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September 6 - 12, 2016

Plotting the Lunar Orbit of July, 2000

The last chapter convinced me that there is a way to calculate how far the moon was from earth at any one second between apogee and perigee, providing that the claimed distances at those two points is correct. I have my doubts on whether they have those distances accurate, but it's all we have to work with. The final thing I'd like, aside from NASA coming clean on the true timing of an eclipse of July 16, 2000, is the lunar velocity 22.333 hours after the apogee position of July 15, 2000. I'm now asking whether I can work out my own velocity, seeing that I can find no lunar-velocity calculator online. I've never tackled this task before due to the difficulty.

The first thing advisable, to find lunar velocity, is the circumference of the lunar orbit for the month of concern. Let's start with the shape of the semi-circumference, a half month. I can now plot the moon at any point of the apogee-to-perigee path using the acceleration-of-the-moon figure (.00005956) discussed in the last two chapters. The trick is to get the shape of the ellipse and then find its distance somehow with some reliable but crude technique that I need to invent. I don't know how the professionals do it, and one can't use an ellipse calculator with calculations from the perfect center of the ellipse, because the earth is not at the center of the ellipse.

For the task, it would be less complicated if we imagined / drew an earth circle with the moon at apogee above the circle's 12 o'clock position, and the half-time of the orbit below the 6 o'clock position. If I recall correctly, the moon orbits the earth counterclockwise when drawn on paper with the north pole in the center of the earth circle, wherefore I'll put the half-ellipse on the left half of the clock.

Half of a lunar month (27.321666 days 23.9333 hours each) is 326.949 hours = 1,177,015 seconds. Entering the latter number in the free-fall calculator (below) along with .00005956 in the g box tells me that the moon had dropped 41,255.5 meters = 25,635 miles by that time, to 252,400 - 25,635 = 226,765 miles from earth. And by the way, 252,400 is not an approximation. An apogee/perigee calculator tells that the moon was 406,199 kilometers away, which works out to 252,400.36 miles. It tells that the perigee distance (July 30, 2000) was 358,378 kilometers = 222,686 miles. As perigee was about 14.67 days after apogee, it was a day after the half-time of the orbit so that a distance of 226,765 miles at the halfway point looks correct.

To find how high the moon was halfway to perigee, enter half of 1,267,200 seconds, which is 633,600, finding the drop distance as 11,955 kilometers = 7,428.6 miles, which is one quarter of 29,714 miles, the latter being the distance of fall at perigee (252,400 - 222,686 = 29.714).

It had been determined, from the data of the free-fall calculator, that any dropping object, whether on an orbital trajectory or not, falls one-quarter of its total fall at its halfway point in time. Therefore, when our clock's full circle represents the lunar orbit of 27.321666 days, the halfway point of the entire orbit is at 6 o'clock, perigee is between 5 and 6 o'clock, and the halfway point to perigee is halfway between the perigee spot and back-around to 12 o'clock (between 8 and 9 o'clock, that is). We can already have a good idea of the orbital shape, and can continue to use the same method to plot as many spots as we would like by simply using halfway points as the guide. It's crude, but it works.

We can find how close to earth the moon will be at the quarter point in time by entering half the 1,177,015 seconds, or 588507.5, in the top box. The fall distance is found as 10,314 kilometers = 6,408.8 miles, which is exactly 4 times less than the fall of 25,635 above. Where on the orbital path do we put the moon at an altitude of 252,400 - 6,408.8 miles? At 9 o'clock. To find the altitude at 10:30, cut the 588507.5 seconds in half, see what fall distance the free-fall calculator gives for it, and subtract the fall distance from 252,400. One might be able to get the average lunar distance by this method. All my life, I've avoided planetary ellipses like the plague, but here I'm seeing a ray of light.

If, between 12 and 6 o'clock on the left side, the moon has a slower average LATERAL velocity (nothing to do with drop velocity) than between 6 and 12 o'clock on the right side, and/or if the lunar distance (from earth) on the left side is longer than on the right, then the moon will NOT be exactly at 6 o'clock when it's halfway the distance to the top of the clock. The halfway point in time is not necessarily the halfway point in distance.

This material confuses me, a rookie at this sort of thing. Although equal-distance units on the clock are equal time periods, not equal distances of drop, I think it may be correct to say that the shape of altitude plots upon the clock are a perfect reflection of the physical shape of the orbit. I was opposed to this at first.

By dividing the number of seconds (1,267,200) between perigee and apogee by the number of seconds in the full month (2,354,139), one can find that perigee, on the time clock, needs to be .5383 the way between apogee and the end of the month. To find where this is on the clock, multiply .5383 by 60 to find 32.3 minutes past 12 o'clock (in the counter-clockwise direction), which lands a little before 5:28. Or, multiply .5383 by 12 (= the numbers on the clock) to get 6.46 hours past 12 o'clock, which works out to .46 the way between 6 and 5 o'clock, which is to say .46 of 5 minutes = 2.3 minutes, same as 32.3 minutes from 12 o'clock (at 5:27.7). However, as the moon was slower on the left half of the orbit, perigee may have been on the left side of 6 o'clock. Certainly, I think, the midway point of the orbit was on the left of 6 o'clock. In other words, don't confuse the time clock with the true picture of the orbit.

The full orbit can be treated as two pie sections, one from apogee to perigee, and the other from perigee to 12 o'clock. The apogee after perigee was on August 11 at 22:25 pm, 12 days, 14 hours and 40 minutes away (301.866 hours). The time between apogee and apogee was therefore 351.23 + 301.866 = 653.096 hours 27.288 days, slightly less than the average month of 27.32166 days. I'm equating the latter with the 12 o'clock position, meaning that the apogee in August was ever-so-slightly on the right half of 12 o'clock. A lunar month is not defined as apogee to apogee, but as one full orbit. Apogee to apogee is not always the same period, but here it's virtually identical to a true month.

It can be assumed that the moon did significantly less distance from apogee to perigee than it did from perigee to apogee. With perigee to apogee taking 50 hours less time than from apogee to perigee, one could probably guess right that the moon's path was slower on the left side, and that the right side represents the picking up of speed for the next few months. At each of the three (or maybe four) apogees after the first one under discussion, the moon will be traveling progressively faster before it slows down again. The moon falls at the same rate regardless of its lateral velocity, wherefore the low velocity in this case explains why perigee was at one of the moon's lowest altitudes. I am learning that, the slower the moon, the lower the perigee, and vice versa.

