
Airman wrote. The Moon Gives up a Secret

Miles wrote:"I have just postulated that the E/M field is proportional to 1/r4"

Is r4 = r^4.  Is r the radius of the planetary body?  If so, then for earth (radius = 6371 km), it would be (1/(6371^4)) ?
Airman. Correct. r4 = r^4. r is usually the distance to the planetary body center, but here, I believe we want to determine the acceleration due to gravity on the moon surface, so r equals the moon’s radius.

But, he also wrote to me that:  

The foundational e/m field strength or charge of a body is determined by its mass and density, simply multiplying the two together, see my papers on axial tilt and Bode's law for the simple math

If I understand correctly, we need only the following things to determine the E/M field and gravity on given planetary body with a know constant (our Sun):

Radius
Density
Mass
Rotational speed (if any)

Airman. I don’t see how rotational speed is a factor, otherwise, OK.

I'm getting referred to papers but I just can't figure it out.  I'm getting lost.  MM seems to have discovered something great, but I need some hand holding to get a formula for trying it out on other bodies.  

Specifically, would someone give an example, step by step, please?  

Example:  "Here is the formula, here is what we need to plug into the formula, here is how we get the numbers we need to plug into the formula."

I'm buried with papers on the why, I just don't know the how.

Airman. Miles shows that the Earth and the Moon have gravitational fields that are exactly proportional to their radii. To determine the gravity, all that’s needed is the radius. However, if you were on the moon you would not feel that calculated gravity, you also ‘feel’ lighter because of the moon’s charge field, (E/M), radiates photons upward, in opposition to the moon’s gravity. You can calculate the charge field directly from density times mass. You then add the moon's gravitational and charge field emission to get the resultant apparent gravity.

PS. Even though we're talking about mass 'expansion', the gravity vector is pointing downward. The charge field vector is 'in opposition' pointing upwards.

I can't do it and get the same results that he did in moon.htm. My numbers are huge and the answers I get aren't even in the ballpark.

That's why I was asking for some hand holding so I can "get it" and then experiment with other bodies.

Thanks for responding and taking time to help. Regarding rotation, MM stated rotation has an effect on E/M .

Rotation isn't affecting the earth's or moon's gravity as far as I know.

You're in the middle of Miles' calculations. Can you show the offending calcs?

Miles likes to work in 'normalized' (relative to say earth) values. That's unusual; I haven't encountered anyone else able to use normalized values so easily.

I'll grant that rotation IS a factor, but it doesn't change true gravity.

Apparent gravity (as opposed to solo gravity) varies over the surface of a planetary body. We must separate out the true gravity portion to see the variations occur in the e/m field.

The main variation of the e/m field is due to rotation. We can expect to see the least amount of earth emissions - e/m, directly above earth's poles.

Maximum e/m occurs over the +/- 30 deg latitudes, or more easily, the equator.

This affects the apparent gravity. Not the solo or true gravity - which is based on radius alone.

P.S. OK I didn't see your e/m clarification till now.

I'm using the closest thing in his papers that I can comprehend to attempt a formula. The problem is, I just don't get it, that's why I'm asking for help. I mean, why refer me to more and more papers?

I wish I could see this being taught as it would be taught to someone in high school. "Here's the formula, here's the preceding calculations, and here's where the numbers come from and how we get said numbers." Maybe my brain doesn't work right, but I truly think I would "get it" if it was presented to me as such.

Regarding rotation, MM told me that: "You can also increase charge by increasing spin, which increased the rate of charge recycling." However, I don't know the math formula to determine some sort of ratio.

I apologize, I'm not very good at explaining things. I'm high school algebra and basic geometry at best. But I can do algebra, when I know the equations and formulas. I'm just lost.

parkham, I think I get it.

How to teach it to kids?

You’re asking for lesson plans with numerical examples. And equation lists.

There are none that I'm aware of.  This site does have a lot of information. Miles gives frequent numerical examples, but there are no assembled book of examples for any particular grade that I'm aware of.

Are you telling me to make some? Are you tasking me Sir/Ma'am? For the kids, no less.

What a wonderful suggestion.

We're trying to learn and grow here ourselves. Something needs to be done. Who shall do it? Maybe we can draw lots.

Thank you parkham, You’ve made my day.

P.S. A correct mental image beats a whole lesson plan.
.

I can't tell if you're being facetious or not.  Serious or not:  I am not telling anyone to do anything.  I am merely asking for help, handholding I'm sad to report.  The only thing I can tell you is that if I eventually understand the math behind this, and if I still believe it then - I will lay it out as simple as I can for others to follow.  I will then follow it up with calculations against other celestial bodies - maybe even against objects on earth, just to experiment.

On a related note - (I was posting this when I saw your last message)  So far, taking this one slow step at a time, I came upon another paper.

