Isaac Newton's Gravity Constant
If you were following the Western-backed invasion on Manbij several weeks ago, it took that long to liberate the city. There were reports that the main army, a Kurd-Arab alliance, were allowing the ISIS fighters to flee in a bid to keep as many innocent lives safer. But as this seems to be a terrible miscalculation, I blame it on the United States. I say the United States convinced the Kurd-Arab alliance that the best thing to do is to let ISIS flee to Turkey. Here's what happened:
[ISIS] militants who fled Manbij after its liberation last week, were able to enter Turkey and strengthen the terrorist presence in the province of Gaziantep, Turkish MP Mahmut Togrul told Sputnik.
On Saturday 54 people were killed and more than 100 injured after a suicide bombing at a Kurdish wedding party in Turkey's southeastern city of Gaziantep. The Turkish government has said that the attack was carried out by a Daesh terrorist...
..."The date of this attack was no coincidence. This explosion is the terrorists' revenge for the liberation of Manbij..." Togrul said
These fighters are going to consider themselves God-blessed for escaping, and will therefore fight another day from another city. This is the bad policy of allowing them to escape. The war has only been prolonged. The liberation of Manjib came with little advantage if the bulk of the fighters were allowed to escape. I say that the Americans will urge these escapees to join the anti-Assad forces not viewed as terrorists by the West. It would be the American, no-brainer will to see them fight along with the "moderate" armies in Syria. This is Western hypocrisy at work in Syria. In the meantime, Sputnik and other Russian media are ever spreading anti-Western sentiments in an effort to topple America's world reputation. Isn't this a world war?
These Syrian Kurds have even been fighting Assad themselves, which may have been the U.S. plan because the U.S. distanced itself from Turkey just as Turkey complained about the Kurd-US alliance. I do not think that the U.S. would sacrifice Turkey's friendship just to have Kurds defeat ISIS, but if Kurds can be turned against Assad, I can see the U.S. sacrificing Turkey in hope that Turkey would continue to remain actively anti-Assad anyway.
The Pentagon has warned Syria and Russia that the US is ready to shoot down their planes, which according to Washington threaten American advisers in Kurdish regions in northern Syria.
The warning came after US fighter jets tried to engage Syrian Arab Air Force aircraft in Syria last week, but the showdown was avoided as government planes left before the Americans arrived.
On Monday, another Pentagon spokesman, Peter Cook, said, "We would continue to advise the Syrian regime to steer clear of those areas."
http://kurdpress.ir/En/NSite/FullStory/News/?Id=14011#Title=US ready to target Russian, Syrian jets in Syria: Pentagon
Ahh, the Americans can now put their own advisors where they don't want the Syrian-Russian alliance to find victory, which is tantamount to using human shields. The Americans are using ISIS tactics. The spirit of ISIS has grown into their own toenails, and so the Americans are becoming animals for seeking to protect animals for an-end-justifies-the-means program. Push-comes-to-shove was yesterday's news. Today, it looks like the animal is ready to fight Russia. If it didn't warn Russia in this way, I might not be calling the Pentagon an animal, but an animal doesn't think as well as a human. The Pentagon spokesman said: "the Syrian regime would be wise to avoid areas where coalition forces have been operating...If [the Russians] threaten US forces, we always have the right to defend our forces..." But wait. The coalition is trespassing in Syria, and aiding the rebels against Assad, and now Assad can't war with the coalition???? Only dumb animals think this badly. Are the Americans the untouchable angels? The article agrees with my charge against the do-nothing Pentagon: "Since , they [American coalition] have been carrying out airstrikes in the war-torn country, but their air raids have done little to stop the Takfiri group's advances there." The Pentagon claims to be in Syria to fight the terrorists, but this is a facade, and everyone knows it. The Pentagon had been bombing hospitals and schools in order to blame it on the Russians and Syrians, and might even bomb their own soldiers to do the same. The Pentagon, lest you forget, bombed New York's twin towers, and even the Pentagon building itself, all in a single day. Don't worry, God has a plan against the Pentagon.
It's not Hillary Clinton alone that we need to worry about; it's the Democrats who would vote to have her be the president. Roughly half the country. What a dismal nation. What spiritual disease grips it. Her woes are continuing, with even the FBI forced to cough up her secret emails...that supposedly never existed, according to herself. Why was she wiping away her nation-business emails? Even the New York Times had to do a write-up against her. Not many would trivialize this anymore. But they are still wanting her as president, shame. The latest scandal is her use of her political position to make money for the Clinton Foundation or Clinton Global Initiative. Who do these rats think they are? What sort of human slime feeds their "global initiative"? One needs to ask whether conflict-of-interest (a nice way of putting criminal activity) was her top priority as a high official rather than being concerned about national business for the good of the people. A normal person would not give her the benefit of the doubt any longer, but Democrats are not normal people. Ask the votes they gave a second time to Obama. He was the one who gave America the Hillary problem, because the Democrats wanted it. For them, it doesn't matter how corrupt or selfish Hillary is, so long as she's a "good" president.
Acceleration of the Moon
I'm still wanting to show that NASA has fixed the data on their lunar eclipse pages. Let's look again at the data for the July 16, 2000 lunar eclipse, telling us that the sun is 15.673 million miles away:
With a shadow path of 5,738 miles, the distance from shadow edge to earth edge is (7,918 - 5,738) / 2 = 1,090 miles. The solar distance using the 1,090 figure is like this: 3,959 / (.004578 - (1,090 / 252,000)) = 15.673 million miles.None of the figures in the math are my own aside from guessing at 252,000. The .004578 figure is their own, though I derived it from the triangle calculator (below) by feeding their numbers into it. Let me explain this before going on. The eclipse page under discussion has the sun's radius (as seen in the sky by the eye) at 15'44.2", which is 15.7367 arcminutes, which is then divided by 60 to find it equal to .262278 degree. Placing that number in the angle-A box of the triangle calculator, with 1 in the edge-b box, tells us the distance (in the edge-a box) that the line spreads out per 1 mile across (e.g. toward the sun). The edge-a box has .004577, and that's why I'm using .00458 in the calculation above.
In the same calculation, the brackets, (1,090 / 252,000), works out to .004325 mile, and this number represents the distance of line spread, per mile toward the sun, of their line (not mine) if it starts from the edge of an earth umbra 5,738 miles wide, and proceeds to the sun while kissing the edge of the earth. That is their lunar-eclipse line, in other words, based on their eclipse data. When that line meets the other line spreading out at .004578 mile per mile, that is where the sun was during the eclipse. It's so simple, yet astronomers have not reported such a gross problem to the world. Why not? Isn't it true that leading astronomers control the "facts"?
So, when I use, (.004578 - (1,090 / 252,000)), it's .004578 - .004325 = .000253 mile. That's the distance that the one line, with more spread per mile, catches up to the other line, with every mile toward the sun. The line with more spread starts at the center of the earth, because that's how they view their angular size of the sun, and their .262278-degree figure is half the angular size of the sun on the day of the eclipse. The lunar-eclipse line, with less spread, rather that staring at the middle of the earth, starts at the edge, and because the distance between the edge and center of the planet is 3,959 miles (their figure), that's why I use: 3,959 / (.004578 - (1,090 / 252,000)) = 15,673,000 miles. That math statement simply means that the line with more spread will spread out by 3,959 miles more than the lunar-eclipse line after they both proceed 15.673 million miles toward the sun.
