Given a planet s radius, its orbital period in days is given by:

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1 Sidereal Messenger by Daniel Browning, April Introduction A mathematician s job is never done. This brief item will attempt to improve on an equation that forges a relationship between the orbital periods of the planets and their orbital distances from the Sun. This relationship is explored in Werner s Way and A Philosophical Extension, and further developed in Three Sheets to the Wind (all in Volume 2 of the Brief Items). Given a planet s radius, its orbital period in days is given by: 2 2 r 3 SOL Eq. 1 The various numerical values are scale factors that generate a day value for some radius. They have been explained as: Luni-solar value (the time it takes the Moon and Sun to conjunct) Number of seconds in an Earth day Seconds in a Moon day, treating Moon diameter as circumference Though the scale factors generate the correct results, they make little sense from a hypergeometric point of view. Our mission is to revise the planetary formula to satisfy two objectives: 1. Use values that make sense within a hyper-spherical or cubic framework. 2. Incorporate a horizon radius from the field equations for hyper-cylindrical spacetime. Part I of this brief item will look at a solution that makes partial sense in terms of the given objectives. Part II will go a little farther and present a solution that meets the objectives squarely. Part III looks at some string theory ramifications of the equation. I. Numerical Reconnaissance Horizon 903 Using the hyper-surface area from the formula presented in A Philosophical Extension, a horizon value can be extracted for the Earth radius r = 92,634, from the following identity:

2 2 (903, ) 3 = (2 2 r 3 ) 1/2 x SOL This identity isn t ideal because the multiplication by the speed of light (SOL, 186, mps) is at odds with the division in Equation 1. This will turn out not to be a problem. Using the new horizon value and terms from Equation 1, (903, ) 3 / SOL = ( x ) x The use of the horizon value requires that the numerator of Equation 1 be similarly scaled x = 157, to cancel out the presence of 5338 in the denominator. These numbers are tuned to the Earth s sidereal period for the Earth orbital radius, so it s no surprise that: 157, / = = (SOL) 1/2 This should be pretty clear from the identities given. Equation 1 can be replaced with 2 2 r 3 SOL [( ) 3 / SOL] = (days) Eq. 2 for the Earth radius. The nice thing about Equation 2 is that it uses an outer-horizon volume in the denominator rather than a superposition of an Earth-day and Moon-day in seconds. But perhaps this is replacing one enigma with another? The outer horizon value is about equal to the outer horizon of the inner compensating system (902, ) as described in the brief item Four Horses. In Part II it will become clear that this value is about right but somewhat off the mark conceptually. At the most basic level, it s a 9 harmonic. An Earth day is encoded in 157,646 since SOL = As expressed above, 2 2 r 3 = (903, )3 SOL

3 3 This is bad news since the horizon value completely cancels out and the hyper-spherical apparatus vanishes, leaving only the components of 157,646. That is, SOL = This is the basic form of Equation 2 for the Earth radius. It makes no sense whatsoever. Operator 157 So far we have ( ) 3 (SOL x x ) = x = 157, The speed of light falls out as 2 2 r 3 SOL [(903, ) 3 / SOL] = Eq. 3 or basically SOL / (SOL) 1/2 = (SOL) 1/2 The reader can trace this back to find out why the result is inverted. The left-hand side of Equation 3 can be inverted to get the non-inverse more on this in Part II. A simple heuristic can be applied to find the final value for Earth. Using the square root of the speed of light as a seed value (highlighted in yellow), and using the result of Equation 3 (for Earth): And, / = 1.0 = m x 157, = 157, , / = (days) This is clearer when applied to Mars. For r = 141,174,780.6, Equation 3 yields: The heuristic for Mars is

4 4 And, / = = m x 157, = 296, , / = (days) The farthest planet and thus the longest diameter is Pluto. For r = 3,653,499,570, Equation 3 gives: Since the result of Equation 3 doesn t drop below unity, it can be said that Pluto is within the bounding sphere of the outer event horizon it doesn t drop off the edge of the universe. (This will be easier to conceptualize in Part II.) The heuristic for Pluto is: And, / = = m x 157, = 39,047, ,047, / = 90, (days) Notice the symmetry between the Mars values of and 229.4, and the reverse for Pluto, and The Trend Here s the whole sequence for all of the planets using Equation 3: Planet Orbit radius A (1/x) B Mercury 35,849, Venus 66,974, Earth 92,634, Mars 141,174, Jupiter 481,976, Saturn 883,639, Uranus 1,776,727, Neptune 2,784,589, Pluto 3,653,499, Table 1