I had read that the building speed of the moon, in the orbital direction, is a consequence of it's building fall speed. Try to figure out how the orbital speed should increase by the fall speed, but I can't see it. Does a cannon ball increase in lateral velocity just because its dropping? If an arrow is shot down at 45 degrees, ignoring gravity, it will have a constant lateral velocity, never increasing or decreasing. The same can be said if the arrow is shot downward at 1 degree (almost horizontal). It can't build horizontal speed just because it's falling, suggesting that all lunar acceleration and deceleration may instead be from the sun's gravity. This develops a question: when we are told of the moon's velocity, do they mean the lateral direction only, or a combination of the lateral and up-down direction? If one divides the orbital distance by the time, only the lateral velocity is obtained.

If it's correct to say that all lateral lunar acceleration / deceleration is due to solar gravity, I could see why astronomers would want to cover this up, for it then allows one to tinker with the size and distance of the sun using the moons acceleration to solar gravity. In fact, I'd like to look at this in the next chapter, if there's anything to say about it.

What should the moon's average distance be, between apogee and perigee? Is it as easy as finding the middle number between 252,400 and 222,686 miles, even if the moon doesn't fall at the same speed? Or should the average be according to where the moon spends its time? For example, it the moon's orbit was steadily at 252,400 miles for 25 days, and 222,686 miles for the remaining 2 days, would it be correct to say that the average was the middle number? No, of course not. Therefore, time spent needs to be part of the calculation for the average distance.

We can divide the apogee-to-perigee period into six equal time slots using seven altitude figures. I'll use six periods spread apart by 211,200 seconds for a total of 1,267,200 seconds, the time between apogee and perigee. I've given the free-fall calculator's corresponding fall distances and the resulting altitudes:

0 = apogee = 252,400 mile altitude
211,200 seconds: 825 mile drop = 251,575 altitude
422,400: 3,302 = 249,098
633,600: 7,428. = 244,972
844,800: 13,206 = 239,194
1,056,000: 20,635 = 231,765
1,267,200: 29,714 = 222,686
(All figures in the left column need to be fed to the calculator with .00005956 in the g box.)

After 633,600 seconds, 1/2 the time to perigee, the moon is supposed to have been 1/4 the way in drop to perigee (29,714 / 4 = 7,428 drop), and this checks out where the corresponding height of 244,972 miles is 7,428 below 252,400. I can now add the seven altitudes, and then divide by 7 to get the average for this half of the orbit; it's 241,670. While this is much higher than the average (236,000 ballpark) I've expected for all orbits combined, I can't see anything wrong with the method or the numbers. It looks full-proof, albeit the second half of the orbit will bring the number down. The problem for NASA is that, the higher the over-all velocity for this month, the wider the earth umbra becomes (according to NASA's eclipse times), and the closer the sun gets where its already measuring only 5 million miles using one of the slowest speeds possible.

I don't know whether there is a formula for finding the velocity at apogee if one first has the average for the month, but this data here will come in handy if a formula comes up. I've yet to investigate Kepler's methods for discovering the secrets of ellipses.

To get the average lunar distance when it's considered to be the middle moon size to the eyes, half way between the apogee and perigee sizes, one would not use the method above. Instead, one uses the middle number between 252,400 and 222,686; it's 237,543, more in-tune with the reported averages. How far in the drop do you think the middle-size should be? Are you stunned by the confusion in all of this? Beware, for I get confused too, and I sometimes find myself making wrong statements that I fortunately catch and correct before the week is up. Some of them get by, and get corrected only after the page goes out to the Internet.

As you can see in the list above, 237,543 is almost 3/4 the way to perigee in time category, but not 3/4 the way in the fall category. It's a confusing thing. We can actually figure out how much time the moon spends between 252,400 and 237,543 miles versus between 237,543 and 222,686. We can know for sure that it spends exactly three times the time between apogee and the midway point in time to perigee as it does between the midway point and perigee. That's because it has fallen 25 percent by the midway point in time, with 75 percent left after that (i.e. a difference of three times). It does 75 percent of its falling after the 633,600 seconds.

We subtract 237,543 from 252,400 to find 14,857 miles (23,910,024 meters) of drop. This number is closer to the drop distance at 844,800 seconds (4/6 the time) than to 1,056,000 seconds (5/6 the time), but don't confuse the percentage of this drop with a midway point between 4/6 (.66) and 5/6 (.83), for the latter percentages belong to the time category.

To find the exact time of drop, we modify the formula, .5 x .00005956 m/s2 x time squared = fall distance. That is, fall distance / .5 / .00005956 = time squared, wherefore time squared = 23,910,024 / .5 / .00005956 = 802,888,650,100. The time is therefore the square root of the latter number, which is 896,041 seconds = 248.9 hours...which is .7071 of the 352 hours between apogee and perigee. The moon has therefore become half its apparent size after .7071 of the total fall, and, indeed, using the calculator with any height of fall (use page below for this) followed by half the fall distance shows that the latter is always .7071 of the total. It also means that the middle-size moon occurs when half the fall between apogee and perigee is arrived to. To prove it, just divide the total fall (29,714 miles) by the 14,857 miles above, to find the answer as exactly half. Half the fall = .707 the time.

The timing (248.9 hours) of the middle-size moon works out to 896,040 seconds, which checks out because its closer to 844,800 seconds than to 1,056,000 seconds. It all means that the moon spends .7071 the time between the altitudes of 252,400 and 237,543, and .2929 (1 - .7071) of the time between the altitudes of 235,543 and 222,686 miles. The first portion works out to .7071 / .2929 = 2.414 of the time of the second portion. This is a handy thing to have, that the middle-size moon between the apogee and perigee distances occurs always .7071 of the time after apogee. It means that roughly 29 percent of the fall time, and 50 percent of the fall distance, has yet to occur after the middle-size moon has been arrived to. Why doesn't anyone mention this? Do astronomers not understand these things? Or, do they think it's unimportant to know? When they speak on average lunar distances, can't they be more charitable with their information? Is giving the proper picture somehow dangerous for them? What is wrong with sharing the knowledge that a middle-size moon is .7 the time between apogee and perigee? This is a law.

The average distance for the perigee-to-apogee side of the orbit is less than 237,543 miles, because the second apogee is said to have an altitude of 405,649 kilometers = 252,059 miles, wherefore the middle number in this half is: (252,059 + 222,686) / 2 = 237,373 miles. I didn't know until writing this chapter that the second apogee fell bang-on one true month after the first apogee, making for an easy way to find the middle number for the full orbit as: (237,373 + 237,543) / 2 = 237,458 miles.

I am not at all sure on whether the rise of the moon in the second half is a mirror image of the fall in the first half, but I think it's bound to be close to it. One can probably say that the height of 237,373 miles is achieved .707 of the time between perigee and apogee. Therefore, while the first half has the moon spending most of its time in the higher altitudes, making for an average of 241,670 miles (number found above), the second half has the moon spending most of its time in the lower altitudes, making the average for the entire orbit significantly lower than 241,670. I don't know of a calculator that can tell us the rise of objects as one uses a free-fall calculator. The best I can do at this time is to assume an average distance of 237,458 miles for the entire orbit. This number is the middle between the total fall and rise.