And, of course, as the two lines make contact at that distance from the earth, they are both upon the edge of the sun. It's so simple, yet astronomers are not telling us concerning this method of solar-distance calculation. It is a thing that would dawn on an astronomer, yet, somehow, they have managed to keep this thing from us, and from amateur astronomers. One way they do it is to report false data in eclipse pages. We assume that they report false data everywhere, of course, but they must do so knowingly, absolutely. This is the mystery, that they have gotten away with this for so long. Surely, it has dawned on Fred Espenak's mind that he could draw two lines on a page, one for his lunar-eclipse line, and one for a solar line as per angular size, and where they meet, that is where the sun sits. Surely, it has dawned on him. But if he reports that the sun is less 16 million miles away, he could lose respect and/or his job, and maybe even his life. This is an important hoax for the enemies of God.
As I said, the only figure I entered in the math was 252,000...because I cannot find what the true lunar distance was on that day, though the apogee calculator page below says that the lunar distance, on July 15, 2000, was 406,199 (not rounded off) kilometers = 252,400 miles. Here is what it says of apogee in that July: "Jul 15 15:35 406199 km" As soon as the moon got to its furthest from earth, at 15:35 pm on July 15, it started to drop toward the earth at 0 mph. The question is how far would it have fallen 22 hours and 20 minutes later at mid-eclipse? Did it fall at the normal rate of gravity (called the acceleration of gravity), or something else due to having a path almost totally away from earth's gravitational pull?
https://www.fourmilab.ch/earthview/pacalc.html I happen to know that a bullet fired from three feet off the ground lands on the ground at the same time as a bullet dropped from the hand at three feet high. The lateral path of the speeding bullet makes no difference at all in the building speed of fall. It means that I me-self can calculate how far the moon was at mid-eclipse, and this is all one needs to know to verify the true solar distance. However, I can't calculate it unless I know how strong the earth's gravity is on the moon at a distance of 252,400 miles. I think I can do this.
Before going on, I want to clarify an error made by astronomy. They follow this rule: force of gravity = gravity x earth mass x moon mass / distance squared (from Wikipedia below). This formula is from Isaac Newton. Although I believe that Newton was wrong with his method, because he had the level of mass directly responsible for gravitational force, it is possible that his formula can be used for something because the gravity x mass numbers within it were made to reflect some reality that was being grasped and worked with. However, if these numbers are used for any other purpose, it will not work because the numbers are wrong to begin with for expressing what is claimed for them.
The moon falls toward the earth, picks up speed for a time, and then decelerates, repeatedly, never falling to the earth completely nor spiralling into space completely. To think that this situation was caused by anything but a Mind who knew what he was doing is the evolutionist's stupidity. You can chalk up most of the leading astronomers into this stupidity category. It is not an exaggeration to say that the moon falls to earth. It literally does so. It falls while moving in a straight line space-ward. Yes, believe it or not, the orbital path of the moon is a straight line. It's the falling of the moon that makes it circular. When it moves away from earth, it's still falling, but not as fast.
To think that blind chance from evolutionary processes got the straight line of all the planets at just the right angle, at just the right altitude, and at just the right speed, so that the falling of the planets can keep circular orbits for thousands of years is worse than stupid. It's rebellion against the Mind. If you're thinking that astronomers are anything but stupid, fine, but at least view them in rebellion against their own Creator. Some would say this rebellion is a stupid choice, and I'm one of them. But astronomers not only rebel, they teach others to neglect God. This is satanism, in the sense of doing satan's will.
Don't you think that One who can put the moon into orbit can be respected? Shouldn't He be the ruler of earth? Shouldn't he have some say in the affairs of mankind? Unlike a man, God doesn't rush to beat on his enemies, but has patience, for in the end, when His goals have been accomplished, punishment will go to his enemies on their resurrection from the dead. For them, it will feel that the moment they die, they will rise to punishment, for they will not feel the duration of death. The very second they die will be the very second they rise to punishment.
The website below confirms that gravity, as per Newton's formula, is a constant. This means that we use the same number (for gravity) in the formula regardless of the specific masses of earth and moon, or the distance between them. It gets easy this way. The writer (Joseph Lazio) says: "Solving for the acceleration on the Moon [everything to do with the fall of the moon to earth], we find a = (G x M) / (d x d)", where M in this case is the earth's mass. From this he finds the lunar acceleration as .0028 meters/s2.
Let's check his math. Here is his full statement changed (by me) for clarification: "Using m = 59.8 x 10 to the power of 23 kilograms [mass of the earth], d = 3.75 x 108 [10 to the power of 8] meters, and G = 6.67 x 10 the power of negative 11 N m2/kg2 [the G constant], we find a = 0.0028 m/s2". By the way, while he uses 59.8 x 10p23, others would use 5.98 x 10p24 (in what follows, I'll ignore the x10p and use p alone). To do his math, it's 5.98p24 x 6.67p-11 / 1.4p17 = .00285. His moon's acceleration toward earth is as per the distance of 375,000,000 meters, which, when squared, is 1.4 to the power of 17 (1.4 followed by 16 zeroes, 17 decimal places in all, or 140,000,000,000,000,000).
There is an online calculator below capable of doing this many zeroes. Do 5,980,000,000,000,000,000,000,000 x .0000000000667 to get 398,866,000,000,000. Then divide by 140,000,000,000,000,000 to get 002849...which apparently matches his .0028.
The writer's 375 million meters (lunar distance), when put into a meter-mile converter, gives 233,014 miles. We want to use 252,400 miles = 406,198,425.6 meters. So, we can re-do the math above, and find the lunar acceleration on the day before the eclipse. But before we re-do it, let's add that Wikipedia says the earth mass is 5.972 to the power of 24. Another says 5.974. NASA has a fact sheet using 5.9723p24 (not quite his 5.98p24).
Therefore, according to NASA, the lunar acceleration was 5.9723p24 x 6.67p-11 / 1.65p17 = .002414257 m/s2, where 1.65p17 is obtained from 406,198,425.6 squared. This thing called m/s2 (same as m/sec/sec) means meters per second per second. "An object experiences a constant acceleration of one metre per second squared (1 m/s2) from a state of rest, when it achieves the speed of 5 m/s after 5 seconds and 10 m/s after 10 seconds." Using this acceleration figure, I figure that one could view the situation with zero speed of fall at perfect apogee on July 16, 2000, then dropping by .002414257 x .5 meter (far less than a meter) after the first second, with a downward velocity of .002414257 x 2 meters after two seconds had passed, .002414257 x 3 meters after the third second, etc. But the truth is, the moon did not start to fall at apogee. It has always been falling ever since it has been in orbit. When it is moving neither away nor toward earth, at apogee, it is still falling, otherwise it would not be on a circular path. However, I'll show that the situation, for acceleration-calculation purposes, can be viewed by ignoring its total fall velocity, and treating only the part of the velocity as concerns its distance from earth.
We need to know how many meters the moon had fallen 22 hours, 20 minutes = 80,400 seconds after apogee, for mid-eclipse was at that time. There is a calculator below that gives the acceleration if you feed it the starting speed of fall, the final speed, and the duration. I have no way of using it unless I start with 0 velocity at apogee.
Ultimately, it may be best to check the results manually from the formulas shown on the pages. A second calculator below tells how to do it manually: "Average Acceleration = (Velocity Difference) / (Time Difference)". It's the same formula shown on the calculator page above. As you can see, the total velocity of the falling object is immaterial; it's the difference in speed that matters between two points in time. Therefore, we can start with 0 velocity at apogee. The problem is, we don't know the fall velocity at the eclipse, nor at perigee, but we do know the time between apogee and the eclipse, and the time between apogee and perigee. The formula we can develop is: mystery speed - 0 speed / 80,400 seconds = mystery acceleration. With two mysteries in the formula, we can't solve for either one.