OK, I might be getting closer to something:  From uft2.html :  Solve, DV is of Earth and dv is of moon

F= G(DV)(dv)/r2

Using MM's formula above, the "average" "force" of gravitational "attraction" between the earth and the moon is:
1.96595E-24

edit:  I got 1.98581E-26 after a recalculation

MM goes on to say:

Once we have density and volume in Newton’s equation, we can assign density to one field and volume to the other. We let volume define the gravitational field and we let density define the E/M field. Both fields then fall off with the square of the radius, simply because each field is spherical. There is nothing mysterious about a spherical field diminishing by the inverse square law: just look at the equation for the surface area of the sphere:

S = 4πr2

Double the radius, quadruple the surface area. Or, to say the same thing, double the radius, divide the field density by 4. If a field is caused by spherical emission, then it will diminish by the inverse square law. Quite simple.

Still reading and trying to understand

Hi Parkham,

I've done some work in this area before (it has been a while since I looked at it though) so I will refactor some programming code I have that calculates gravity and charge vectors for a given body.

Let:
r = radius of body;
g = solo gravity of body;
d = density of body;
m = mass of body;
e = charge of body;
v = resultant length of gravity vector;
SOLO_GRAVITY_EARTH = 9.82;
EM_FIELD_EARTH = -0.009545;
RADIUS_EARTH = 6378000;
GRAVITY_EARTH = SOLO_GRAVITY_EARTH + EM_FIELD_EARTH;
DENSITY_EARTH = 5515;
MASS_EARTH = DENSITY_EARTH  * ( 1.3333333334 * PI * RADIUS_EARTH^3 );

Given:
r, d;

Then:
// we know g of earth, so assume g is proportional to radius and calculate from earth numbers
g = SOLO_GRAVITY_EARTH * ( r / RADIUS_EARTH );
// note: PI = 3.14, not 4
m = d * ( 1.3333333334 * PI * r^3 );
// find mass as a ratio to mass of earth
mr = m / MASS_EARTH;
// find charge strength at surface of body
e = EM_FIELD_EARTH * mr * ( d / DENSITY_EARTH );
// add solo gravity and charge to get resultant gravity
v = g + e;

v is the length of any vector pointing out from the surface of the body in any direction. In the case of the Earth it would be 9.81. That is a general solution and does not take charge recycling of the body into account. The gravity value is only applicable at the surface of the body.

I used this to show the relative length of the solo gravity, charge and resultant gravity vectors in a little applet I developed years ago. You can vary the parameters to see how the vectors were affected.

If we want to determine the interaction of 2 bodies, then we need to take charge dissipation into account. Charge drops off with an inverse r^4 relationship (ignoring time differentials). I can't find the relevant papers at the moment but you need to find the charge density of the primary body at the surface of the other body. Find the charge of the other body and determine the distance between the 2 bodies where the 2 fields equal. I think Miles paper on the orbit of Mercury will help.

I agree with you that Miles could be a bit more explicit in his math. I'm sure it all looks so obvious to him but as an amateur physicist it can be really hard to figure out what equations he is using. It helps to have the equations of motion in view somewhere and then try to figure out which ones he is using in a particular problem. Look at the values he is using and try to figure out what they are and where they come from. The related text often tells you what he is using but the math just dives in and starts calculating.

I should create a new section on my site called 'Mathis for Dummies' (no insult intended, I put myself into that group) where it provides better descriptions of the equations and what other equations he is using for a given problem.

Anyway, I hope that helps a little bit.

Nevyn, Hello. Thanks. We all appreciate a simplified view. Reviewing the above I see I would benefit greatly from a 'dummies' section.

I'll try to expand on these other questions.

parkham wrote:How exactly does the E/M field relate to magnetism and / or electricity?

It depends on what you mean by E/M field. In Miles theory, we don't really care too much about the E/M field, we really care about the charge field which is made up of charge photons (usually IR photons). The motions of the charge field directly cause the motions of the E/M field (which is electrons and protons) just by collisions. Our machines only measure the E/M field and not the charge field directly and that is why mainstream science thinks it is the E/M field doing all of the work. They just don't have any underlying theory for the E/M field so they never look below it.

Given a real field with real particles, as Miles has described, we can get a better understanding of what electricity and magnetism really are (or at least a working theory). Each charge photon has a linear velocity and a set of spins. We don't really need to know too many details about those spins but it is important to remember that they do have spin and the outer spin is the one we do care about in this problem. The linear velocity is electricity and the top-level spin is magnetism.

To get a good picture of how that works think about a current carrying wire. We know that the magnetic field around that wire causes a torque (or tangential force) which is caused by the spinning of the charge photons being emitted from the wire as the current runs through it. The charge photons are moving out from the wire in all radial directions (that is, from the center of the wire directly to the outside of it and it keeps going). So it is easy to see that a force exists that would push things away from the wire but not quite as easy to see how it creates a torque.

