captcass

Hi folks. I finally found my formulas. Thank you, Mordred, for your early help and decent, courteous, replies. What is happening is this: Time evolves space, and spatial events, forward at c in the forward direction of time. Densities in space, masses, slow that evolution and their velocities are how time keeps them up with the rest of the continuum. Since orbital velocity, VO, = √(GM/R), the time dilation formulas for orbital velocities are derived from the gravitational time dilation formula, T0 = T√1 – (2GM/Rc2), which contains the velocity formula, by substituting VO for GM/R, i.e.: T0 = T√1 – ((2/c2)(VO)), resulting in: VOp = √2*√((Tc2 – T0c2)/2T)), for planetary and moon velocities, where T = 1 is the distant observer’s invariant rate of universal time, T0 is the rate of time of the coordinate point, and √2 is the gravitational acceleration factor, due to the fact that the multipliers for the planets are in the range of Mercury = 1.41474412726671 and Neptune = 1.41379959891788, very close to √2 at1.4142, and what follows: Einstein's Fundamental Metric (with the x,y,z,t,coordinates), describes the diagonal of a cube. (image in the paper). Assigning 1 km for the length of each side, we see that the diagonal, y = √3 and the hypotenuse of a side x = √2, and we find that using VOg = (√((Tc2 – T0c2)/2T))/((√3)/(√2)) = (√2*√((Tc2 – T0c2)/2T))/(√3), using the Sun’s surface dilation factor, gives us us the actual galactic rotational velocity of 231 km/s for the Sun. Again, looking at the cube, Y is the Fundamental Direction of Evolution. The planets appear accelerated in the plane of the ecliptic, along X, at √2. The stellar system as a whole, centered on the Sun, is also being accelerated by a factor of √2, but it is along Y, the FDE, so the planetary result is divided by √3. As in the cube, this keeps the planets in the same plane of evolution as the Sun. So, the equations are: VOp = √2*√((Tc2 – T0c2)/2T)), for planetary and moon velocities and: VOg = (√2*√((Tc2 – T0c2)/2T))/(√3), for galactic velocities. Since T=1, these formulas reduce to: VOp = √(c2 – T0c2), for planetary and moon orbital velocities and: VOg = √((c2 – T0c2))/(√3), for galactic rotational velocities. For cometary velocities, it is necessary to derive the time dilation velocity formula from the Vis-Viva equation, which is: VOc = √((2*(c2 – T0c2) – (c2 – T0αc2)) Where T0 = the rate of time factor for “r” in the Vis-Viva equation and T0α = the rate of time factor for a distance equal to the length of the semi-major axis, α. Since this is the derivation for the Vis-Viva formula, it is the exact solution for any stellar system body when using the dilation gradient of the central mass of the stellar or planetary system. The Force in Time Now that we know that √(GM/r) = √(c2 – c2T0), we can make that substitution in other equations. Deriving the gravitational force, in Newtons, through time dilation, we use Newton’s equation: F = G(M1m2)/r2 = (GM1/r)*(m2/r) = (√(c2 – c2*To)2*(m2/r) = ((m2)(c2 – c2To))/r, where To is the rate of time factor for the coordinate mass, m2. Of course, the equation works in the obverse for the other mass, M1. As of today, 75% of the astrophysicists/astronomers in the English-speaking universities worldwide have downloaded my paper. The latest version is at thetruecosmology.com. If you have downloaded it before, please hit "refresh" when the paper loads so you get the latest version, and not your cached version.

Actually, it does. You just aren't understanding what you are seeing. The continuum isn't hidden from view on some tiny scale. We are looking at it all the time. Soooooo....after re-reading the original UCSB answer, I, too, think it was an error on the part of the person answering the question. Sooooo. Thanks to all anyway, but it looks like it is my bad. I was excited because 3.2 km/s would have made sense to what I am working on in one respect. My calculations based on the curve of the orbit gives me only 5.97042055×10−6 km/s, which seems way too small. So. Sorry and thanks.

km/s. In the answer originally quoted from UCSB, it is stated the Earth moves "sideways" towards the center of the Sun at a rate of 3 km/s. If the Earth suddenly began to continue straight along its current tangent line, at what rate would the orbit drop away below it?

I used that as that is how people commonly see objects, as "falling" due to gravity. I don't see them as falling, but as evolving through the continuum. The Earth is always accelerating towards the center of the Sun. Hence it is in free "fall" in space. One of the problems we have is that we still refer to events in classical terms, when it is a quantum world. Things do not fall "thru" anywhere. They evolve within the continuum. We currently have to use external forces to shift masses within it. We need to be focusing instead on the continuum and how to evolve masses through it. This is the only way our "visitors" can maneuver the way they do in the air and water. I think this could possibly be a gravity drive based on virtual dilation gradients.