The way I understand the .002414257 (or .002414 for short) figure obtained from NASA's earth-mass figure, an object dropped from the lunar distance of 252,400 miles from the earth's core will have a drop rate (toward earth) of .002414257 m/s2. We can drop the moon or a cannon ball, it's all the same. Using this number, one may, at first glance, think that the downward velocity at the eclipse is obtained like this: .002414257 x 80,400 seconds = 194.106 meters per second (you can verify these numbers at the distance-of-fall calculator below this paragraph) = 434.2 miles per hour. This cannot be correct, for, after the first 22.33 hours (= 80,400 seconds), there remained 322 hours until perigee, at which time the moon had fallen to an altitude of 222,686 miles, according to the apogee and perigee calculator. That's a fall of 252,400 - 222,686 = 29,714 miles over 14 days, 16 hours = 352 hours, but if the moon was already falling at 434 mph after the 22 hours, it would have fallen at least 434 x (352 - 22) = 143,220 miles after the 352 hours. That cannot be correct.
I cannot see anything wrong with my numbers, wherefore I see the problem lying with the .002414257 figure obtained from the Newtonian gravity number and/or the earth-mass figure. I'll deal with this in the next chapter.
I don't know what Isaac Newton was thinking when he tried to make the mass of the earth part of the gravitational force on the moon, but he was dead wrong. Atoms do not attract atoms. That's what the air in the tires of your car are telling you. If the air atoms inter-attracted in your tire, the air would form a liquid and not keep the tires inflated. It is a no-brainer that air atoms repel, therefore. Newton himself would have told evolutionists that an object in motion would cease to move if it was smacked in a direction opposite the motion by a force equal to the force of the motion. That is, if two billiard balls come directly at one another with the same force, they would both stop in their tracks. They would not bounce off of one another and maintain their respective velocities.
In other words, Newton knew that two objects, when they made contact with one another from any direction, would lose some of their energy, more-and-more so with continual contacts. But evolutionists, lying beasts that they are, took the position, and then taught the fantasy to your children, that atoms are in constant motion, never able to lose energy when making contact. Therefore, they say, that while air atoms in your tires are attracting one another, they are flying about at hundreds of mph, and thus keeping your tires from going flat by continually smacking up against the inside of the rubber. The atoms never stop moving while you sleep, never stop moving while you drive a million miles, because they never lose energy when they make contact with one another; they just transfer all their energy to one another...exactly what Newton's law opposed. He, and the evolutionist both, knew that an object moving in pure space cannot slow down unless something makes contact with it, but evolutionists so-badly needed atoms in attraction that they sacrificed this reality for a fantasy.
We have quacks guiding our education because they need atomic attraction to form galaxies and stars from the big bang, and without a Creator in the picture. This is a fact, as surely as we exist: quacks are guiding our physics education. If the math above doesn't work, and it doesn't, I would suggest it's because Newton was wrongly interpreting gravity. It's not atoms attracting atoms; it's the negative charge of gravity particles (free electrons in the planets) that attract all atoms. But even if you can't agree with my view of gravity, we can still reject Newton's for something else yet to be discovered.
I have found a distance-of-fall calculator (below), and it uses none of the Newtonian numbers. Feeding it 80,400 seconds along with the acceleration figure (.002414257) above gives the result of 7.8 million meters = 4,848.6 miles. This would bring the moon from 252,400 to less than 247,600 in just one day, which is far too much. This shows that we cannot use that acceleration figure for merely the part of fall that we see, the part that brings the moon closer to the earth. There is the invisible part of the fall that needs to be figured into the math.
The calculator page gives the formula used as: height of fall = 1/2 x m/s2 x time squared. The page gives the formula for velocity of fall as m/s2 x time, easy. Here is what we get for the height of fall using the first formula: .5 x .002414257 x (80,400 x 80,400) = 7,803,071.76456 meters = 4,848.604 miles, same as what the calculator gave above.
By the way, when the free-fall calculator gives the velocity (works out to 434.2 mi/hr), one might assume it to be the velocity at the end of the fall, not the average velocity. I can get the same fall distance (4,848.6) using half of 434.2 (which is what I see as the average velocity over the 80,400 seconds): 434.2036 x .5 x 22.3333 hours = 4,848.6 miles. This appears to prove that 434.2 is the speed at the end of the fall, but does not prove that the average velocity is exactly midway during the fall. Is it, or isn't it? In free-fall, most of the fall is near the end so that, half way through the fall, one would naturally predict less than half the final velocity. It proves to be a wrong idea.
If we enter half of 80,400 seconds (with .002414257 in the g box), the free-fall velocity is given as 97.05 m/s, whereas it's given as 194.106 (434.2 mi/hr) for 80,400. It's cleanly half the velocity with half the time of fall, which I thought was incorrect. But, checking, it proved to be correct. In the same way, using 1/4 of 80,400 gets 1/4 of 194.106 m/s, giving me the impression that the calculator gives a smooth free-fall picture. It is made plain everywhere that free fall has a compounding fall over time, and I was under the impression that there should be much more than 2 times the velocity after 20 seconds as compared to after 10 seconds, but, the truth is, there is much more than twice the distance-of-fall after 20 seconds as compared to 10. In case you get the same impression, test by putting 10 in the top box and 1 m/s2 in the g box, then use 20 in the top box. The velocity figures will be 10 m/s versus 20, a difference of 2 times. But the fall distance will be 50 versus 200, a difference of 4 times.
If you need to get a picture of what goes on (otherwise skip this paragraph), enter 1 meter in the top box and 1 m/s2 in the g box. The velocity is correctly given as 1 m/s after 1 second, with a fall distance of .5 meter over the first second. It sounds reasonable. Using 2 seconds in the top box then gets a correct velocity of 2 m/s, and a fall distance of 2 meters. It sounds logical. Using 3 seconds in the top box, the fall distance is now 4.5 meters while the velocity is correctly given as 3 m/s. Sounds good. With 4 seconds in the top box, the distance of fall is 8 meters. Sounds good again. Entering 8 seconds gets a fall distance of 32 meters, showing how the fall distance compounds with added time. The fall distance shows as four times as much for every doubling of time. This is true when we use 2 meters in the g box.
Confident that the calculator is working fine, we can proceed, but, for now, all I can do is find the acceleration for the visible part of the lunar drop. It may mean that, after apogee, the moon was dropping at half the velocity when it was half way in time to the eclipse, but by half way in time, the moon had dropped only one quarter of the full drop achieved by the moment of eclipse. By the time that it was at half the velocity, it was only a quarter of the way. This is useful information for the eclipse task at hand. Let me remind you, we are not talking about orbital velocity or distance.
However, I'm not sure whether we should view the moon in this way starting at apogee i.e. with most of the net fall near perigee. This gets complicated where the moon is falling continually much faster than the velocity by which the moon closes in on the earth between apogee and perigee. If the moon is constantly dropping, does it mean that it continually increases its drop velocity, forever? Of course not. In that case, it's drop velocity must be slowed repeatedly in cycles so that it does not ever double its total drop velocity. In that case, it's not strictly free fall. It's as though a thing in free fall had a string attached to its back, pulled up to slow the fall speed, then let go to increase the fall speed, repeatedly. The situation can be viewed as a free fall that has reached terminal velocity, but with cyclical deviations -- less drop speed and more drop speed.
By the way, I've been switching to mi/hr from mph because I want to differentiate between meters per hour. If you ever read mph, it is never meant as meters per hour.
Here is another free-fall calculator, in case you need it.
I have read more than once that a projectile moving horizontally across the earth does fall at the same rate as an object dropped straight down. Motion parallel with the earth's surface doesn't change the fall speed. Therefore, the moon's acceleration in the downward direction is not affected by its orbital path parallel with the earth surface.
One can work backward using 352 hours = 1,267,200 seconds as the time between apogee and perigee. Their math is .5 x .002414257 m/s2 x 1,267,200 seconds x 1,267,200 seconds = 1,938,401,923.6 meters = 1,204,467 miles. We can do it backward using 29,714 miles of visible fall, which converts to 47,820,048 meters. This is great because it gives us an acceleration number to work with, one of the two mystery items mentioned above. Their math backward is 1,204,467 / 1,267,200 / 1,267,200 / .5 = .002414257. To find the acceleration figure for the visible part of the fall, we therefore do: 47,820,048 / 1,267,200 / 1,267,200 / .5 = .00005956 m/s2.