5 5 Column A starts with 1792 and ends with 1.742, roughly 1790 / Likewise, column B starts with and ends with 247.6, roughly x 1000, so there s a nice cross symmetry between columns. Column B is simply the year count for each planet s sidereal period. This intriguing sequence leads to a more meaningful sidereal equation. II. Numerical Redaction Sun and Earth We assume that the Earth, a small sphere with a circumference of about 21,600 miles, is surrounded by a series of event horizons upon which the motions of the planets are reflected. This is similar to the old Greek conception of celestial spheres. The Copernican view of little round balls orbiting a big round ball is still valid, since the planets are as close as they appear, and indeed the accepted distances of the planets from the Sun are used as a starting point for the calculations. The Sun is about 864,000 miles in diameter. In the hyper-cylindrical framework this number is treated as a radius, in line with all of the other horizon work in the Brief Items. Since we see the planets from the surface of Earth, we start with the Earth s circumference, comprised of miles per degree of meridian arc, as explained in the brief item How Far To Sirius? The hyper-cylindrical world and the geodetic world are one and the same, but the numbers in How Far will be very slightly tuned for hyper-cylinders. Thus, 360 (degrees) x (miles per deg of meridian arc) = (miles circumference, Earth) Neither the Earth circumference nor the measure of 360 is of immediate interest. The key number is the miles in one degree of meridian arc: = = n 3 = 216, (hyper-miles) The cubed measure is close to the Earth s circumference times 10. However it s not a circumference. It s an angular measure, perhaps related to angular momentum J of the black hole system: J = / J = 183, ~= SOL

6 6 After much experimenting it was found that the solar diameter of 864,000, when treated as the inner horizon of the canonical black string system, generates an event horizon of: 864,000 = n 3 = m x (ratio) = 3 x = 898, (event horizon) This is a very easy formula to remember since the inner horizon is simply 864,000. The canonical ratio between the event horizon and the inner horizon, nine-eighths, is used. (See Einstein s Fractions for the formulas.) Next, the hyper-surface area kernel of Equation 3 from Part I is used: 2 2 r 3 SOL = k This generalizes the equation since each planet orbits at a different radius from the Sun. Given a radial distance for a planet in miles, the units are: miles (2 2 r 3 ) mps (SOL) = seconds The surface-area kernel is a measure of the number of seconds it takes for light to travel the distance to the planet but not in unit space, in cubed space. In other words, the equation for a hypersphere describes a surface area, and the seconds are an area measure. That s why the whole thing is square-rooted: it turns a surface area into a side or a hyperline, and a side in cubed space is the same as an orbit in unit space. The new sidereal equation is simply: (eh) 3 3 k Eq. 4 Everything is solved in cubed space. For the Earth radius the hyper-line time in seconds is, 2 2 r 3 SOL = 9,177,770,521 and the full solution is: 898, = n 3 = x = / 216, (hyper-miles, 3 ) = x = / 9,177,770,521 (line-seconds, k) = (days)

7 7 The event-horizon distance, which can be thought of as an incipient circular measure, is boosted into cubed space, which in the very simplest interpretation is a volume. Then, this extent of the universe is first divided by another mileage value, being the number of miles in a degree of meridian arc around the Earth, but this too boosted into cubed space. This adds angular momentum, but it will be factored out for the day value. The result can be interpreted as a reduced volume. The volume is further divided by the hyper-line describing the planet distance, being a measure of seconds at the speed of light. All of this is happening in cubed space. On a more physical level, the solar period of the Earth is being reflected off of the solar event horizon which surrounds the Earth at a conceptual distance of 898,596 / 2 = 449,298 miles from the center of the Sun. This distance can be multiplied by 1000 to roughly approximate the radial distance to Jupiter. Inversion As shown in Table 1, the Earth has an AU factor of unity. What happens if Equation 4 is applied to Mars, which has a different AU factor? Applying the Mars radius, 2 2 r 3 SOL = x = k and / k (line-seconds) = (days) This is not the Mars sidereal period of days. A thought experiment is in order. Suppose the problem is that the sphere of Mars, like the sphere of Earth, needs to be taken into account to get the right value. Thus, Mars radius = x 2 = 12, = c c / 360 (degrees) = (miles per degree of meridian arc) = = n 3 = 38, (hyper-miles) Then / 3 = x = / k (line-seconds) = (days)