For that we just need to look at which way the charge photons are spinning. Imagine you are sitting next to the wire so that it runs from your left to your right (direction is not important just that it is parallel to you). Now imagine a single charge photon leaving the wire and travelling directly at your eyes (so perpendicular to the wire). The torque we are looking for will cause a force up or down from your perspective. So we just need our charge photons to be spinning in such a way that it creates that force. If each charge photon is spinning about an axis that is still parallel to the wire, then it can easily create that force up or down depending on whether it is spinning clockwise or counter-clockwise (when looking down the spin axis of that photon, not from your perspective).

To get a better picture of that, imagine you have a basketball in your hands and are holding it out in front of you. You put one finger of your left hand on the left edge and a finger on your right hand on the right edge. You can either spin that ball up or down, not left or right and not front to back. The line between your fingers that are touching the ball is the rotational axis. If you could look from the side, so that you could only see one of your hands, then you would see that the ball is spinning clockwise or anti-clockwise from that perspective. Now imagine you have a little pea-shooter in your mouth, if you shot one directly at the spinning basketball then it will be deflected up (assuming the closest edge of the ball is moving up from your perspective). That is magnetism in a nut-shell.

parkham wrote:Could I increase the E/M field density on a sphere using magnetism and / or electricity?

No, it would be the reverse. If you increased the charge density, then that would increase the electricity and it can change the magnetism. It is difficult to say how magnetism would be affected. Magnetism, as we measure it, is really a sum of the magnetic component of individual charge photons. So if the existing charge was primarily spinning a given direction and we added more charge that also spun in that direction then we would increase the magnetism. However, if we added charge that spun in the opposite direction then it would decrease magnetism, possibly even reversing it. The values also matter. Say we define the magnetism as a value between -1 and 1 inclusive, where a positive magnetism gives a torque to the right and a negative magnetism gives a torque to the left (or up and down or front and back, direction is not important). If our initial magnetism had a value of 1, say, and we introduce an equal amount of opposite spinning charge photons (so they will sum to a magnetism of -1), then the resultant magnetism value will be 0, which is no magnetism. If we added in twice as much charge as we started with then it would sum to -0.5 which means we have reversed the magnetic direction (we would have to remove all of the existing charge in order to reach -1).

parkham wrote:What experiments can I do to test MM's theories?

Tough one. Miles is really about providing explanations for the experiments we have already done. He did perform an experiment relating to the bending of star light which is explained in one of his papers. But I think you want something that doesn't require travelling all over the globe to take measurements, something that you can build in your home and I don't really have any ideas in that regard, sorry. But if you come up with something, feel free to post about it on this forum and we can try to help with it.

parkham wrote:If everything is constantly expanding, would this explain "Pangea"?

No, that is a different kind of motion, although the expansion could be a cause of it. Miles has written a geology paper but it has been too long since I read it.

Thank you all very much for taking the time to respond.  I'm going to digest this information when I have some quiet time and respond in more detail later.  One thing MM stated about charge though:

[..] because the charge field from the Sun is a constant.  But you could increase charge in any body by increasing the ambient charge field it is recycling.

 does he mean magnetism / electricity too?  Maybe I'm wondering again.. is it all just about photons.  I know this is probably a dumb question, but what is the relation between the Sun's charge (photons?), electricity and magnetism?  (OK, I'm going to add charge.html charge2.html to the reading list)

 In other words, if you increased the size of our Sun, it would emit more charge, and therefore all bodies would recycle more charge.  You can also increase charge by increasing spin, which increased the rate of charge recycling.  But in most cases you don't have to worry about that, because it isn't happening and can't happen.  You are dealing with givens.

He adds:

Charge comes from the ambient field. It is recycled by all bodies. The Earth gets its charge from the Sun, the SUn gets its charge from the galactic core, and so on. And no, I don't know where charge comes from beyond that. It just is. Photons exist.

I am not sure what to make of the statement that the Sun gets its charge from the galactic core - but that's another big question not quite related to what I'm attempting to determine at this time.

Nevyn:  

DENSITY_EARTH = 5515;

You're using the density as 5515.  How did you get this number, what is it measured in?  

The number I'm using is 0.00005514 (g/km3)

For MASS_EARTH, I got 599359882867371x10^10 kg (or 5,993,598,828,673,710,000,000,000.00 kg - my scientific notation might be off)

Is this correct?

For Mercury I got:

r = radius of body = 2,439,700.00
g = solo gravity of body = 3.7563270617748500
d = density of body = 5,427.00
m = mass of body = 330,109,261,861,959,000,000,000.00
mr = mass as a ratio to earth = 0.0550769698303292
e = charge of body = -0.0005173211998630
v = resultant length of gravity vector = 3.7558097405749900

Long story short, gravity on Mercury is 3.75 m/s² which is .333 to infinity of what earth is?

I'm not sure where the calculations went wrong: I did Jupiter and its results came back 107.0286 m/s²

Sorry, I should have shown the units. Everything is in meters, seconds and kilograms. So the density is kg/m^3, velocity is m/s, acceleration m/s^2, mass is kg.