When we put .00005956 into the free-fall calculator along with 80,400 seconds, we get 192,502.685 meters of fall = 119.6 miles (that's more like it), and a velocity of 4.79 meters per second = 10.71 miles per hour. We can do the calculation manually: .5 x .00005956 m/s2 x 80,400 x 80,400 = 96,251.34 meters = 119.6 miles. We now have the average-acceleration formula, velocity 2 - velocity 1 / time, fulfilled like this: 4.79 m/s - 0 m/s / 80,400 = .00005956 m/s2. This assumes that the moon had 0 fall velocity at apogee.
By the way, the regular online calculator that I'm using, shared above, does not always work. The problem may be pasting numbers with commas. It has been an enormous pain. It should be re-loaded from time to time.
The .00005956 should be changed because it represents the acceleration rate between apogee and perigee. The formula, over just a 119.6-mile distance, will get another acceleration level, but we can't find it until we have the correct distance. It's not exactly a drop of 119.6 miles, in other words, as that is obtained with .00005956. For example, if the true distance was 150 miles (241,401.6 meters), the acceleration would become: 241,401.6 / 80,400 / 80,000 / .5 = .0000747 m/s2. As I don't have the true distance, I can't find the more-accurate acceleration.
[I'm going to leave the paragraph above there even though the premise is wrong. Yes, one can find the distance by that formula, but it doesn't necessarily mean that the .00005956 is wrong along with 119.6. The latter will be verified in the next chapter. I assumed wrong here.]
While the free-fall calculator (and its formula of acceleration x time = fall velocity) automatically starts all falls at 0 velocity, the formula, velocity 2 - velocity 1 / time = acceleration, allows any starting velocity. Let's, for example, do: 4.848 m/s - .05956 m/s / 80,400 seconds = .00005956 m/s2. That's my number. Those two velocity figures were obtained from the free-fall calculator, one with 81,400 in the top box, and one with 1,000 in the top box (in both cases, .00005956 was in the g box). I did this to see whether the same .00005956 number would come up if the drop time were calculated between 1,000 seconds and 80,400 + 1,000, and it proves to be correct. In fact, when we subtract .05956 from 4.848, we get 4.788624, which is the velocity figure with 80,400 in the top box. In other words, whether we begin at 0 and end at 80,400, or begin at 1,000 and end at 81,400, the fall velocity at the end of either period will be 4.848 m/s more than at the start.
Therefore, it seems, it doesn't matter what velocity we use as the starting point of the moon at apogee, so long as we claim the acceleration amount over 80,400 seconds. It doesn't matter that the moon was dropping invisibly several hundreds of miles per hour at apogee, we can still find its acceleration over the next 80,400 seconds by fulfilling these formulas normally...if, that is, we had the lunar distance at the eclipse.
The moon's fall must decelerate and accelerate; the questions are: when does deceleration begin; when does acceleration begin? (This is not orbital-path acceleration and deceleration). If the moon were doing a perfect circle, it would remain parallel with the earth's surface constantly, and it's fall velocity rate would never change. Therefore, as soon as perigee is reached, and it starts moving away from the earth, the moon is beginning to decelerate in the up-down direction. When it reaches apogee, it starts to get closer to earth, meaning that apogee marks the point where acceleration begins.
We can do the same math over the apogee-to-perigee period of 1,267,200 seconds if we had the acceleration velocity over that time period. The .00005956 figure, with the free-fall calculator, has a velocity of 75.474 m/s when we use 1,267,200 seconds instead of 80,400. Therefore, we check: 75.474 m/s - 0 / 1,267,200 = .00005956. As the velocity figures given by the calculator are twice the average speeds, the average velocity is 75.474 / 2 = 37.737 mi/hr. The moon fell 29,714 miles = 47,820 kilometers (over the 1,267,200 seconds), wherefore the math can be done: 47,820 / 352 hours = 135,852.3 meters per hour = 37.7367 m/s on average, which is virtually the 37.737 above.
We can't use 37.737 as the average velocity if we're just doing the 22.333 hours to the eclipse. As the calculator gives 10.7 mi/hr when entering 22.33 hours (80,400 seconds), we use half of it over the 22.333 hours to find a lunar drop to the eclipse: 22.333 x 5.356 = 119.6 miles. Assuming the apogee calculator has the correct distance, 252,400, at apogee, the moon was about 252,280 miles away at the eclipse. Having this, there is just one more thing needed, the correct umbra diameter at the eclipse. I'll re-visit this in the next chapter.
Having the velocity of drop toward earth doesn't give us the moon's orbital velocity. The latter is the thing to know to get the correct umbra diameter.
Correct View of Gravity
I have not read how modern science explains the fact that all objects have the same free-fall rate. I have an explanation: all atoms weigh the same when they are the same distance from the gravity source. It doesn't matter whether a piece of steel has twice as many atoms, gravity attracts each atom separately, and thus both objects fall at the same rate (when dropped from the same distance from the gravity source). I cannot conceive of a second explanation. Modern science cannot adopt this apparent fact because it has different weights for all of the different atoms.
Weight is defined expressly as the force of gravity on a material. If one atom has more weight than another, it is pulled more-strongly by gravity. Therefore, iron balls should fall faster than plastic balls, if all atoms have different weights. While I keep reading on how fantastic it is that all objects should fall at the same rate, no one has ventured to explain it, so far as I've read over 20 or more years.
I tackled this question: how can all atoms be attracted by the same level of gravity force? I came up with a logical answer, but only after realizing, independent of this problem, that gravity was a negative force, having free electrons within the planet as the source. In this theory, heat is defined as free electrons and gravity force all in one. The internal heat of the planet has a negative charge that acts as gravity, and heat is the physical entry of electrons into the atomic pores of a material. The electrons that act as heat are free to roam, and always "rub" (probably not physically) against the captured electrons that make up the outer shell of atoms.
Electrons within atomic pores hover because they repel one another. They crowd around atomic shells, crushing inward on the shells, and thereby adding themselves to the outer layers of the shells. With increasing density (= increasing temperature), they crowd in all-the-more. They thus add negative charge to the atoms, causing atoms to repel one another increasingly with increasing temperature, explaining material expansion with increasing temperature.
The negative wave of force extending outward from the earth's hot core assures that some of an atom's captured electrons will be blown off the atom, for electrons repel electrons. You understand: the negative force of the great pool of interior, planetary electrons blow electrons off of atoms. You wouldn't have it any other way, if you knew what you were talking about. The outer electrons of atoms are barely held on with any protonic force; wouldn't you expect a large negative charge from the earth to be capable of affecting some of the outer electrons? What happens to an atom's charge when some of its electrons are blown away?
The closer you bring the atom to the gravity source, the more that electrons will be blown off, and the more positive the atoms becomes, thus being attracted all the more by the great pool of negative charge = gravity. That's why every material weighs more when brought closer to gravity.
How many electrons will be blown off? While I can't tell you the numbers per different type of atom, I can tell you, as a logical fact, that each atom, regardless of type, will only be able to retain the electrons that its proton holds stronger to itself than the gravity force repelling them away. This is simply too logical to view as a mere theory; it has got to be a fact. What force level should we assign to the outer layer of electrons if it is held to the proton by roughly a force equal to gravity? We can say that the outer shell is held to the atom by g force, not quite repelled away, and hovering dangerously, in danger of being lost from the atom if someone should come along and rub a material. That's called friction, and it produces heat because outer electrons are being rubbed away from atoms. Heat is the freedom of electrons from atoms. You cannot feel heat by touching an atom unless electrons are being freed to enter your skin. There is an enormous amount of heat energy on atoms, but unless freed, it poses no energy loss. Only the outer layers are freed, in normal processes, until someone causes a nuclear explosion by destroying the protonic cores. All the captured electrons then come flying out under their strained inter-repulsion. It's strained because the protonic core crushes them inward even while they seek to inter-repel away. It's like a closed spring wanting to shoot out. Scientists are daft to teach that protons can never be destroyed. Why can't protons be destroyed? They are so small.