The circumference is now attainable tentatively using 237,458 x 2 x 3.14159 (pi) = 1,491,991 miles, and the average velocity for the full orbit works out to 1,491,991 / 27.32166 / 24 = 2,275.34 mi/hr. NASA's mean orbital velocity for the moon in all orbits combined is currently 2,286 mi/hr. The month of concern here is on the slow side, wherefore the 2,275 looks not bad in comparison with NASA's number.

So, where do we plot the moon when it's 237,543 miles from earth (this concerns the first half of the orbit only)? It depends on where perigee is plotted, and I don't yet know it. If perigee were exactly at 6 o'clock, the .7071 spot would be back at 7:45 (the .707 spot in the second half should be a half clock later). All we know is the formula to find the plotting of any altitude between apogee and perigee, but we cannot put any altitude's position on the distance "clock" until we have established where perigee is upon it.

So, when the moon in its orbit is .7071 the way in time from apogee, the distance between it and the center of the earth is identical with the distance from apogee to the center of the ellipse...where the particular ellipse is the one formed after the moon has reached perigee in that particular month. But where on the physical orbital path is the middle-size moon to be plotted??? This is a tough question, but if one follows the clues of the list above, it can be plotted to a close approximation. As the middle-size moon is after the 633,600 seconds (= mid-time point), which is itself at the point where the moon has dropped 1/4 of its distance, the middle-size moon is somewhere past 9 o'clock. The reasoning here is that 1/4 the drop is half way in time to a little past 6 o'clock, making the 1/4-drop point a little past 9 o'clock. The middle-size moon is also after the 844,800 seconds (lunar distance of 239,194 miles), itself 2/3 the way in time = a little past 8 o'clock.

If we take NASA's average lunar velocity and add it to NASA's slowest velocity, then divide by 2 to get the middle ground for tentative use in the first half of this apogee-to-perigee period, we find 2,222.7 mi/hr. If we multiply that by 176 hours, half the apogee-to-perigee time, we find the moon traveling only 391,200 miles, about .26 of the total orbital distance. So, the orbital distance laterally is likewise 1/4 of the way per half the time. It seems that the drop-acceleration is at least nearly proportional with the lateral acceleration, unless NASA's velocities are not correct.

We may as well plot the eclipse of July 16, right now, using the same method as per discovering the time of the middle-size moon. In other words, this puts the eclipse on the time circle. It was established (last chapter) that the eclipse occurred when the moon had dropped 119.6 miles (192,477.5) from the apogee height, wherefore we do this: time = square root of (192,477.5 / .00005956) = 80,395 seconds. I've been using 80,400 in the past couple of chapters, as per 22.3333 hours = 80,400 seconds between apogee and mid-eclipse. We can thereby see that the formula works. The plotting of the eclipse is 22.333 / 351.23 hours = .063586 of the way (in time) between apogee and perigee. On the time clock, the latter was put at 5:27.3 (6.46 hours past 12 o'clock), wherefore the eclipse should be at .063586 x 6.46 = .4107, meaning .41 of 5 minutes (there are 5 minutes on the clock between two hour positions) = 2.05 minutes after 12 o'clock = 11:58 on the clock.

If you check the apogee-to-apogee time periods for the year 2000, you'll see that they are greater than the orbital period by a few to several hours. In a couple of the months, it is more than 28 days. BUT, for the month that has the eclipse, the month under discussion, the apogee-to-apogee period is actually a little less than the orbital period. In fact, the difference is so small that we can say they are identical. Is this a God-send for anyone wanting to prove the true solar distance? Is this a God-send to help the man on the street shame the evolutionists? I would have been completely baffled had apogee, at the end of July, been a day or so past 12 o'clock. I would not have been able to find the average lunar distance / velocity, had that been the case. Having the apogee-to-apogee time period match the lunar month in the very month that has the best eclipse I could find, to shame the astronomers, seems like a God-send. It's not my will only to shame NASA. Shame NASA. When will you fix the times on this eclipse??? The world will be watching as news gets around. Why did you change the times of this eclipse??? What was the motive? Put your hands up, NASA, you are under arrest. God, your enemy by your choice, is taking you to jail.

I've just loaded the year 2018 to check its central-path eclipse on July 27, which is on the same day as apogee. It turns out that, from apogee on June 30, at 2.44 am, it's only 27 days, three hours from apogee on July 27, at 5:45 am. That's 27.125 days, some four hours short of the lunar period (sidereal month, as they call it). The 2018 eclipse is nigh.

Wikipedia's Moon article does not say that 27.32166 is the average orbital period. And it uses even more decimals than I have included, suggesting that while apogee-to-apogee changes month-by-month, the orbital period may stay exactly the same. However, Fred Espenak says "mean sidereal month," though he doesn't give any figures aside from 27.32166. On his "Eclipses and the Moon's Orbit" page, he speaks on the various types of months but leaves out discussion on the sidereal month. Hmm. Is there something about the nature of the sidereal month that causes them concern? We would like to know why the orbital period stays the same even when the orbital velocity and distance is ever-changing? What rule keeps the orbit equal from month to month while the moon bounces around from altitude to altitude?

Apparently (I'm guessing), the lateral velocity is inversely proportional to the drop distance. At least, if the moon speeds up while the orbital path gets proportionately smaller, and vice-versa, it can predict the same orbital period, time and time again. As the middle of the apogee-to-perigee drop arrives .707 the time between them, does this mean that the lateral velocity increases by .707 over that drop (or decreases by .707 over that rise)? The free-fall calculator reveals that the drop velocity of an object dropped over one distance is .7071 the drop speed of an object dropped over twice the height. It is perfectly proportional to the drop time. You need the calculator below (asks for the distance) to verify this, not the calculator that asks for the time.

In other words, if you take two pages at the calculator with a drop-distance difference of 2, both the fall times divided into one another, and the drop velocities divided into one another, equal .707. It means that the drop velocity of the middle-size moon is .7071 the drop velocity of the perigee moon, for the middle-size moon is at half the fall of the perigee moon. To find the drop velocity at perigee as compared to the middle-size point (in any lunar orbit), multiply the latter's velocity by 1 / .7071 = 1.414227 times.

Let's say that an object is dropped from 5 feet, 10 feet and 20 feet. When dropped from 5 feet, it falls for .557 seconds before hitting the ground at 12.23 mi/hr. Dropped from 10 feet, it hits the ground after .788 second at 17.296 mi/hr, and from 20 feet it hits the ground after 1.11 seconds at 24.46 mi/ph. The idea here is to discover what number to use, in multiplication, between the object dropped at 5 and 20 feet. The answer is 24.46 / 12.23 = 2.0. An object falling four times as far has twice the velocity.

