Based on the staff statement, that refusal to defend a point of view in light of specific questions would lead to a site ban, I'm going to challenge you on last time on this issue!

I snipped of all the unintelligible and/or irrelevant stuff.

johnlear wrote:According to the Astronomical Almanac for the Year 2009 published by the United States Naval Observatory in Washington D.C. the gravitational force of the moon is equal to 1.540260256 x 10^{13}. The equatorial radius of the Moon in 1,737,400 m. The area of the Moon's equator is 9.48308205 x 10^{12}. So, the acceleration at the Moon's equator is 1.624219059 m/s^{2}. This is about one-sixth of the acceleration at the Earth's equator or about 16.57%. And these alleged numbers have been allegedly confirmed by the Apollo astronauts during that alleged space program. If true this certainly would not be sufficient to hold a breathable atmosphere.

Let's ignore the fact, that you didn't provide a unit of measurement for half of your numbers, which shows that you have no idea how to present scientific data, and also shows that you apparently didn't really unterstand what you wrote anyway. Terms like "area of the Moon's equator" don't make sense at all, because the equator is a

line, which doesn't have an

area. But that's not my point.

So, to calculate the moon's surface gravity, we need:

1) The

mass m of the moon

2) The

radius r of the moon

3) The formula to calculate the gravitational acceleration g from

m and

r:

g = G *

m/

r^2 (G = Gravitational constant)

And here are my specific questions to you:

Q1Do you agree, that the 3 items above are sufficient to calculate the moon's surface gravity? If not, state what else would be needed, and why!

Q2Do you agree that the moon's mean radius is (approximately) 1737 km? If not, give evidence for your negative answer (e.g. pointing out which specific errors were made when measuring the moon's diameter by, say, observing it from earth and calculating the true size using the known earth-moon distance).

Q3Do you agree that the moon's mass is about 7.35*10^22 kg? If not, please provide evidence, why the measurements of the moon's mass using astronomical observations (see

http://www.mathpages.com/home/kmath469.htm) are wrong. Such measurements have been made with increasing precision since hundreds of years.

Since the moon's mass is critical to your argument about a higher-than-generally-assumed surface gravity, you can

not simply say that the "astronomers got it wrong"! Provide

supporting evidence from a reasonably up-to-date and scientifically valid source, if you claim that earth-based observations of the moon's mass are invalid.

Q4Do you agree that, given mass m and radius r, the moon's surface gravity can be calculated using the formula in item (3)? If not, provide

strong evidence for your assumption, that Newton's Law of gravitaional attraction is

grossly wrong (as opposed to

very slightly off, because of non-uniform mass distribution, relativistic effects, unknown quantum effects etc., all of which is completely negligible for our general argument about a 0.16ge vs. 0.64ge moon surface gravity). Be aware, that Newton's Law has been shown by innumerable observations and measurements in the solar system to be

very accurate, so in order to dismiss Newton's Law, you would have to explain why it

appears to be so accurate!

=> If you either answer all 4 questions with "yes" (and a

failure to answer any of the questions would be regarded as a "yes"!)

or can't provide any solid evidence for your "no"s, then I ask you to retract your claim, that the moon's surface gravity is significantly different from the generally accepted value of about 1/6 of earth's.

Now, as a free service, I'll show where the error lies in your argument ...

johnlear wrote:Using the Bullialdus/Newton law of Inverse Square which states that some physical quantity or strength is inversely proportional to the square of the distance from the source of that physical quantity" and using the 'neutral point' of 43,495 miles given to us by Wernher Von Braun and approximated to that value by Apollo 17 astronaut Gene Cernan in his book, "Last Man On the Moon" and Reginald Turnhill in his book "The Moon Landings" and using generally accepted diameters of the Moon and Earth and distances between these two bodies we come up with a value of .64 which means that the calculated gravity of the Moon using Earth as a value of 1 would be 64% that of Earth.

{remaining irrelevant gibberish snipped ...}I know from the past that your argument goes essentially like this:

A) You look at the problem in the frame of reference (FR), where the earth and moon are fixed, and assume that the Apollo spacecraft travelled a straight line between earth and moon in this FR.

B) You (implicitly) assume, that your FR is an

inertial FR, i.e. a non-accelerating and non-rotating one.

C) With this assumption, you can calculate a point on the straight line between earth and moon, where the gravitational attraction between two bodies cancels out. Using the quoted value of 43495 miles (69600 km) distance from the moon, you arrive at a moon surface gravity of .64ge.

However, your assumption (B) is

false! The FR, where earth and moon are fixed, is a

rotating FR! This makes calculation of points where earth's and moon's gravitational pull cancels out (for a body, which moves at relatively low velocities within this FR), much more complicated. In fact, such points are known as "Lagrange Points" - see

http://en.wikipedia.org/wiki/Lagrange_point. The Lagrange Point, which presumably comes closest to what you regard as a "neutral point", is L1. L1 is about 61500 km from the moon, which is already rather close to the quoted "69600 km". It's important to note, that the position of L1 was of course calculated using the generally accepted values for the moon's mass and gravity!

So, where could the remaining difference (61500 km vs. 69600 km) come from? Simple - the Apollo trajectories did

not follow a straight line in the fixed-earth-moon FR, and as such did

not pass through L1! On the Wiki page

http://en.wikipedia.org/wiki/Lagrange_point , there is a map of the "effective potential" in the fixed-earth-moon FR, and you can see that this is not a very simple structure. For an "off-center" trajectory, the closese thing to a "neutral point" is the point where a (comparatively slow) spacecraft goes over a local maximum of the potential. These maxima are

very roughly(!) on a smooth curve from L4 over L1 to L5. Therefore, for trajectories not on the "center line", the distance of these "neutral point" from the moon increases significantly.

And now, I await your answers to the four emphasized questions above! And don't dare to evade them!!

yf