As one goes deeper into the atom than the outer sphere, the lower levels of electrons will be held with progressively greater force than gravity force, a no-brainer. It's exactly the same in the atmosphere, with the lower air atoms held to gravity with more force than the outer air atoms. This is due only to the distance between the air atom and the gravity source. In the same way, the outer electrons are further from the protonic core that has captured them. The atom is really a mini-earth situation with an atmosphere of captured electrons (they do not orbit, of course, except in the mind of a quack).
Again, what is the level of gravitational pull on all atoms having their outer sphere of electrons held by the gravity force? One can begin to see the potential for all atoms to weigh the same. The better question is: does identical positive force go outward from all atoms when their outer spheres of electrons are held by an equal protonic force? It sounds logical to me. The pull of gravity is purely upon the positive force extending outward from the protonic core (I see one proton per atom, the quacks see otherwise). In all likeliness, all protonic cores of the different atoms have different levels of positive force, but, the question is, how much of that force gets out and away from the atom so that gravity can bite into it? All of it gets out, yes, but not all of it, really.
One can demonstrate that not all the positive force of an atom gets out past the outer sphere of electrons. For this, we take gravity out of the picture. The proton is now able to attract electrons forever, right? Wrong. It can't continue to attract electrons forever because with each electron captured, the negative force upon the atom builds, radiating outward to counteract some positive force. So, while the proton attracts free electrons in the vicinity, captured electrons repel them. Therefore, a protonic core can only capture or load electrons until the combined negative charge upon the atom equals the outward positive force of the proton. The atom is now neutral in charge. It has no net positive charge radiating outward. Yes, all the positive radiates out, but it is counter-balanced by the negative charge that it has brought to itself.
Enter gravity. What happens now? Does gravity fail to pull the atom because it is no longer net positive? Yes, and no. Yes, that would be true, except that gravity will blow some electrons from the atoms, and thereby open a net-positive charge upon them, and so gravity will get its bite on all atoms, pulling them to itself. It seems fair to say that, if all atoms lose all electrons aside from those held to the protonic core by the force of gravity, they will all radiate the same net positive charge. That's because every atom will have the same level of negative charge radiating outward after gravity has had its effect upon them. So, the pull of gravity on an atom can be described as: neutral or fully-loaded condition minus depth of electrons blown away. End of story.
As gravity force is not based on atomic mass, the formula used by science for free-fall acceleration has got to be faulty. It's possible that the Newtonian formula (or call it what you will) can be correct when the mass of the earth and moon are chosen (call it "fixing") to conform to hands-on experimental results. But the level of earth mass used by the calculators is clearly not working to express the free-fall reality, and that's because astronomers fixed the earth mass and gravity constant to correspond to orbital heights and velocities, not planetary free-fall situations.
It is wildly incorrect to view solar gravity based on the mass of the sun, which is a lot lighter, for unit volume, than the rock called earth. The sun, though having less mass per unit volume, has wildly greater gravity than earth per unit of volume. Here is how they find the orbital velocity of earth using Newtonian theory:
Mass of the sun (2x10p30 kilograms) x Newton's constant (6.7x10p-11) / solar distance (1.5x10p11 meters) = about 2.8 kilometers per second [= about 66,700 miles per hour as the earth's orbital speed].
Clearly, they have the Newtonian formula fixed so that it supports the 93-million-mile hoax. They simply chose a mass for the sun that works mathematically with the gravity constant to get the speed of earth as predicted by the 93-million-mile distance. So, they assigned the sun the wrong mass. In other words, they work backward, starting with 93 million, and then enter the solar-mass figure needed so that the math gets the expected (but wrong) orbital velocity of the earth. If the earth is only 5 million miles from the sun, the earth's orbital speed works out to 10 million x 3.14159 (pi) / 365.24 / 24 = 20,092 miles per hour.
Here is how we can test their solar-mass figure, as per finding the volume of a sphere: V = 4/3 x pi x r³. Their sun has therefore got 4/3 x 3.14159 x 695,991 kilometers x 695,991 x 695,991 = 1.4122x10p18 cubic kilometers = 1.4122x10p33 cubic centimeters. When we divide the solar mass of 2x10p30 kilograms by 1.4122x10p33 cubic centimeters, we find .0014 kilograms = 1.4 grams per cubic centimeter, which is heavier than water (1 gram per cubic centimeter). We need to ask whether the sun, made mainly of hydrogen (9 times lighter than water by volume), can be 1.4 times heavier than water, or whether their solar mass is just incorrect. There is so much heat in the sun that hydrogen atoms are predicted to be very far apart from one another (makes the sun less dense), as compared to the distance of water molecules in a vapor (weighs 9 times that of hydrogen) at normal temperatures here on earth. The density of the sun is given by them as 1410 kilograms per cubic meter, which is 1.41 grams per cubic centimeter.
The figure (.00006236) I used for the moon's acceleration is more than 38 times smaller than their figure (.00214). Why should this be so?
Here is an online statement not elaborated upon: "A series of measurements of Newton's gravity constant, G, dating back as far as 1893, yielded widely varying values, the variation greatly exceeding the stated error estimates (Gillies, 1997; Quinn, 2000, Mohr et al 2008). The value of G is usually said to be unrelated to other physics..." I wonder what warning is meant by the latter statement? Is it like saying a round shape doesn't fit a square hole, or like false doesn't fit true?
Let's look at the Newtonian formula, which is essentially gravity force / distance x distance. Suppose there is a magnet 3 centimeters from a nail. He would have said that the force of the magnet was: magnet force / 3 x 3. If the nail was 4 centimeters away, he would have said: magnetic force / 4 x 4. The latter gets a lower number, and reflects a weaker force. If the nail were 6 centimeters away, we would use 6 x 6 = 36, which is FOUR times 3 x 3 = 9. That is accurate because his use of distance x distance was intended to describe / reflect the so-called inverse-square law of magnetism, meaning that an object twice as far (as in 3 versus 6 centimeters) is attracted with FOUR times less force, and vice versa.
The problem is, how does one assign a number for the magnet force? If there was only one magnet in the world, all calculations would use the same number for the magnet, and that's why they can assign earth gravity one number. But how does one go about giving gravity a number? What does that number mean? They are correct to say gravity / distance x distance, but when they give gravity a number, that's where it's debatable. Hasn't anyone argued with Newtonian club? Have we all become stupid with them? What did Newton, know, anyway? He had so little to work with. His gravity = mass was just a guess, wasn't it?
The formula for finding acceleration is actually 6.674 x mass x mass / distance. What does it mean by mass x mass? Why not just one mass instead of mass x mass? That's because Newton saw any two objects as magnets, attracting one another. Mass x mass expresses attraction between two objects. In every gravitational situation envisioned by Newton, there was always mass x mass, and the size of the masses (or their weight to be precise) mattered, wherefore mass x mass needs to be expressed in the specific weights of the two masses. That's how mass got to be generally synonymous with weight. To be sure, they are not identical at all, for the electron has mass yet weighs zero. It weighs zero because it is repelled by gravity. Weight requires attraction to gravity.