While we're at it, let's do 40 feet of fall: hits the ground in 1.577 seconds at 34.59 mi/hr. An object falling eight times as far has 34.59 / 12.23 = 2.828 times the velocity (twice the 1.414 above). The 119.6-mile drop at the eclipse is almost 8 times less than the 825-mile drop at 211,200 seconds.

Let's bring the list above back down:

0 seconds and 0 drop = apogee = 252,400 mile altitude
211,200 seconds: 825 mile drop = 251,575 altitude
422,400: 3,302 = 249,098
633,600: 7,428. = 244,972
844,800: 13,206 = 239,194
1,056,000: 20,635 = 231,765
1,267,200: 29,714 = 222,686

You can see where the drops are four times as much (e.g. 825 versus 3,302) so that there is now revealed a doubling of velocity between them. There is a four-fold increase in drop between the one at 633,600 seconds (half-time point) and the one at perigee, meaning that the velocity was 2 times slower at the mid-time point. But this is the drop velocity. Might the lateral velocity use the same .7071 scheme? Is there a connection between the drop and lateral velocities? I don't think so. The more I delve into this topic, the more I realize that all changes in lateral velocities, both lunar acceleration and deceleration, is due to solar gravity, zero of it due to earth gravity.

I've repeated several times that NASA is stone-cold in giving the lunar velocity at its eclipse pages. The reason became obvious when learning that it's not giving the correct times of the eclipse under discussion, and the best way to protect that lie is not to give the lunar velocity, nor to give us a way to find it for ourselves. I have never read from anyone that the orbital speed of the moon is due to solar gravity alone. But as I learn more and more on free-fall issues, I realize that the earth cannot cause a change in lateral motion, for the earth is always inside the lunar orbit, and can only pull the moon to itself at all times. A magnet pulling an object cannot give it lateral motion. Why didn't I think of this before? Well, I did, when I was very young, but I read from others that the earth's gravity changes the moon's orbital speed, and I bent my brain agreeing with that. It was bending because it really didn't seem correct.

Yes, it seemed above that the lateral velocity is proportional to the drop velocity, but this can, perhaps, be a coincidence of solar versus earth gravity. I delved into free-fall studies in the past few weeks to learn how the moon accelerates to earth's gravity, but am now realizing that I need to see how the moon accelerates to solar gravity.

We are told that the slowest velocity is at apogee, and the fastest at perigee. From now on, never assume that the acceleration of the moon between apogee and perigee has anything to do with its lateral path. View that acceleration in the drop direction only, and do not assume that the lateral velocity builds in any proportion with the drop velocity. If astronomers gave us the tools to find the orbital velocity at any point of the orbit, we might be able to see that the speeds are not perfectly in-tune with drop velocities. Each time the moon passes by apogee, it's essentially beginning a new fall all over again, We can say that the lunar orbit is repeatedly in about a eight different positions in respect to the sun, and in each case, the sun will pull the moon differently.

For example, when the apogee end of the orbit is closest to the sun, it will pull and accelerate the moon throughout the perigee-to-apogee track, then slow the moon by an identical total force throughout the forth-coming apogee-to-perigee track. This is predicted to make the apogee tip longer and the perigee tip shorter. If there is never any net gain in lateral velocity due to solar gravity, then the sun can only cause changes in the lunar path. But there is a way for the sun to increase lateral velocity.

What could possibly be the problem if we were all to learn that the sun causes all orbital accelerations and decelerations? Unlike the earth, the sun is not inside the lunar orbit, and can therefore change the orbital speed. The sun's force on the moon is ever-changing upon the many faces of the moon. The earth's force is always on the same face of the moon, always pulling it toward one direction.

The earth's distance from the sun changes, and one might argue that the solar gravity is causing this variation, wherefore the earth can cause the variations in lunar distances. I have no problem with the earth and sun changing distances of planetary bodies, but there is a marked difference between the earth and moon orbit. The earth always keeps to the same ellipse with only very-minute variations. The sun is not able to speed the earth up, apparently, or slow it down. The way that the Creator tossed the earth (so to speak) around the sun, that's where she stays. We can't say that the sun caused the ellipse in the first place, though evolutionists are just that stupid. They can do great math, otherwise they are stupid. I am stupid at math, but I acknowledge the Creator.

The planets can play a small role, only, in changing the earth's orbital velocity, but have you ever heard that one year is longer than another? The moon's orbital period is always changing because there is a large outside-the-orbit body always yanking on the moon from various directions. There is no-such comparison for the earth.

Wikipedia has a formula for finding orbital velocity, but there is nothing in it that includes the sun: "For the precise orbital speed of a body at any given point in its trajectory, both the mean distance and the precise distance are taken into account: velocity = the square root of (standard gravitational parameter x ((2 / radius) - (1 / semi-major axis))." This is another case where the writer defines "mean distance" as the semi-major axis; I don't see how they can be the same thing.

Clicking on standard gravitational parameter, we come to an article telling: "In celestial mechanics, the standard gravitational parameter of a celestial body is the product of the gravitational constant G and the mass M of the body." This is bad news. It then says that the standard gravitational parameter is identical to G x M (M is the mass of the body of concern). A thing similar was discussed a couple of chapters ago where the moon's acceleration is said to be found using G x earth mass, yet their number didn't work to explain the moon's rate of fall from apogee. I've briefly already explained why these formulas, using mass for gravitational computations, are bogus.

I want to know the lunar velocity at apogee of July 15, 2000. The standard gravitational parameter given for the moon is 4,904,869,500,000. The semi major axis, for the month of concern, is half of 252,400 + 222,686, or 237,543 miles (382,288 km). We can now do the lunar-orbital-velocity formula above to see what we get, using kilometers:

square root of (4,904,869,500,000 x ((2 / 406,199) - (1 / 382,288))
=square root of (4,904,869,500,000 x ((.0000049237) - (.0000026159)))
= square root of (4,904,869,500,000 x (.0000023078))
= square root of (11,319,457.8)
= 3,364.4

We now get a result that's equal to 2,090.5 mi/hr, not bad but too low (the number doesn't change even if the moon's gravity number is increased to 4,904,869,549,999). This velocity number is lower even that NASA's lowest lunar velocity of 2,156.4 mi/hr. I've noted that there is nothing to indicate, to the formula, that the planet is off center, meaning that the formula may be for circular orbits only. When an orbit is not round, it has so-called eccentricity. "The eccentricity of the Earth's orbit is currently about 0.0167; the Earth's orbit is nearly circular."