Your calculations may be off if you are mixing units. Make sure everything is the same. You can't compare kg to g/km^3 without a conversion. I haven't checked your values so I don't know if this is the case, but it could be.

parkham wrote:One thing MM stated about charge though:

Mathis wrote:[..] because the charge field from the Sun is a constant.  But you could increase charge in any body by increasing the ambient charge field it is recycling.

 does he mean magnetism / electricity too?  Maybe I'm wondering again.. is it all just about photons.  I know this is probably a dumb question, but what is the relation between the Sun's charge (photons?), electricity and magnetism?  (OK, I'm going to add charge.html charge2.html to the reading list)

Yes, it can increase electricity and magnetism but those properties are not as important as the charge density in this particular problem.

Yes, it is all about photons. Everything is photons. All interactions are between photons. Electrons, protons and neutrons are just photons with more spins. In Miles theory, the only thing that actually exists is the BPhoton, which is just a little bit below what science calls a photon. Give the BPhoton an axial and an X spin (you'll have to study stacked spins to understand this) and it becomes a photon. Give that photon a lot more spins and it becomes an electron, a few more and it becomes a neutron or proton.

parkham wrote:

Mathis wrote: In other words, if you increased the size of our Sun, it would emit more charge, and therefore all bodies would recycle more charge.  You can also increase charge by increasing spin, which increased the rate of charge recycling.  But in most cases you don't have to worry about that, because it isn't happening and can't happen.  You are dealing with givens.

That quote, taken out of context, is a little bit misleading. Increasing the size is not enough to increase charge emission. It is the density that affects charge emission and size is related to gravity. Miles is still correct but he is implying that the density of the sun remains the same as the size is increased which would increase charge emission because there is more mass recycling it. However, if you just took the existing matter in the sun and stretched it out into a bigger sphere, then the charge density would not be affected because the density of the sun would decrease in proportion to the size increase. Conversely, if you kept the sun at the same size and increased its density, then the charge emission will increase which we would measure as a decrease in gravity.

parkham wrote:He adds:

Mathis wrote:Charge comes from the ambient field.  It is recycled by all bodies.  The Earth gets its charge from the Sun, the SUn gets its charge from the galactic core, and so on.  And no, I don't know where charge comes from beyond that.  It just is.  Photons exist.  

I am not sure what to make of the statement that the Sun gets its charge from the galactic core - but that's another big question not quite related to what I'm attempting to determine at this time.

This is not too important to Miles theory, he is just extrapolating upwards. Photons must come from somewhere so we look around the sky and see a lot of energy in the galactic core. We see galaxies with arms and jets coming out of them. We see solar systems in the galaxies that try to stay within those galactic arms. We know that the earth and sun has variances in their properties like magnetism and Miles takes this to mean the sun is moving in and out of different charge (and anti-charge) densities from the galactic core. Miles doesn't concern himself with cosmology too much and this is just a little venture into that field.

parkham wrote:Is this correct?

For Mercury I got:

r = radius of body = 2,439,700.00
g = solo gravity of body = 3.7563270617748500
d = density of body = 5,427.00
m = mass of body = 330,109,261,861,959,000,000,000.00
mr = mass as a ratio to earth = 0.0550769698303292
e = charge of body = -0.0005173211998630
v = resultant length of gravity vector = 3.7558097405749900

Long story short, gravity on Mercury is 3.75 m/s² which is .333 to infinity of what earth is?

I loaded Nevyn's data and formulas into an excel doc. Your numbers - the significant digits that is, agree with mine. Here's an image. The numbers are in scientific notation.

.

Nevyn wrote:Sorry, I should have shown the units. Everything is in meters, seconds and kilograms. So the density is kg/m^3, velocity is m/s, acceleration m/s^2, mass is kg.

Your calculations may be off if you are mixing units. Make sure everything is the same. You can't compare kg to g/km^3 without a conversion. I haven't checked your values so I don't know if this is the case, but it could be.

The reason I was confused here is I got an accurate reading for Mercury and then plugged the Jupiter values (using the same units) in and it was 4 times larger than what is commonly reported.

I'll convert everything you have to the appropriate units as well. Thanks!

There is something funky going on in my math. Some planets work fine but most do not.

Here is the data I get, compared to values retrieved from this website:

Body	Calculated Value	Value	Ratio
Sun	276.8390157047	274	1.01
Mercury	3.75627145	3.7	1.01
Venus	9.3106980148	8.87	1.05
Earth	9.810455	9.79	1.002
Mars	5.2279784001	3.71	1.4
Jupiter	109.2967918347	24.92	4.386
Saturn	92.6677175362	10.44	8.876
Uranus	39.3197904317	8.87	4.433
Neptune	38.0790428609	11.15	3.415
Pluto	1.8398964695	0.58	3.17

Note that the value for the Earth has changed, all of this math is based on a value of 9.81. This is not enough of a difference to explain the discrepancies.