So, we have one problem exposed already, that his mass x mass is wrong, which provides an acceleration figure too high every time. But even if we knock off one mass and use only one, the formula is still erroneous because mass is not in correlation with gravity force. In dealing with earth attraction to small items like satellites, they do use only the earth mass, and neglect the satellite mass, which is why the math shows only the earth mass. When it's the acceleration of the moon that they calculate, they should be using both masses in the math. However, let's go back to Joseph Lazio: "Solving for the acceleration on the Moon, we find a = (G x M) / (d x d)". What? Only one mass? Why isn't the lunar mass involved too? One person explains it in this way, that the actual formula is GmM/r2 = mass x acceleration, and then says that the small m cancels out, wherefore the formula becomes GM/r2 = a. I'm not sure whether this claim is a correct way to see things.
There is no graviton in atoms, stupid. Try electromagnetism as the source of gravity. There is no proof that the electron weighs anything. Whenever heat comes out of a substance, electrons are known to issue forth, yet the object does not decrease in weight. In fact, they got rid of the caloric model of heat (virtually my theory) partly because heat added to a material, they argued, did not increase its weight. They were then of the opinion that everything had weight, wherefore caloric, defined as material heat, had to have weight. Wrong. If gravity is an electromagnetic force (i.e. either negative or positive), it must repel either negative or positive particles. There cannot be protons by themselves, but there can be electrons by themselves. It is clear to me that heat rises, even through a metal rod in a vacuum, because gravity repels electrons.
I conducted an experiment. I called up a light-bulb manufacturer, asking for a bulb that still had a vacuum inside of it. I think it was Sylvania, and the person said that their 25-watt bulbs still came with vacuums (they started to use gases in them to prevent implosions when bulbs were broken). I got a bulb, dipped it fully in wax, and then turned the bulb on. No matter how the bulb was positioned, the wax melted only in a small spot directly ABOVE the filament, proving that electrons rise from gravity. It is known that a light bulb issues electrons from the filament as it burns. According to the evolutionist's theory of heat, it consists of no particles, but rather they view heat as the motion of atoms. Yet, there are no atoms, virtually, in a vacuum. But even if there were atoms in the bulb, the evolutionists say that they fly around in all directions equally, wherefore their prediction is that the wax should melt equally all around the bulb. I don't think it's fair to say that there is hot gas rising in a vacuum, as a means to explain why the wax melted only at the top.
Heat rises up a metal rod, and no one can argue that this is due to hot air rising around the rod. That's because the air around the rod doesn't get very hot at all, certainly not hot enough to make the rod too hot to touch inches up from where the heat source was applied. How can we explain heat too hot to touch six inches above the spot where the rod is heated? Even if we place the rod horizontally while heating it with a torch, so that rising hot air does not pass along the longitude of the rod, yet, when we turn off the torch and turn the rod vertical, heat will begin to rise through the rod. Why? Because heat was forced into the rod as free electrons from the flame, and they rise repelled by gravity. A small fraction of the heat goes down the rod because electrons inter-repel one another (in all directions equally).
Wikipedia says: "The dimensions assigned to the gravitational constant are force times length squared divided by mass squared..." There you have the mass x mass concept. It was built into the gravity-constant figure. When they say that the gravity constant is 6.674×10-11 N x m2 / kg2, the N represents the force, the meters squared represents the length x length, and the kilograms squared represents the mass x mass. The force x length x length (= total net force) portion is fine in theory, so long as the force figure is accurate, but the formula divides the total force by an irrelevant mass x mass acting as the attractive forces upon one another. At the top of the page below, you can see a diagram of two masses attracting one another over a distance that they call, r, and they give the attraction force between the two masses the formula: Gravity constant x (m x m) / r squared. Clearly, this is erroneous because the two masses do not attract one another.
Let's assume that the two masses did attract one another, as in two magnets. Why can't we just use magnet x magnet / distance x distance? This is correct. Yet, in the formula, they throw in G = the gravity constant. Why do they multiply by G too? It may be too early to guess, but taking a stab, I'd guess it's because the two masses do not acquire the truth by themselves, wherefore they use the gravity constant as a fix, as best as they could swing it.
Using mass would be acceptable if the heat of all planets were proportional to their masses, but that's not likely. Newton's theory might be accurate if he simply changed "mass" to "heat of attraction" (the internal heat of the planet), in which case I see no need for a gravity constant in the math. The heat body A attracting the atoms of body B x heat of body B attracting the atoms of body A / distance x distance. That's all that's needed for math purposes. What would be the use of a gravity constant if the gravity force between the two bodies is already involved in the math? If mass x mass already expresses the attraction between the two bodies, why do they throw in a gravity constant? Is this like a pill for a sick man? Is it the witch's concoction hoping to fix an irreparable situation with magic? You can read on the Wikipedia page above that N x m2 / kg2 is equivalent to meters3 x kg1 x sec2, the latter being a hard one to wrap the head around because I don't see why time should be involved in attraction levels between two objects. Is that more hocus-pocus? Do they introduce a time number as a fixer-upper?
Here was the problem:
The gravitational constant appears in Newton's law of universal gravitation, but it was not measured until seventy-one years after Newton's death by Henry Cavendish with his Cavendish experiment, performed in 1798 (Philosophical Transactions 1798). Cavendish measured G implicitly, using a torsion balance invented by the geologist Rev. John Michell. He used a horizontal torsion beam with lead balls whose inertia (in relation to the torsion constant) he could tell by timing the beam's oscillation. Their faint attraction to other balls placed alongside the beam was detectable by the deflection it caused...In modern units, the density that Cavendish calculated implied a value for G of 6.754×10-11 m3 kg-1 s-2.
What this means is that the gravity constant was determined by the electromagnetic forces between lead balls. Probably, all objects express some electromagnetic force, but this is not the same as gravity. Electromagnetic force from atoms is not gravity. If the Cavendish method was generally the way in which the gravity constant was derived, it has zero meaning. It is completely daft. The free electrons within a material, which define its heat volume, is gravity. The heat within a lead ball at room temperature is not expected to send out a measurable gravity force on another led ball. If materials heated to the extreme attracted other materials noticeably, we would have heard about it. The problem is, heated materials pour out free electrons that create a current (wind) against neighboring items, which to a viewer would look like repulsion between two objects. The repulsive wind probably has a greater force on neighboring items than the attraction from the increased gravity within the heated object. However, if a wind shield (a heat-resistant sheet of material) were used between a heated and unheated ball, some attraction between the two might be measurable before the heat gets through the shield.
Until now, I thought that, the heavier the planet, the greater the velocity of orbit needed, if for example two planets were viewed as orbiting at the same distance from the sun. But there is a calculator below giving the orbital speed and orbital period of any satellite, with weight being immaterial, when one merely feeds the top box with the altitude of the satellite. It tends to verify that gravity does not affect a planet by attraction upon of the accumulation of atoms -- meaning it does NOT attract the entire batch of mass as one unit -- but on the law I've discovered that gravity attracts each atom individually. In this picture, each atom on earth can be viewed in orbit around the sun independent of all other atoms on earth, as though they were disconnected from one another. This has never occurred to me before. In other words, it doesn't matter if we double the atoms on earth, the planet will maintain its orbital height, not becoming closer to the sun.
This calculator is very interesting for finding some of what I've been after. It's not until we go down from their lunar average (ballpark 238,700 miles) to 235,475 miles of altitude that the calculator givers an orbital period of 27.32 days (this is the average period of one lunar orbit). We cannot use this calculator for earth around the sun, as it's programmed with the earth's gravitational force. I don't think we can enter 252,400 miles of altitude to find the orbital speed of the moon at apogee in July of 2000, because the altitude entered is taken (by the calculator) either as a perfect circle, or the average height. In any case, I'll record that it gives a velocity of 2,198.72 mi/hr for 252,400 miles up.