Why didn't the Wikipedia writer inform the reader as to whether the formula was for round versus ellipse shapes? Is it because a conspiracy exists not to give the man on the street the means to find the lunar velocities at any one place of the orbit for any one month whatsoever? If the formula is so simple, where are the automatic calculators for lunar velocity? Whom has demanded that we not have access to them, and why is Google supporting this conspiracy?? Do the goons rush to control certain Wikipedia articles? Of course they do. The goons are the control freaks, the ones who engineer society; they never stop working together, of the same mind, to maintain the anti-God program of this wicked generation.

They are feeding us garbage and/or keeping us from knowing the true lunar velocities. This formula promised to get "the PRECISE orbital speed of a body at any given point in its trajectory," exactly what I've wanted for weeks. And here it gets an impossible speed on the low side, playing better to NASA's impossible 2,000 mi/hr in the eclipse of July, 2000.

When it came to devising the formula for the earth figure, they made up the gravity constant to fit that need, and for all the planets, which were assigned the same gravity constant, they assigned fictitious masses according to the need(s). It's playing with numbers.

Let's try the formula with the question on the velocity at perigee (222,686 miles = 358,378 kilometers):

square root of (4,904,869,500,000 x ((2 / 358,378) - (1 / 382,288))
=square root of (4,904,869,500,000 x ((.0000055807) - (.0000026159)))
= square root of (4,904,869,500,000 x (.0000029648))
= square root of (14,541,955.7)
= 3,813.4 k/h = 2,369.5 mi/hr

Someone might say, well, it's close to the claims. But for anyone wanting serious investigation / analysis, this is not good. They say that perigee velocities are in the neighborhood of 2,400 mi/hr, give or take 10, but here the figure is shy by far too much. No near-precision is possible with this formula.

The article for Standard gravitational parameter has three formulas to calculate the standard gravitational parameter for a circle, and one formula for an ellipse. The latter is: 4xpi squared x semi-major-axis cubed / orbital period squared. That seems easy enough in order to check whether the 4,904,869,500,000 number used above is accurate (use the large online calculator below). The semi-major axis (382,288 kilometers in meters) cubed is 55,869,141,413,711,872,000,000,000, and 4pi squared is 157.909; the two multiplied get 8,822,240,251,497,827,995,648,000,000. The orbital period (2,354,139.14 seconds) squared is 5,541,971,090,480, so that we can end with 8,822,240,251,497,827,995,648,000,000 / 5,541,971,090,480 = 1,591,895,754,680,618. Nope, no good. Should we try the orbital period in hours? No, because the number is already too large. Should we use kilometers? I tried that, and got the same result with nine less decimal places, a number way too small. Why isn't it working?

Let's go to Wikipedia's "Orbital period" article to find the Kepler-based formula (for an ellipse): 2pi x square root of (semi-major-axis cubed in meters / standard gravitational parameter). Using the earth's standard gravitational parameter (398,591,302,000,000), the math is: 6.28318 x square root of 55,869,141,413,711,872,000,000,000 meters / 398,591,302,000,000, or 6.28318 x 374388.15 = 2,352,348 seconds = 27.226 days, though it's 27.30 if one uses 23.9344 hours to a day, which is close enough to 27.32166 days to be remarkable. However, if one uses the average semi-major axis (384,400) reported by NASA instead of the semi-major axis of in July of 2,000 (used in the math above), the math is: 6.28318 x square root (56,800,235,584,000,000,000,000,000 / 398,591,302,000,000) = 2,371,868.8 seconds (27.45 or 27.5 days). It's not close enough to make Kepler's third law perfect (the law is "The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit"). One can spot here that they have the G x M for the earth well-calibrated to get the lunar orbital period, but it doesn't quite work for them no matter how they fashion their GM combo. I suggest adjusting (changing) the standard gravitational parameter until the formula gives the correct number of seconds before doing the formula below.

The same page gives the formula for finding the semi-major axis: cubed root of (standard gravitational parameter x orbital period squared / 4pi squared). One should be able to use the latter formula for finding the distance to the middle-size moon in any orbit, not only because the orbital period is known (I have yet to find anyone telling of the variations in the lunar period), but because the semi-major axis is equal to the distance of the middle-size moon. In other words, this formula is fairly reliable for finding the middle-size moon in any month where the apogee and perigee distances are known. The period number obtained from the formula is 2,354,139 - 2,352,348 = 1,791 seconds = .5 hour different than my figure. Mine used days of 23.9344 hours, but if we use 24 hours, the lunar period is 2,360,591 seconds, which is now more than 8,000 seconds off from the result in the formula.

After giving the formula for orbital period, the writer says: "For all ellipses with a given semi-major axis the orbital period is the same, regardless of eccentricity." In other words, the formula gives a period that is unchanged whether an oval of one length is skinny or fat, a head-scratcher. Doesn't a skinny oval of the same length have less circumference and therefore less velocity per any one period? The early astronomers probably needed to devise a formula for finding the periods of planets while not knowing their velocities or orbital shapes.

To find eccentricity, it's: (distance of planet from the long side of the ellipse minus distance of planet from the opposite side) / (distance of planet from the long side of the ellipse plus distance of planet from the opposite side). The problem in figuring it out for the moon is that apogee is not exactly opposite perigee. That is, while we can use the perigee distance for the formula, we can't use the apogee distance, except to get an approximation. We might be able to use the apogee distance to find what the ultimate distance is to the end perfectly opposite from perigee.

Lunar-Orbit Pattern

Fred Espenak tries to give us a great idea on the lunar-orbit pattern, which I'm going to share and record here in case it can somehow get us the velocity of the moon at any apogee. "Major axis" is a fancy phrase merely for the long distance of an oval or ellipse as opposed to its width. The major axis will always have perigee at one "tip" of the oval, and apogee at the other tip. Fred has a drawing (Figure 4-2) on the page with some ellipses. In his four lunar situations (same drawing), a line from the earth through the moon at perigee ("direction of perigee") always points in the same direction exactly (to the right of the page), but don't let this fool you into thinking that this is the case in every month of the year. In fact, he's not even showing four different months by the four situations, for the perigee position always hops from point to point from month to month, and therefore points in all directions eventually. One could get the (wrong) impression, as per his drawing, that every 1/2 year exactly, the direction of perigee has come round to point in the same direction again.

If Fred had another lunar situation a little below situation C, at about 3:30 on the clock, a line from perigee to apogee would point directly at the sun. This spot on the clock is where the earth would be where he says, "At certain times, the perigee of the lunar orbit and the perihelion [= earth's closest to the sun] of Earth's orbit can have the same ecliptic longitude...the major axes are then essentially parallel to each other and point in the same direction." I've done the thinking for you, so let me explain. He's starting off with the earth nearest the sun, when the earth's perihelion tip (same as perigee but with a different name) pointing directly away from the sun. He's saying that the moon's perigee tip is roughly pointing in the same direction, meaning that the moon is then situated in a straight line with the earth and sun, with the moon further from the sun than earth. Just keep in mind that this is roughly at 3:30 on the clock of his drawing. And, by the way, the earth's tips are always pointing in the same direction, more or less, from year to year (perihelion always occurs in January).