Looking at the ratios we can see that the inner bodies are fairly close. Mars starts to jump out a little bit more but the Jovians are really different, especially Saturn. I would tentatively suggest that it is Jupiter and Saturn that are causing this. You see, planets don't exactly orbit the sun, they orbit the sun plus all planets on smaller orbits. This is especially true for the planets outside of Jupiter and Saturn, which sort of act like a second sun. Notice how after Saturn the ratios start to come down again. There isn't enough data to know if that means anything or not but it is suggestive.

I think Miles papers on Axial Tilt and Bode's Law will help to understand this.

I attempted to factor in equatorial rotation speed of the planets in an effort to see if things would line up.  By coincidence(?) only, Jupiter worked right off the bat.  I thought "great!".  Sadly, the rest of the planets did not give the expected results though.

Here is what I'm working with:

Mercury - 3.76
Venus - 9.32
Mars - 5.22
Jupiter - 107.00
Saturn - 89.64
Uranus - 39.06
Neptune - 37.90
Pluto - 1.83
SUN - 260.68
Probably not of any importance?  But here are the rotational speeds at the equator of each planet, in meters per second:

Mercury - 3.03 m/s
Venus - 1.81 m/s
Mars - 241.17 m/s
Jupiter - 12,444.44 m/s
Saturn - 9,690.00 m/s
Uranus - 2,590.00 m/s
Neptune - 2,683.33 m/s
Pluto - 13.11 m/s
SUN - 1.91 m/s


How did you calculate earth?  I though earth was the constant on which all the other planets were measured.

Here are my numbers.

Nevyn wrote: Looking at the ratios we can see that the inner bodies are fairly close. Mars starts to jump out a little bit more but the Jovians are really different, especially Saturn. I would tentatively suggest that it is Jupiter and Saturn that are causing this. You see, planets don't exactly orbit the sun, they orbit the sun plus all planets on smaller orbits. This is especially true for the planets outside of Jupiter and Saturn, which sort of act like a second sun. Notice how after Saturn the ratios start to come down again. There isn't enough data to know if that means anything or not but it is suggestive.

The calculations depend on known densities; but the planetary densities are calculated from faulty orbital velocity equations, and Miles knows what else. We expect deviations.

I agree, we are using the earth as the standard.

I would argue that we should use the much better known earth data to predict more accurate planetary densities.
.

Does that mean that Nevyn's math calculations is correct and reported densities are truly off, sometimes extremely so?  If so, then reported gravity is also off as well, yes?  It would have to be in order for MM's theory to be true, correct?

It seems the biggest divergence is with the gas / ice giants.  I wonder why that is?  I thought that it might be something to do with radius, but the Sun's radius is huge and its results aren't horribly off. Also, the Sun is negatively off while everything is positively off (if that makes sense).

I am standing on shaky ground, but for some reason I can't "shake" the hunch that rotational speed * has * to have some part it in all this.

The e/m field is balanced against gravity - producing the apparent/actual gravity differential. I don't believe variations in the e/m field - such as rotation - are necessary for determining the 'rough' resultant planetary gravity in these examples. The difference between these calculated numbers and the official numbers are much greater than the body's e/m field variations. On the other hand, along with the 'knowns' come mainstream assumptions that we are trying to replace. We do need to consider additional factors such as planetary tilt and orbital inclination; that's why Nevyn gave additional references.

I tend to believe Miles' or Nevyn's work, unless I'm practicing being thrown to the mat.
.

Nevyn:

Why do we have to use constants based on the earth in order to determine gravity and e/m field of other planetary bodies?

parkham wrote:Does that mean that Nevyn's math calculations is correct and reported densities are truly off, sometimes extremely so?  If so, then reported gravity is also off as well, yes?  It would have to be in order for MM's theory to be true, correct?

It is important to remember that the mainstream numbers are calculated based on assumptions that Newton's equations are perfect and the only force in play. We have never measured the gravity of the sun, we just assume some things and back-calculate. Same with all of the planets except Earth.

However, Miles makes a similar assumption in Newton's gravity equation. He assumes it is correct and then goes further and pulls it apart to show how size and density work individually.

It is also important to realise that we don't live in a gravity only universe and the way charge plays its part is complicated. We are talking about how to calculate the surface gravity of bodies which is very different to determining orbits. The mainstream only has gravity so they stop there and assume everything is fine and dandy. We have charge and gravity and both are required for an orbit to balance. This is why Jupiter and Saturn are so important. They are huge and have large charge fields that affect the rest of the field. This will push all of the planets further out from them further away and possibly the inner planets will be pushed further in, although the Sun's charge gets denser the closer you are to the Sun so it is a reduced effect than that imposed onto the outer planets where the charge from Jupiter and Saturn combine with the Suns charge.

parkham wrote:It seems the biggest divergence is with the gas / ice giants.  I wonder why that is?  I thought that it might be something to do with radius, but the Sun's radius is huge and its results aren't horribly off.  Also, the Sun is negatively off while everything is positively off (if that makes sense).