Another webpage verifies that the weight of the orbiting object is of no relevance: "The equation is independent of mass. If the moon rather than the artificial satellite orbited at 400 miles and you could ignore air friction and collisions with the Earth, it would have to go at the same speed as the satellite in order to preserve its close orbit (which would make for some pretty spectacular moonrises)." Yet, in his first paragraph, he contradicts the above: "Thanks to physics, if you know the MASS [caps mine] and altitude of a satellite in orbit around the Earth, you can calculate how quickly it needs to travel to maintain that orbit." He now says that mass is necessarily involved when reckoning an orbital speed, yet the calculator gives an orbital speed for any satellite weight. At the bottom of the page, you will see the formula used by the calculator (above), and that it uses the gravity constant as well as the earth's mass, which should explain why the calculator has the moon's average distance too close to earth.
Here is more inconsistency, starting at Wikipedia's Geostationary-Orbit article: "A geostationary orbit...is a circular orbit 35,786 kilometres (22,236 mi) above the Earth's equator and following the direction of the Earth's rotation. An object in such an orbit has an orbital period equal to the Earth's rotational period (one sidereal day) and thus appears motionless, at a fixed position in the sky, to ground observers...The orbit...is sometimes called the Clarke Orbit. Similarly, the Clarke Belt is the part of space about 35,786 km (22,236 mi) above sea level, in the plane of the equator, where near-geostationary orbits may be implemented." It's saying that the satellite needs to be 22,236 miles above the earth's surface, not that far above the earth's core. The dummies.com page above says: "You have to use the distance from the center of the Earth, not the distance above Earth's surface, as the radius."
The calculator page says: "This calc determines the speed and orbital period of a satellite at a given height above average Earth sea level" (italics mine). The calculator is supposedly working from the earth's surface as the starting point of the orbital height or radius, and it even has a drawing to make this crystal clear...yet it is lying. Amazing. You will see later that it is taking the altitudes from the earth's core.
The Newtonian law on acceleration definitely has basis at the core. Entering 22,236 miles in the calculator gets a period of .997 day, and it gives the orbital speed at that height of 6,878.4 miles per hour. It agrees with the one-day period of a geostationary satellite, yet I'm suspicious because it's based on the Newtonian numbers (that's why it's important that it is lying). The dummies.com page above claims that these calculations are based on a gravity-constant formula.
The period of one day exactly is given by the calculator with a satellite height of 22,289 miles = 35,870 kilometers. There is disagreement on the height of geostationary satellites. Why? Shouldn't it always be the same height? Another reports: "The height of a geostationary orbit is about 35,900km above the surface of the Earth." Did this person get data from one of these calculators? I don't trust this calculator nor its program...meaning that when they put up geostationary satellites, they are not really 22,289 miles up. That's just what the calculation says, and therefore what everyone has been taught to say. Plus, the calculation is really 22,289 miles from the core, just 18,330 off of sea level. What is going on?
When we go back to the 235,475 figure that got an orbital period of 27.321 days, we being naive would naturally add one earth radius to it, bringing the figure up to 239,434, near to a typical claim for the average lunar distance out there in Internetland. In other words, the Newtonian numbers, coupled with the dishonesty of calculators like the one under discussion, may be wholesale-responsible for the too-high average lunar distance figures floating around the world.
I have just gone back to the calculator, and found that, instead of 235,475, it's now saying that an orbital period of 27.3218 days is with an altitude of 234,140 miles. That's funny. I was very meticulous when getting that same orbital period with 235,475. Now, however, 235,475 gets a period of 27.55 days. There is no way that I mistook 27.55 for 27.32. Either someone re-programmed this calculator in the past few hours since I used it last, or it's not reliable. Wikipedia: "The sidereal month is defined as the Moon's orbital period...It is about 27.32166 days." It sounds as though the moon keeps to that orbital period quite closely, for no other figures are given for a month-to-month basis, and the one figure given is not an approximation.
The orbital speed for the 234,140 number is given as 2,281.5 mph, which is close to NASA's average lunar velocity of 2,286 mph, yet 234,140, even when given an extra 3,959 miles, is a distance of 238,099, while NASA has an average distance of about 237,675. The calculator says that 2,289 is the velocity when the moon is 232,540 miles up, which could at-best be construed as 236,499 miles above the core. Things are still not shaking out correctly, especially as the calculator is actually taking measurements from the earth's core so that adding 3,959 additional miles is uncalled for.
In other words, the calculator, and Newtonian theory, are not giving us the lunar facts. It has also become clear to me that they do not, as claimed, have speed-of-light measurements to the moon that can be relied upon. No one is mentioning them as I've sought the lunar facts over a period of weeks. I still cannot make any sense of the average lunar distances given by NASA or others.
If mass matters on finding satellite velocity, why doesn't the calculator ask for the satellite's mass? Perhaps the writer of the dummies.com article has had the same misconception that I've had until now, that the weight of the satellite matters. He claims that the satellite velocity is obtained by the square root of gravity constant x earth mass / the square root of the distance, but there is no satellite mass involved in this formula. I am now convinced that the weight does not matter. However, I can see that leading astronomers, in covering for the grand hoax, do not want the public to know that mass is immaterial, for we can now use their calculators for lunar assessments. Any satellite of any weight, including the moon, at lunar distances, is supposed to conform to what the calculator is saying, and the calculator (or the manual calculations) does not conform to what astronomers have told us. I can see why they would be scrambling to confuse us on how this calculator works.
Steven Holzner, the writer at the dummies.com page ends by saying: "Human-made satellites typically orbit at heights of 400 miles from the surface of the Earth (about 640 kilometers, or 6.4 × 105 meters). What's the speed of such a satellite? All you have to do is put in the numbers:" After giving the formula with the numbers, part of it showing the earth radius added to 400 miles, he says: "This converts to about 16,800 miles per hour." Then, when we put 400 miles into the calculator's top box, it gives the altitude of 16,862 miles (same figure). Are we therefore to think that the formula started the measurement from the core, while the claim on the calculator page concerning sea level is wrong? Clearly, 400 miles up is a lot different than 400 + 3,959, yet both the calculator and the formula get about 16,800 mph. Who's right, who's wrong?
This is important because, if the calculator page is wrong with its sea-level claim, then one can begin to realize that it's deliberate disinformation to make us think that the moon's average distance is not around 234,140 miles, but 3,959 miles on top of that.
To find out who's right, who's wrong, let's go back to where he says that the satellite velocity is obtained by the square root of gravity constant x earth mass / the square root of the distance. For the distance, he has 3,959 miles + 400 miles (albeit he gives it in meters), a clear sign that he's measuring from the core. Let's do the formula with those numbers: .00000000006674 x 5,972,300,000,000,000,000,000,000 / square root of 7,015,130.496 (the latter is 400 + 3,959 miles in meters). The multiplication of the first two numbers is 398,591,302,000,000, and its square root is 19,964,751.49. The square root of 7,015,130.496 is 2648.61, wherefore our math is finalized with 19,964,751.49 / 2648.61 = 7,537.82 meters per second = 7.5378 kilometers per second = 16,862 miles per hour. Therefore, as the calculator has that velocity when entering 400 miles, the calculator is indeed using the core of the earth as the starting point! The claim on the calculator page is false. It leads people to believe that the moon's average height is in the ballpark of 3,959 miles above 234,140 (or 235,475).
Hmm, I wonder whether the calculator is programmed to change the 27.321 from 234,140 to 235,475, back and forth repeatedly over a certain period, to create deliberate disinformation so as to sow confusion concerning the lunar-distance average.
There are other ways to get the average lunar distance aside from this calculator. The question is: how could they create a formula, that they could trust, if it didn't provide what they truly believed was the average lunar distance, when the formula provides an orbital height having an orbital period of 27.32166 days? Shouldn't the lunar distance provided by the formula, above all else, coincide with the lunar fact of 27.32166 days? What else was their priority if not this? What else did they make their formula conform to? Does it mean that they secretly believe in an average lunar distance of 234,139 miles while using other definitions of average distance in order not to convey that figure to the masses?