He then continues: "As time passes, the major axis [perigee and apogee together] of the lunar orbit slowly rotates east with respect to Earth's major axis until it becomes perpendicular to it 2.21 years later. Each passing month, the tips of the lunar orbit shift over (hop) a certain distance, in the same direction, so as to go around in circles, forever, and this is what he means by saying, "rotates." Each monthly hop is not the same distance, but I'll work out the average distance below. It takes 2.21 years of these hops to get the apogee-perigee line 90 degrees (perpendicular) off from the line at the 3:30 position. He then says, "In another 2.21 years (4.42 years from the start), the major axes of the orbits are again parallel to each other, but the perigee and the perihelion are 180 apart as they point in opposite directions. After an additional period of 2.21 years, the axes are once more perpendicular. Finally, the Moon's perigee and Earth's perihelion again share the same ecliptic longitude after a total interval of 8.85 years." This shows a pattern, with perigee returning to the same position every 8.85 years. By 180 degrees, he's referring to half way around the circle; it's just to say that perigee, after 4.42 years, is on the left end of the orbit rather than on the right.

I'd like to know what the average monthly hop is, when a span of 8.85 years (118.308 months of 27.321677 days) sees the perigee position hopping to make one full "rotation" around the lunar path. The problem is, every lunar orbit has a different circumference. They say that the average lunar orbit is nearly 1.5 million miles, whereby the math is 1,500,000 / 118.308 = 12,678.77 miles. Divide the latter by 1.5 million to find that it's .0084525 of one lunar orbit, which, when multiplied by 360 gives 3.0429 degrees. Therefore, every month of 27.32166 days, perigee rotates 3.043 degree on average. To check the math, we start with 365.24 days / 27.32166 days = 13.36815 months, and then find 1 year / 13.36815 = .074805 of a year, the same as saying .0748 of one orbit. That latter number divided by 8.85 is .0084525 of one orbit (same number as above) = 3.0429 degrees. It checks out.

In this context, what does it mean when we divide one lunar month by 8.85? Why does that much of the month get the amount of perigee shift per month? It's because 8.85 years gets successive perigees to hop one time around the lunar orbit = one month. We can now say that one perigee to another is one full monthly circle, plus 3.043 degree on average. We can check the correctness of this because they say that the so-called anomalistic month -- perigee to perigee -- is 27.55 days on average (Fred has an average of 27.55455). When we convert 3.043 degree to time, we first divide it by 360 degrees to get .0084525, and we multiply the latter by 27.32166 to get .23094 of one day. It means that perigee to perigee is .23094 day on top of 27.32166 days, or 27.5526 days. It works, though it's not quite Fred's number, perhaps because I'm using 1.5 million exactly to derive mine, while astronomers don't likely use a perfect 1.5. Fred defines the anomalistic month as perigee to perigee, but does not say that it includes apogee to apogee. He says that the anomalistic month can vary, give or take less than a day, typically.

Before going on, I want to say that Fred is likely rounding 8.85 off. As he uses 2.21 as 1/4 of 8.85, it seems that the correct figure lies between 8.845 and 8.85. By using the lowest possible, 8.845, the math above gets 27.5527, not much different than my 27.5526, which tends to reveal either that 1.5 million miles exactly is the reality, or that astronomers were using that number when they weren't sure of the true distance. Which option sounds more likely?

As I just got the anomalistic month with math using their average lunar distance (it's what gets 1.5 million), astronomers can do the same, meaning that Fred's number can be as wrong as their wrong average lunar distance. I can do the math in reverse to get the correct average lunar distance, if Fred's number is correct. The math would start with his number, 27.55455, minus 27.32166 days, to get .23289. To convert the latter to degrees (on a circle), we first divide it by 27.32166 to get .008524, then multiply the latter by 360 to find 3.068642 degree (I had 3.043 using a perfect 1.5 million). Now what? Well, for starters, 360 / 3.068642 = 117.3, not the 118.3 that Fred is required to use according to his own statements. Fred's anomalistic month is looking a little wrong.

To finish the math using .008524, let me repeat: "They say that the average lunar orbit is nearly 1.5 million miles, wherefore the math is 1,500,000 / 118.308 = 12,678.77 miles. Divide the latter by 1.5 million to find that it's .0084525 of one lunar orbit, which, when multiplied by 360 gives 3.0429 degree" As we can see, in continuing to work backward, we need to multiply .008524 by the lunar distance, yet the latter is what we want to discover. Are we stuck? From this problem, I was able to see that .008524 x lunar distance = lunar distance / 118.308. We can do hit-and-miss until we find the correct number for lunar distance, except that the math doesn't work with those two numbers (see below).

Plus, .008524 x 118.308 = 1.008457. This is incorrect. Instead, my number above, .0084525, is obtained properly with 1.0 / 118.308. This is necessary because .0084525 of 1.0 lunar orbit is equal to 118.308th of a lunar orbit. This again means that the number used by Fred for the anomalistic month is looking wrong, unless his 8.85 number is wrong.

I probably wouldn't have included this section, where its seems I'm being petty on Fred, except that a couple of coincidences crept up. At first glance, I don't know how to explain (coincidence?) that 1.008457 is equal to 1 1/118th. That is, 1 / .008457 = 118.245. As each perigee hop has worked out to 118/th of the lunar orbit, this is like tacking on one month too many over 8.85 years. Why would they tack on an extra month? Maybe they didn't; maybe his wrong number happens to be 1/118th too large by coincidence.

When we subtract his number from mine, we get 27.55455 - 27.5526 = .00195 of a day, which, over 118.308 months is .2307 day, which is virtually the difference (= .2309) between a true month and 27.5526 days. Why are these coincidences cropping up? Is the math crying out some underlying thing? If it is, I can't grasp it with the time I'm willing to devote to it, which isn't much at the moment. Believe it or not, I have other things to do in life more pressing.

Although it may seem irrelevant to multiply 1.008457 by 27.32166, it gets 27.5527, which is virtually my number of 27.5526 obtained from the perfect 1.5 million. Actually, my number became 27.5527 when 8.845 years (118.241 months) was used instead of 8.85 (118.308 months). This is amazing or perplexing or revealing, that his number, 1.008524, that goes with his number for the anomalistic month, gets my number for the anomalistic month. It might start to appear that astronomers, when they didn't know the true average lunar distance, or when they were not yet committed to a number, were using a perfect 1.5 million for the circumference. The idea coming to mind is that, upon using 1.5 million in their math, they also came up with 1.008457 as the difference between the true month and their wrong anomalistic-month figure.