I am standing on shaky ground, but for some reason I can't "shake" the hunch that rotational speed * has * to have some part it in all this.

The Sun works because it is the center of the solar system. It does not depend on any other bodies. It sets the field and everything else is determined with respect to it. You can almost say the same about Jupiter and to a lesser extent, Saturn. The combined charge fields of Jupiter and Saturn basically reset the field. Saturn and Jupiter can sometimes be thought of as a single entity. They play an intricate game with each other because Saturn is smaller than Jupiter so it wants to go below it. A stable solar system has smaller planets orbiting inside of bigger planets. We don't live in such a system and the Jovians are the main deviation. Everything above Jupiter wants to move below it in order to find its stable position. Saturn tries to do so and has enough size to make some headway but ultimately, Jupiter overrules it and pushes it back out. There may come a day when it doesn't and the whole solar system will be thrown into turmoil for a while.

The rotation rate of the Sun and each planet does have a part to play because it can affect the charge density of that body. In this problem though, they are constants so we can ignore them. We don't have an equation to determine the charge density that includes the rotation rate so we couldn't use it anyway. This will need to be developed at some stage. Prime area to work in if you want to submit a paper to Miles for publishing. I'm not quite sure where to start with that. Probably create a spreadsheet containing the basic properties of each planet such as radius, density, gravity, etc. Now that I start to think about it, I've already started in my post above with the table of gravity values. Add in the known rotation rates and any other properties you think are applicable and see what you can calculate from it all. Use Miles approach of calculating relative values and see how they relate to each other. There may be fame, but probably no fortune, waiting for anyone that wants to put the time into it.

However, given all of that and probably some more I have forgotten, my calculations could still be incorrect. Miles equations may be on the right path but not quite complete yet. Or it all could be just plain wrong, but I don't tend to believe that. The equations I have used mostly came from the 'Moon gives up a secret' paper and that was only concerned with the Earth and Moon system. We are extrapolating up to the whole solar system and so much more comes into play that I think it feasible to forgive some discrepancies as long as there are reasons for it. If you think about it, the equations have worked for those situations that are most like the Earth/Moon system and fail when other entities have affects large enough to change the situation.

parkham wrote:Nevyn:

Why do we have to use constants based on the earth in order to determine gravity and e/m field of other planetary bodies?

Because we have measured data for the Earth but no other planets.

I can give a slightly better answer than that.

Miles likes to use relative values in his calculations. Well, I'm sure he would prefer to use concrete values (ie measured), but we often don't have that data available, or at all. He is often looking for relationships between things so relative values are a good way to work. Sometimes the absolute values don't look like they are related, as Miles points out in the 'Moon gives up a secret' paper, but the relative values show a different story and with a little bit of work you can find out why the numbers don't seem to add up to what you want.

Miles has shown, time and time again, how the mainstream gives up too early. They might look at the gravity of the Earth and Moon, say, and compare the values, look at their radii and compare those and find that the relationship doesn't work. But Miles has the charge field and this changes the assumptions and gives you something to fill the gap which he seems to have done quite nicely. Instead of seeing the numbers that don't match and moving on to some other problem, Miles asks why don't they match, what could cause them to be different. He then tries to find answers to those questions, and the math to show the gap. If he does so, then we get a paper on it, but I'm sure there have been cases where he didn't.

Thank you very much. I'm going to re read your responses again before responding.

OK.  I've decided to go back to uft.html  and really, really, really read it.  I thought I understood but, I still don't.  

Would someone please clarify H = m(A + a) where we would be dealing with the earth and the moon?   Starting at " Let us use the letter H.  "

I assume that m = mass of moon 73.50, M = mass of earth 5,973.60, A = acceleration of earth (9.81?), a = acceleration of moon (1.62?)

.
parkham said. OK.  I've decided to go back to uft.html  and really, really, really read it.  I thought I understood but, I still don't.

9. The Unified Field Theory. http://milesmathis.com/uft.html Newton's gravitational equation is shown to be a compound equation that includes the E/M field. I mathematically separate the two field equations, do transforms on them, and create simple and useful Unified Field Equations. 24pp.

Would someone please clarify H = m(A + a) where we would be dealing with the earth and the moon?   Starting at " Let us use the letter H.  "
Airman says. I welcome any opportunity to reread papers and answer questions. I didn’t see your last post till a couple of hours ago. Maybe a halfway reread will do before I quit for the day.

I assume that m = mass of moon 73.50, M = mass of earth 5,973.60, A = acceleration of earth (9.81?), a = acceleration of moon (1.62?)