Again, the first time I used the calculator to find the lunar distance when the orbital period was 27.321 days, I had found 235,475 miles. A few hours later, it changed to 234,140. Is 235,475 closer to the truth? I had provided (see two and three chapters ago) an argument for an average lunar distance of 235,896 miles.
The satellite-velocity math needs 398,591,302,000,000, which is obtained with their numbers, .00000000006674 x 5,972,300,000,000,000,000,000,000. I'm suspicious of their gravity constant. Whatever it is, or whatever its purpose, it should be called something else. A constant number can perhaps be justified in the math for finding orbital velocities, but they do not use the gravity constant alone as the constant. Instead, they use the square root of G x M as the constant. That combination never changes in the math. The only thing that changes is the orbit's altitude. They just take their constant number, the square root of 398,591,302,000,000, and divide it by the square root of the orbital altitude to get velocity. The square root of the latter number is 19.964 million. I'm thinking that they got their magic number in something to do with distance from a sphere. Why should the square root of 398.59 trillion be the magic constant for finding orbital velocities? Let's check it out to see if it's true. They use: 19,964,751 / square root of altitude = velocity. Simple enough, but where did they get this number, really? Why didn't they use 19,954,751?
To work in reverse, we do velocity x square root of the altitude = 19,964,751. For example, for 10,000 miles up = 16,093,440 meters, the velocity is said (by the earth-orbit calculator) to be 4,212.29 meters per second, wherefore we multiply 4011.663 (= square root of 16,093,440) x 4212.29 = 16,898,288. It's not even close to their constant. What went wrong? The lower we go from 10,000 miles, the worse it gets. By the time that we get to typical satellite heights, it looks like a joke. But no one this pointing out, on the first pages provided by Google, anyway, that this formula doesn't work. The masses are happily to believe that their wonderful astronomical wise men have all things slam-dunk. Four centuries after the telescope, the wise men are yet fools. You can tell because, instead of seeing the handiwork of God, they see billions of years of Nobody. The universe becomes a dark and eerie place with Nobody as the master. Astronomers could start to think that their children should become high-level agents of a Universal Master.
However, with ten times the 16,0934,400 meters, or 160,934,400, the calculator says the velocity is 1,543.51 meters per second. If we now multiply 12,686 (square root of 160,934,400) by 1543,51, we get 19,580,967.9, much closer to their constant. We can begin to see that they chose their magic number for far-out distances, and this was Newton's forte, after all, as is that of the astronomer. You can test any of their altitudes against how far off they are from their not-so-magic constant by simply entering any altitude in the top box, then multiplying the velocity in the second box by the square root of the top box. As one goes up from 160 million, the number gets continually closer to their 19.964 million, but as distance goes up, the number goes more slowly toward their constant. By 300 million meters, the number is 19.756 million.
400 million, 19.8 million; (this is 248,548 miles out)
700 million, 19.875;
1,000 million, 19.90;
4,000 million; 19.949:
9,000 million, 19.9577
200,000 million, 19,964,509
400,000 million, 19,964,660
800,000 million, 19,964,777
We finally pass their 19,964,751 at almost 500 million miles from earth. One can see that for distances out as far as the moon and more, the math happens to work fairly closely. They have the magic number working perfectly somewhere between Jupiter and Saturn, which one can construe as the center (or average ballpark) of the solar system. If Newton knew what he was talking about, shouldn't this Newtonian scheme work better for the lower realms, since this is where gravity acts at its best? It's where Newton lived, after all. Instead, they have the magic number way out there, beyond Jupiter, and this scheme has to do with earth gravity, remember, not solar gravity. How much earth gravity do you imagine out Saturn way?
So, it just so happens that when one multiplies the square root of these high numbers by the velocities given in the earth-orbit calculator, roughly the same number results at nearly 20 million. The question is: are the velocities given by the calculator correct, or are they artificially canned by the Newtonian formula in the first place? No one can prove that the velocities given are correct. And no one should trust an astronomer. Con-artists and astronomers are virtually synonymous. Both waste everyone's time. Neither have good for humanity. As with con-artists, it's the astronomer's job to make false claims seem true.
How much slower do you predict a satellite should move when it's attracted by four times less gravity? Let's try 20,000 miles up, because it's twice as far as 10,000 and therefore at a level where gravity is four times lower. At 20,000 miles, the velocity is given as 3,215.2 meters per second, as compared to 4,212.3 at 10,000 miles. There is a difference of .763 between the two velocities. To put it another way, there is .763 times the lateral-direction force to keep a satellite in orbit where it is affected with .25 times the gravity. This is what their numbers are telling us. Does it feel right?
Let's compare 100,000 miles with an orbital velocity of 1543.51 m/s, and 200,000 with 1101.97 m/s, a difference of .714. Why isn't it the .763 above? Something wrong? There doesn't appear to be a gravitational constant here. According to the calculator, it is not true that gravity force is 4 times lower on objects twice as far. Did they sacrifice the inverse-square law of gravity when fixing their magic number?
The difference between 1,000,000 miles, with 496.69 m/s, and 2,000,000 with 351.58, is 7.078, not far from .714. It looks like some constancy will be achieved eventually, way out there where earth gravity has virtually no bite whatsoever, but let's go closer to earth, where the difference grows lower. We can test 200 miles up (from the core), with 7717.19 m/s, versus 400 miles up with 7538.07 m/s, with a velocity difference of just .977. It's virtually the same velocity (lateral force) where gravity is four times as weak / strong. That makes no sense? Comparing 20 miles (7890 m/s) with 40 (7870 m/s) has the velocities even closer. Or, comparing 2 miles (7908 m/s) with 16 miles (7,902 m/s) has virtually the same velocity even though gravity acts 64 times as weakly 16 miles up as compared to 2. This is a good way to expose that the formula behind the calculator isn't to be trusted, nor the men behind the hoax.
In the calulator below, one can first click "Orbital Radius," then enter the earth-mass figure, 5973600000000000000000000 (20 0's), as well as 27.32166 days, to find that this method gets a radius = lunar distance of 238,111 miles. This tends to explain, again, their faulty, average lunar-distance figure in the ballpark of 238,000 miles. The calculation is based on Kepler's third law, which calls into question all, claimed lunar distances if this is the methoid used to acquire them.
Finally, let me caution you on the nuttiness of the graviton particle. You understand that, if evolutionists are going to maintain their theory that gravity is different, neither positive nor negative, they have got to show that this something-different is in every atom. And so they call it the graviton particle. On top of electrons and protons, they say that atoms have graviton substance. Does this seem to you like baloney? Shouldn't gravity merely be either the positive or negative force? Can we establish that there are free electrons in the center of the earth? Yes, for NASA says that solar-wind electrons were found flying out in space having the burning sun as their source. Wherever there is heat, there also are the free electrons. Is there heat in the center of the earth? There too you will find free electrons and their negative force.
How are the gravitons stuck to the atoms? Super-glue? Are they held to protons? Wouldn't that make gravitons negative in charge? Are they held to the captured electrons? Wouldn't that make gravitons positive in charge? How are gravitons held to atoms if they are neither negative nor positive? Maybe with the mustard on a baloney sandwich. At least its an answer. Evolutionists are silent on gravitons. It's best not to bring up the topic, and instead just say Newton's name over and over again, because everyone respects him. He's the apple-on-the-head guy, so smart and lovable all at once. Evolutionists would have us believe that, until the apple fell in the 17th century, no one had ever before figured that something in the earth pulls things down. Newton owns gravity, that is, according to evolutionists. Newton and gravity-by-mass are just facts of life together. We are not supposed to debate this, even if mustard can't really hold gravitons to atoms.
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