Let's go back to, "The way I see it, the distance is found with: .008524 x lunar distance = lunar distance / 118.308. We can do hit-and-miss until we find the correct number for lunar distance, except that the math never works with those two numbers no matter what we use as the lunar distance." The formula can be re-arranged as .008524 x lunar distance x 118.308 = lunar distance. Here is what I get:

.008524 x 1,600,000 x 118.308 = 1,613,530
.008524 x 1,500,000 x 118.308 = 1,512,686
.008524 x 1,400,000 x 118.308 = 1,411,840

The answer is always more than 10,000 miles off what it should be, and there is no match until one gets down to under 1 million (no one thinks the lunar circumference is under 1 million). The number, .008524, is required by Fred if he insists on using 27.55455 days as the anomalistic month. He is clearly wrong with 27.55455, according to his own 118.308 number. It's not a big deal, except that there's a reason for it. Let's try my number obtained from a perfect 1.5 million:

.008452513 x 1,600,000 x 118.308 = 1,599,999.85
.008452513 x 1,500,000 x 118.308 = 1,499,999.86
.008452513 x 1,400,000 x 118.308 = 1,399,999.87

Perfect. If the 188.308 was tweaked a little, not forgetting that it changes the .008452513 figure, the math should be even more perfect. However, it has surprised me to find that different lunar-distance numbers all work in this equation. I was hoping that it would reveal the true lunar distance. There is nothing magical or revealing here because one orbit divided by 118.309 = .008452513 of an orbit. Nor does this corroborate that 118.309 months is proved correct for Fred's claims. One could divide an orbit by 5 and then say, 1/5th x lunar distance = lunar distance / 5. That's all we're doing above.

I have no doubt that astronomers can measure the apparent size of the moon to find, with high accuracy, where each perigee and apogee is thereby. Keeping tabs on the lunar sizes as they appear to our eye / telescope, over a period of years, will likely prove that Fred is conveying the truth concerning one full rotation of perigee over 118.3 months. In his Figure 4-7, he shows a string of perigees and apogees over three years from the start of 2008. There is a perigee pattern as follows: 1) three months back-to-back where perigee draws progressively nearer to earth, followed by 2) a fourth month that has perigee virtually at the same distance as the third, followed by 3) three months back-to-back where perigee moves progressively further from earth, followed by 4) a fourth month that has perigee virtually at the same distance, followed by 1), 2), and 3), but where 4), this time, is not followed by a fourth month at virtually the same distance. That's the end of the pattern.

Perhaps the most notable feature in this chart is the way the apogee distances are more uniform than the perigee distances. It's the perigee distances that go into big swings ((356,000 to 370,000 kilometers), while apogee distances stay more or less the same (between 404,000 and 407,000 kilometers). As with the perigee changes, the apogee changes are in swings (smooth roller-coaster shape), not erratic. The moon is definitely controlled, not bouncing around in danger of getting away from us, or worse. Something is causing the pattern to repeat itself, and of course it can't be the earth alone because there is a giant neighbor.

It becomes a lot easier to envision what the sun might do to a lunar orbit if every lunar month is drawn on the earth orbit with the sun in the center. But this would be helpful only if we know which way the lunar ellipse points. We now know: perigee points progressively 3.04 degree counterclockwise, per month.

In the pattern of Figure 4-7, the furthest apogees always coincide with the nearest perigees, and vice-versa. This is appreciated information, thank you Fred. He says, "the extremes in the length of the anomalistic month are 24.629 days (2.925 days shorter than the mean) to 28.565 days (1.011 days longer than the mean". This is a tidy mess, but a mess just the same.

If you have the time to compare the chart above with Figure 4-4, see how the latter shows an extremely short anomalistic month every seven or eight months. I find this unexpected when the sun is not part of the situation. Why should perigee to perigee step on the gas every seven or eight months? What puts the moon in over-drive speed at those times? If you compare charts, it always takes place when the perigee moon is nearly at its farthest. There is nothing in Figure 4-7 to indicate why these short months should occur, but Fred alludes to the cause when he says: "But once or twice every 7 to 8 months, the anomalistic month is significantly shorter than the mean by 2 to nearly 3 days. The difference in longitude of the Sun and perigee show that the shortest anomalistic months are correlated with values of 90 and 270, when the line of apsides is perpendicular to the Sun's direction.". It suggests that solar gravity plays a role.

Let me explain in the way I see it, starting with the fact that apsides is defined by Fred as the perigee line. He (in Figure 4-4) has the perigee line from the earth to perigee, and the earth is at the center of the ellipse in this particular direction, meaning that his apsides line divides the ellipse perfectly in half, as it should because an apsides line is defined as cutting an oval perfectly in half.

We are getting to the point. The earth is closer to perigee than to apogee so that earth is not centered on what one may call the anti-apsides line. This one is drawn through the center of the oval too, but perpendicular (90 degrees) to the apsides line. This is simple; the two lines form a cross upon the oval, both crossing its perfect center.

The anti-apsides line is the shortest-possible line across an oval, providing that the line goes through the perfect center. The anti-apsides line is not the shortest-possible line crossing an oval, but is if the line must pass though the perfect center. A line passing in this direction through the center is the so-called minor axis (the perigee line is across the major axis, simple enough).

So, this is important: the anti-apsides line crosses the widest part of an oval between perigee and apogee. The point is, in the case of the lunar orbit, it can't be an oval. Due to lack of any other explanation for the occurrence the extremely-short months (Fred doesn't attempt an explanation more than was said above), I'm claiming that the lunar orbit is an egg-shape rather than an oval. The lunar orbit, every seven months or so, is longer at one end, just like an egg. This may not be news to astronomers, but I have never read that the lunar orbit can be an egg-shape.

To put this another way, I'm claiming that the widest part of the lunar orbit's minor-axis region is not perfectly midway between the perigee and apogee ends (Fred uses an oval with the widest part dead-center). In this picture, "wide" refers to the minor-axis region, and "long" to the major-axis direction. If the widest part is perfectly centered from end to end, the situation above, every 7 / 8 months, cannot occur, in my opinion. It can occur only if one "half" of the ellipse is longer than the other end, with the widest part off-set from the center. Thereby, the ellipse starts to tighten (become less wide) before it reaches the anti-apsides line i.e. at the perfect center.

So what? Well, let the moon orbit until the apogee-perigee line points in both directions as far as possible from the sun, when the anti-apsides line is pointing toward the sun. What happens at this time? If the moon has reached the widest part of the orbit before it's half way across the length of the orbit, it starts to near the sun before it gets to the halfway point, meaning that it will fall toward solar gravity for longer than half the lunar orbit, and therefore rise away from the sun for less than half the orbit.