Why are you using numerical examples for H?  
H = m(A + a) is not a formula for determining field strengths. Miles created a gravitational form intended to be used as a means of expanding Newton’s formula,
F = GMm/r^2
In order to derive

the E/M field equation that was buried in Newton’s equation
E = (m/R2 )[GM – AR2 – aR2]

I'll finish reading, but that’s my tentative answer.
.

.
parkham said. OK.  I've decided to go back to uft.html  and really, really, really read it.  I thought I understood but, I still don't.
Airman. One of Miles' defining characteristics, in my opinion, is his ability to continually astonish me. I read, and reread and never see any of his papers the same way twice.

The Unified Field Theory is a particularly rich paper. I'm far from understanding it. For example, even with Miles simplifying things, relativity still throws me.

Starting at " Let us use the letter H.  " is a bit confusing. At first it appears you're asking for an interpretation, but following up with numbers makes me think you want a verification. Your number assumptions look good to me, especially considering our previous discussion. Miles uses them in an example on the next page and elsewhere in the paper.

I'm sorry. No disrespect intended. What's your question?
.

Miles Mathis, uft.html wrote:

gE - EE = 9.8 m/s2
gM - EM = 1.62 m/s2
RE/RM = gE / gM = 3.672
gM = .2723 gE

EE /EM = 1/3.6724 = .0055
EM = 181.81 EE

But that last equation is assuming that the Earth and Moon have the same density. So I must now correct for density. Notice we are correcting the E/M field for density, not the gravitational field.

DE /DM = 5.52/3.344 = 1.6507 = 1/.6057
EM = 110.12 EE

How did he get:  gM = .2723 gE  & EM = 181.81 EE

.

parkham said. How did he get:  gM = .2723 gE  & EM = 181.81 EE

The Unified Field Theory references another paper.
8. The Moon Gives up a Secret. http://milesmathis.com/moon.html Numbers from the Moon are analyzed to show that "gravity" is a compound field. 4pp.

For this reason, a large spherical E/M field will vary as 1/r^4, if measured from a distance...

Gravity varies ONLY as the radius of the object, and no longer as the distance of separation.

Given these two postulates we can proceed directly to the math. Let us first make a prediction, using the postulates above. I am claiming that that I can show that the gravitational fields of the Moon and the Earth are directly proportional to their radii. Let us do the math to show what the Moon’s gravitational field would have to be if that were true.
gE / gM = 3.672
9.8 m/s2 / gM = 3.672
gM = 2.669 m/s2
     But the current number for gM is 1.62 m/s2. That seems like a huge amount of acceleration to make up, and I can understand your doubts. When I first did the math I thought there was little chance the numbers would work, to be honest. I was just following an idea. But watch closely:

We know that the total field of the Earth at its surface creates an acceleration of 9.8 m/s2 and we hypothesize that this is the gravitational field minus the E/M field [the gravitational field is an attractive field and the E/M field is a repulsive field]. And we know the same for the Moon.
gE - EE = 9.8 m/s2
gM - EM = 1.62 m/s2
     I have also postulated that the gravitational part of this acceleration should be proportional to the radii.
gE / gM = 3.672
gM = .2723 gE
     And I have just postulated that the E/M field is proportional to 1/r^4.
EE /EM = 1/(3.672^4) = .0055
EM = 181.81 EE
     But that last equation is assuming that the Earth and Moon have the same density. So I must now correct for density.
DE /DM = 5.52/3.344 = 1.6507 = 1/.6057
EM = 110.12 EE
     So, we just substitute:
.2723 gE - 110.12 EE = 1.62 m/s2
gE - EE = 9.8 m/s2
.2723gE - .2723EE = 2.6685 m/s2       [subtract the two equations]
-109.85EE = -1.0485 m/s2
EE = .009545 m/s2
EM = 1.051 m/s2
gM - EM = 1.62 m/s2
gM = 2.671 m/s2

You can see that the math bore out my prediction exactly. Once we correct for the presence of the E/M field, the Earth and the Moon have gravitational fields that are exactly proportional to their radii.
     We did not get an exact match in the third decimal place only because we used 9.8 m/s2 for gE in the first equation. We must now add .009545 to that, and if we do we get 2.671 m/s2 in the first equation as well.

The two numbers are based on a relative comparison of the earth and moon radii.

I figured out gM = .2723 gE  

rM = radius of moon = 1,737,100.00 m
rE = radius of earth = 6,371,000.00 m

rM / rE = 0.2727 gE - All he had to do was put that in his steps, but he didn't. I guess he just assumed we all could follow along?
1,737,100.00 / 6,371,000.00 = 0.2727


Still not sure how he got EM = 181.81 EE

.
parkham, Look at my previous post. You showed it too, again based on the radius ratio 3.672.