8 8 That s wrong too! One can try setting alpha to the inner horizon, or even the redshift horizon, but none of it works. Therefore, one can compute a whole new horizon system based on the ratio between the diameters of Earth and Mars (don t worry, this won t work either) (Earth) / (Mars) = (ratio) = 1 x = m m x 864,000 (Sun) = 487, = q q = n 3 = m x (ratio) = 3 x = 507, (event horizon) The reason this might work is because equivalent ratios in unit space will produce equivalent ratios in cubed space. The Earth-Mars ratio applied to the solar diameter will produce an equivalent ratio in relation to the Mars radius. Predictably, 507, = n 3 = x = / 38, (hyper-miles, 3 ) = x = / k (line-seconds) = (days) The answer is still wrong. Since the horizon was calculated by ratio the result is exactly the same as for the physical Earth sphere. However, information has been lost. Equation 4 is meant to be used with an event horizon based on the solar diameter, and this can only be achieved with the Earth radius. Taking stock, a couple of things have been tried: 1. Changing the meridian measure to the Mars sphere. 2. Using the Mars meridian measure with new horizons based on the Mars radius. Neither approach worked. There s only one more thing to change: the AU value for Mars. The results of Equation 3 shown in Table 1 are inverted. Column A in Table 1 is a reciprocal listing. With Equation 3, the cubed horizon value of 903,645 is divided into the surface area. With Equation 4, the surface area is divided into the cubed horizon value. Since Equation 4 is inverted relative to Equation 3, then to compute non-inverted values the natural thing to do is to invert the AU measures. This should result in a sequence of non-inverted results. Thus, the AU table from the brief item A Philosophical Extension, which is the basis for all of the sidereal orbit radial distances, can be modified as follows:

9 9 Planet AU 1/AU Distance Mercury ,365,298.3 Venus ,124,993.7 Earth 1 92,634, Mars ,783, Jupiter ,804, Saturn ,711, Uranus ,829, Neptune ,081, Pluto ,348, Redshift NA 1,263, Event NA 898, Turning-point NA 864,000 Table 2 The black-string event horizons have been included to show that no inverted value falls within the boundaries of the Sun, although in one sense, all of these distances fall within the boundaries of the Sun if the inner planets are thought of as the Sun. The Mercury distance is a harmonic of the distance to the Moon, 239,000 miles; the Pluto distance is another Moon harmonic. These two planets bound the sequence. If the regular distance to Jupiter, r = 481,976,629.4, is used in Equation 4, the result is days, which is close to the solar axial and lunar periods of and days respectively. It s also close to the Jupiter solar opposition time of days if multiplied by x 13 = (days) The Jupiter radius applied to the inner horizon produces the lunar period: / (ratio) = / 13 = (days) In other words, you can t go much farther than the boundary of Jupiter without generating numbers below the solar limit. The Jupiter-line is a barrier in this inverted scheme. From another perspective the regular orbit of Jupiter can be thought of as the surface of the Sun but outside the orbit of Earth, not inside. Clearly the Sun is as it is and nothing has changed, but mathematically the parameters of the Sun can be described with an exterior scheme. Applying Equation 4 to the inverted AU value for Mars, 60,783, (miles) = n 3 = m x 2 2 / SOL = x = 4,878,201,596 = k

10 10 Using the regular values of and alpha for Earth (which will serve as constants for all of the planets), / 3 = x = / k (line-seconds) = (days) A double check using the Jupiter radius yields: k = 773,315,038.0 / k (line-seconds) = (days) The inversion works! The New Trend In Part I the speed of light is used as a divisor for both the event horizon distance and the planet orbital distance. In Part II the speed of light is used only for the planet orbital distance, as appropriate since it converts miles to seconds. The event horizon distance, by contrast, is divided by the arc-meridian circularizer, 216,450, which as we have seen may be related to the speed of light through angular momentum J. When Equation 3 is applied to the Earth radius, the result is the square root of the speed of light, but for Equation 4 the result is the Earth year. The Earth year thus becomes the new seed value for the heuristic, although since Equation 4 generates the day value directly, no heuristic is needed. But to illustrate the parallelism of Equations 3 and 4, the heuristic will be applied anyway. Thus the heuristic used in Part I can be modified to utilize the Earth year instead of the square root of the speed of light. As well, operator 157, which is the square root of light times the Earth period, simplifies to the Earth period squared, but inverted, i.e.: ( ) -2 = x 10-6 The heuristic for Earth is, And, / = 1.0 = m x x 10-6 = x x 10-6 x = 1 x = (days)