Falling to solar gravity significantly longer than rising away amounts to more acceleration of lunar velocity. Hence, the month is extremely short at this time.

In the year 2000, these short months are a few months on either side of the lunar eclipse i.e. not near the eclipse. At the link above, you can verify that they take place when the apogee moon is amongst its closest distances. One might therefore predict that, because the moon's fastest velocities occur in relation to these short months, the moon's slowest velocities occur midway between them i.e. smack at the lunar eclipse. This eclipse was on the day after the apogee moon was at its furthest distance on this particular swing. It is roundly claimed that the moon is at its slowest precisely at apogee, but here we have the eclipse on one of the furthest-possible apogees. It means that NASA has no choice but to claim a lunar velocity (for the eclipse) very near 2,156.4 mph, the number it claims for the slowest-ever lunar velocity.

I can see why evolutionist astronomers might not like to report the lunar picture I've just shared, where the sun plays a significant role in altering the orbital shape every seven or eight months. Certainly, if the moon is accelerated more than normally at these times, the shape of the orbit will be changed too. We therefore learn that there is a greater balancing act taking place, than if this situation was not the case, to keep the moon in orbit . It screams all the louder for a Creator knowing exactly what he was doing. It seems to me that, every seven / eight months, the orbital shape is made rounder, less egg-shape (the apogee end is not as long / pointed).

Between extremely-short months, something needs to slow the moon down to normal, otherwise the said added acceleration due to solar gravity would continually build lunar velocity until the moon spirals out of orbit. There are two counteraction possibilities: 1) earth's gravity; 2) ether friction. In an effort to find the counteracting force, it's not correct to assume that the moon is falling for a longer period to solar gravity in one extremely-short month, and then rising for a longer period away from solar gravity in the next extremely-short month. In fact, the moon never rises from the sun more than it falls because it can be gleaned that equilibrium has been reached midway between the extremely-short months. That is, with the apsides line pointed at the sun, the moon falls to the sun as much as it rises.

Checking Figure 4-4, you can see that the month after an extremely-short month is another shortish month, suggesting that the moon is still falling more than rising (to the sun). Let put the egg to the test. In case you need a drawing, let's put the egg representing the shortest month at the 12 o'clock position of the earth orbit. The anti-apsides line goes vertically through the sun at the center of the clock. You can put two ticks on the egg, one a little to the right of the line at the top of the egg, and a second one a little to the right of the line at the bottom of the egg. The two ticks represent the widest part of the egg.

The next month occurs roughly at the 11 o'clock position, but to find exactly how far, we divide 365.24 days by 27.32166 days to find 13.37 months per year. We divide 360 by 13.37 to find 26.9 degrees of moon shift per month (works out to 4.45 minutes on the clock). You can draw / imagine a second egg at the appropriate spot near 11 o'clock, having a line from its center and running 26.9 degrees to the center of the clock. Your 11 o'clock line is different from the 12 o'clock line by 26.9 degrees. We need to keep in mind that, in each month, the line of perigee hops 3.04 degrees counterclockwise around the egg, making the anti-apses line shift 3.04 degrees too.

When will the moon start to fall to solar gravity as concerns the second line? Will it be as soon as it reaches the top of the line, or will it be a little before it reaches the line? Shouldn't we put little ticks to the right of the second line too, and start the moon's fall (to the sun) at those ticks? I say, yes, we should have the ticks again, but not as far from the second line as the first line, because the longitudinal aspect (opposite of round) of an egg is greatest at its middle while the ends of an egg have more roundness. It's the longitudinal aspect of the central part of the moon-orbit egg that increases the length of time for the moon to fall / accelerate. There should be no ticks on the fourth line near apogee.

If we draw a third line 26.9 degrees from the second line, the roundness only increases at its top and bottom (i.e. we expect less lunar acceleration and therefore longer months). By the time we are at the fourth line, with a top 26.9 x 4 = 107.6 degrees from 12 o'clock, it has probably passed apogee, and is midway to the next extremely-short month. The latter is 26.9 x 7 = 188.3 degrees past the 12 o'clock position, and 8.3 degrees past the 6 o'clock position. What do you see? I see perfection according to my theory. For while the seventh month starts at 8.3 degrees past the bottom of the clock, and ends 35.2 past the bottom, thus having its center at 21.75 degrees, the anti-apsides line, after 7.5 months, has shifted from the vertical by 3.04 x 7.5 = 22.8 degrees! It's close enough of a match to see the picture. In other words, the middle of the seventh month has the anti-apsides line virtually straight with the line to the sun so there is a repeat of the situation at 12 o'clock! Excellent. Now we know the secret to the seven/eight month period, which, I can now gather, is really a repeated period of slightly less more 7.5 months, not either seven or eight.

Let me fix the above so that both scenarios work out to the same angle, and let's use the more-accurate 26.93. When the moon as orbited 7.5346 months past 12 o'clock, it has gone along by 26.93 x 7.5346 = 22.906 degree past 6 o'clock. When the moon has orbited 7.5346 months, the anti-apsides line has shifted 22.905 degrees.

We can do the math for double that amount, or 15.0692 months after 12 o'clock. It works. After 15.0692 months, the moon has progressed 15.0692 x 26.93 = 405.81 degrees, and that's equal to 405.81 - 360 = 45.81 degree past 12 o'clock. Meanwhile, after the same amount of time, the anti-apsides line has shifted 15.0692 x 3.04 = 45.81 degrees, a match. The line through the center of the egg is once again perfectly in line with the sun's center. Why didn't Fred tell us this amazing thing? Why would he want to hide such grand evidence that the cause of the extremely-short months is due to solar gravity working in this way?

Figure 4-4 has the eight-month interval every-other short month. That is, there is a repeat of 7, 8, 7, 8, 7, 8, assuring that it's really 7.5. Comparing Figure 4-4 with 4-7, it can be seen that the eight-month periods always come after the two, back-to-back perigees at virtually the same distance (from earth), wherefore an eight-month period and a seven-month period both occur seven months after a perigee is the furthest from earth.

We are also learning here, in case we need it, that the apogee-perigee line was pointing more or less at the sun during the eclipse (July 2000). The egg shape poses no advantage to solar gravity when the apogee-perigee line points toward the sun, wherefore the sun could not make the moon fly faster at this eclipse.

Fred's eclipse page has the moon at 2,000 mph, and it's not a wonder, therefore, that Fred doesn't mention any lunar velocities in that very large article that I've been discussing in this chapter. An entire article on lunar months, but no velocity figures given, and virtually zero on the nature and details of the sidereal month, which is the true month. How about that. He says that "Although the Moon's position and velocity can be described by the classic Keplerian orbital elements..."


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