And I have just postulated that the E/M field is proportional to 1/r^4.
EE /EM = 1/(3.672^4) = .0055
EM = 181.81 EE

.

eM / eE = 1 / (.2723 ^4) = 181.81 eE

OR

gE / gM = 3.672

EE /EM = 1/(3.672^4) = .0055
EM = 3.672^4 = 181.81 eE

For adjustment relevant to density:

DE /DM = 5.52/3.344 = 1.6507 = 1/.6057
181.81 * .6057
EM = 110.12 EE

I'm attempting to reason through this now, it all seems so circular to me though:

edit:  been looking at this, and I think I'm missing the obvious again.  Would someone help me along with explanations of how he's getting these numbers?

.2723 gE - 110.12 EE = 1.62 m/s2  = -109.85 EE  I can see that.

.2723gE - .2723EE = 2.6685 m/s2 hunh?  9.81 / 2.6685 = 3.672  which is rE / rM how does that fit in here?

     So, we just substitute:
.2723 gE - 110.12 EE = 1.62 m/s2
gE - EE = 9.8 m/s2
.2723gE - .2723EE = 2.6685 m/s2       [subtract the two equations]
-109.85 EE = -1.0485 m/s2
EE = .009545 m/s2
EM = 1.051 m/s2
gM - EM = 1.62 m/s2
gM = 2.671 m/s2

Hi Parkham,
Mathis also steps through this in this paper:

http://milesmathis.com/third7.html


The Third Wave -  Part VII
A Redefinition of Gravity
 Part VII

Mathis wrote:We know that the total field of the Earth at its surface creates an acceleration of 9.8 m/s² and we hypothesize that this is the gravitational field minus the E/M field. And we know the same for the Moon.

gE-EE= 9.8 m/s²
gM-EM= 1.62 m/s²

I have also postulated that the gravitational part of this acceleration should be proportional to the radii.

gE/ gM= 3.67²
gM= .2723 gE

And I have just postulated that the E/M field is proportional to 1/r4.

EE/EM= 1/3.672⁴ = .0055
EM= 181.81 EE

But that last equation is assuming that the Earth and Moon have the same density. So I must now correct for density.

DE/DM= 5.52/3.344 = 1.6507 = 1/.6057
EM= 110.12 EE
So, we just substitute:
.2723 gE-110.12 EE= 1.62 m/s²
gE-EE= 9.8 m/s²
.2723gE-.2723EE= 2.6685 m/s² [subtract the two equations]
-109.85EE= -1.0485 m/s²
EE= .009545 m/s²
EM= 1.051 m/s²
gM-EM= 1.62 m/s²
gM= 2.671 m/s²

Once we correct for the presence of the E/M field, the Earth and the Moon have gravitational fields that are exactly proportional to their radii. We did not get an exact match in the third decimal place only because we used 9.8 m/s² for gE in the first equation. We must now add .009545 to that, and if we do we get 2.671 m/s² in the first equation as well.

Also this paper gives some background:

 
An Update on Weight in which I falsify F=GMm/R²
by Miles Mathis

Mathis wrote:To find what it must be, we will solve as above. My gravity pseudo-field obeys the inverse square law, but only due to Relativity. So the apparent attraction at one Earth-radius will be 2.68 times 1/3.67², or .198 m/s². The E/M field drops off with 1/d⁴, however, so if the distance increases by a factor of 3.67, the field will decrease by 3.67⁴ or 181.4 times. Since I have shown in another paper that this field has an acceleration of 1.051 at the surface, at one Earth radius it will have an acceleration of .0058. We subtract that from the expansion pseudo-field to get .1924 m/s². That is 1/50 that of the Earth, which is roughly the volume ratio, not the mass ratio.
 
Which is to say that the standard model believes the number will be about .12 in that situation. 9.8/81 = .12. I predict it will be almost .2.

http://milesmathis.com/weight.html

There is a relatively simple experiment to show charge, and there are also some videos on youtube.

You should do this (if I remember correctly), use an axis that is free to rotate at the center, and put 2 heavy objects at the ends. Then place 2 very dense objects on the ground big enough to be taller than the objects on the axis, one at the left and one at the right of the objects on the axis.

Wait and you will see the axis starting to rotate towards the dense objects on the ground.

This phenomenon is not explained by gravity between objects, because it is so much stronger than it should be. It is due to the dense objects shielding part of the surrounding charge wind. This shielding allows the other side wind to be the only part to operate on the heavy objects and on the axis that is free to rotate, and so it rotates until blocked.

This needs to be done with all the due measurements and calculations (and predictions) to be called an experiment of course.

I can't post links but one of the youtube videos has this code: https://www.youtube.com/watch?v=euvWU-4_B5Y
- title: Gravity Experiment - Author: Nars Guzman


EDIT: and this is probably the same as the Cavendish experiment - but watch this video https://www.youtube.com/watch?v=6Lzd86ZYf_o where an other experiment is described wich is against mainstream gravitational theory (put 3 blocks near each axis ends, 1 at one sid, 2 at the other with a gap between them to let charge pass. Gravity expects the axis to move towards the 2 dense objects but it still prevers the direction of the single dense objects)