11 11 For Mars, And, / = 1 / = m x x 10-6 = x x 10-6 x = 1 x = (days) The complete sequence is: Planet Orbit radius A B (1/x) Mercury 239,365, Venus 128,124, Earth 92,634, Mars 60,783, Jupiter 17,804, Saturn 9,711, , Uranus 4,829, , Neptune 3,081, , Pluto 2,348, , Table 3 Column B, which is the year value, is listed in inverse form, whereas in Table 1 the inverse values are listed in column A. Meanwhile, column A is the non-inverted result of Equation 4, which is simply the sidereal period day value. Like Table 1, the columns are bounded by a factor of In column A Mercury s 87.9 x 1000 is about 90,469 for Pluto. Since column A in Table 1 is the denominator of a fraction, i.e. it s the inverse, the trend increases from Mercury to Pluto, as does the trend in Table 3. III. Galactic Reconnaissance String Theory The hyper-cylindrical field equations solved in Einstein s Fractions apply to a black string, which in turn applies to atomic theory, not celestial mechanics. It may be that string theory can be applied at the cosmic level as well. The main idea is that the Sun and Earth may be the compact fibration of the galaxy as a whole, in the sense that the surface of the Sun is like a torus that s wrapped in light coils, or what mathematicians called fibers. If one thinks of a fibration of space and of coils of light being unwound along an arc, then when the tiny point of the Sun is unwound, its string may reach clear across the galaxy. In other words, the Earth and Sun may be a

12 12 compactification of the galaxy as a whole. The Sun would exist as one reality, and the galaxy as another, and yet the two realities would be complementary. If the Sun s fundamental diameter is dictated by the galaxy as a whole, then the solar event horizon, which is the extent of the universe in cubed space, can be treated as a linear distance rather than as a fibration or surface area, and as such should turn out to be an interesting number. An Earth lightyear is defined as: 3600 secs x hrs x days x 186, mps SOL = x (miles) The conversion of cubed horizon lengths to lightyears is shown in Table 4. Horizon unit cubed ly tp 864, x , eh 898, x , rs 1,263, x , Table 4 These are very nice results since the diameter of the Milky Way galaxy is thought to be about 100,000 lightyears. The inner or turning-point horizon is right on the mark, taking into account the use of a diameter as a radius in the field equations. The numbers point to a fundamental relationship between each star in the galaxy and the macrostructure of the galaxy as a whole. Is it just a coincidence that the diameter of the Sun unwinds as the diameter of the galaxy? Not all stars can make that claim because not all stars are the diameter of the Sun. Perhaps the math itself is playing tricks, because one cannot remain blind to the fact that all of this is happening in cubed space. What is it about a cubed number (whether treated as a volume, a surface area, or something else) that supports all of these cosmic realities or do the realities support the numbers? Is our description of reality too narrow? Conclusion An improved formula for computing sidereal periods from orbital distances has been presented. Unlike the work in A Philosophical Extension, the formula is firmly entrenched in cubed space. As well, a fundamental Earth value miles in a degree of meridian arc ties the results to the Earth in a definite way. All of the sidereal activity is bounded by an event horizon surface area based on the incredibly simple number 864,000, the diameter of the Sun.

13 13 The inverse scheme presented in this paper puts the orbits of the planets between the regular orbit of Jupiter and the solar center. These distances produce sidereal periods identical to the normal scheme. The new formula is consistent with results from The Long Cycle of the Sun, which found that solar values are associated with the black-hole event horizon, and the lunar period with the inner horizon. Since the sidereal formula is strictly cubic, then the results are cubic, which means that the things we see in the heavens are cubic, and yet the cubicity is transparent to us. All of these insights are based on numbers, or more to the point, on geometry. However the geometry is hard to visualize so one should treat the various interpretations with caution. The simplicity of the math is a point in its favor. Clearly the diameter of the galaxy, or in unit space the diameter of the Sun, is just the right length to generate the sidereal periods and distances observed for all of the planets. A different boundary would produce an entirely different set of numbers. References Kaku, Michio. Hyperspace, Anchor Books, 1995 Thurston, William. Three-Dimensional Geometry and Topology, Vol. 1, Princeton University Press, 1997 Previous brief items as called out in the text.

14 14 Appendix (Figure) Figure 1. The Inverse Orbits (from Table 2) The Earth is shown by the blue circle. Pluto is shown near the center in red. The event horizons are inside of Pluto. The outermost circle is Mercury.