How about this: The Earth is always falling towards the Sun. For each km it moves forward in its orbit, how many km does it fall?

Thank you, I knew that. But I am looking for the other aspect i described. Consider that I am correct and light appears to bend around a massive object because it is evolving down the gradient, not moving through curved space. At what velocity is the light evolving down the gradient?

I realize that. It is a hypothetical to show there is a gravitational velocity component acting orthogonal to the orbit. Borrowing from StackExchange: "It (the body) does keep accelerating. Its velocity in the direction of the object being orbited keeps increasing. But this direction keeps changing. The reason the satellite's total speed doesn't increase, at least in the case of a circular orbit, is that while its velocity towards the object increases, its tangential motion moves it forward so that that direction is always perpendicular to the direction of motion. Thus while the satellite is undergoing constant acceleration, that acceleration is always perpendicular to the direction of motion and the speed of the object never changes." In other words, the acceleration, which is always towards the center of the Sun, manifests as part of the total orbital velocity.

(Can't seem to get this image to display - just click on it) for the Earth at perihelion. I don't see how this will help. It is just comparing different speeds at different points in the orbit. I am looking for a gravitational component. Consider the problem this way, The Sun and Earth are infinitely far away and there are no other planets around the Sun. The Earth is moving at a steady velocity relative to the CMB. As it enters the Sun's gravitational field it appears to accelerate. It continues to accelerate until it makes its closest point of approach and then, because its new momentum equals the force of the gravitational field, it "falls" into orbit. It is still accelerating towards the Sun, but it gets no closer. This means the acceleration is manifesting as part of the overall velocity, the part added as the Earth entered the field..

What do you get for Earth on an equinox?

Everyone I found referred to the velocity along the orbital path as "sideways" as it parallels the sun's surface, i.e., orthogonal to a line from the planet to the center of the Sun. I've looked at the radial component and it isn't what I am looking for. The 3 km/s in the original post is the range I am looking at confirming, and then solving for the other planets will nail it all down. But I can't do that until I know how they computed the 3 km/s. That is not a typo, either, which is clear from the whole context of the answer. (Also, the earth's average velocity is not 30 km/s, as someone noted above, but 29.79 km/s, often referenced as 29.8, as on this NASA site https://nssdc.gsfc.nasa.gov/planetary/factsheet/)

You are looking at space and time wrong. What we are looking at is the evolving energy field, not a fixed space. Time evolves space, and spatial events, forward in the forward direction of time, which has no depth. So even though we see dimensions, there is no depth, just an evolving energy field. As time evolves the continuum forward, we see the apparent curvature of motion as events evolve down dilation gradients. My paper was reviewed and published, not paid for. As for the JofC, after I was published, I acted as managing editor/webmaster for nearly a year and saw the inside workings. All papers are reviewed and fees are even waived for authors who cannot afford them. There is a vanity press aspect for the owners of the journal, but not for others they publish. They haven't published anything in about 6 months due to mental "end of life" issues of the Executive Editor. Until that is resolved I suggest people do not submit to them. If you check Vol 27 you will see what the problem is. The reference was to planetary orbits, not circles. Thanks for that, but it doesn't really help.

🤣

You know, one of the reasons I hate asking questions in this particular forum is the snide remarks i get. I come back from time to time because I have had things clarified for me here. But the rude comments really show a lack of professionalism and make it difficult to come here. I am doing time dilation cosmology. Whereas you see objects "falling thru" space. I see objects evolving along time dilation gradients in an evolving energy field. For instance, the reason light bends around massive objects is because it is evolving down the time dilation gradient. It is not because space is bent. It is because the forward direction of evolution is down the dilation gradient. This is happening to all particle events, regardless of size or complexity. I am just wondering if anyone here can tell me how to compute the gravitational component without getting into working EFE's? It should be a simple Newtonian calculation, but damned if I can find it.

This is why it has to manifest as just a velocity. Otherwise we'd spiral in. Velocities are accelerated in a gravity well. You are not looking deeply enough at the contributing factors. I understand what you are saying, but I am not asking that. I am a published cosmologist. Thanks, though.

Here is a link to the site. It is the top answer. http://scienceline.ucsb.edu/getkey.php?key=770 I should mention that I found statements like, "the sideways velocity is almost entirely due to the forward momentum...", or some such, "sideways" meaning parallel to the Sun's surface, without explaining the other factor(s). The Earth is always passing tangentially to the Sun, but gravity is accelerating the Earth downward. In a stable orbit, the acceleration due to gravity manifests as a steady velocity towards the Sun's center, orthogonal to the sideways motion. As the orbit is stable, that downward velocity combines with the tangential velocity and gives us the total free fall velocity along the curve of the orbit. I am trying to figure out how to compute that gravitational velocity component.