<p>Thanks for all the posts so far, guys. </p><p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Ask them to describe the moon's orbit and its tidal effects with it being 4 times as massive. I'll check their posts later... i gotta run right now. <br /> Posted by derekmcd</DIV></p><p>That's a good point, but the only problem there as I see it, is there is no benchmark to compare with how tidal effects or the Moon's orbit would be, if the figure of .64% [do they mean 64% btw? Still wrong of course,] were correct.</p><p>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I have to run RIGHT NOW so I didn't get time to check the following over but hey, let's peer review it now. What I did instead of work for the last 45 minutes .... I'm not sure what his (the poster over at OM) math proves other than he can do the wrong equations ...Newtons law of gravitation is : F = (G*m*M)/r^2 where:G = gravitational constantm = mass of object #1, which for our purposes will be my mass, the mass of me (Mac)M = mass of object #2, which for our purposes is either the Earth's (Me) or the Moon's (Mm) massr = the distance from the center of Object #2 to the center of Object #1So lets see where the force of Earth's gravity (on me) = the force of the Moon's gravity (on me) using the law above.Fe = Fm (Force due to Eath = that due to Moon)(G*m*Me)/de^2 = (G*m*Mm)/dm^2 (Me = Mass Earth, De = distance from Earth to me at equal gravity point, Mm, Dm apply to Moon parameters)Obviously the G*m is the same on both sides and can be dropped leaving us with ;Me/de^2 = Mm/dm^2 Now rearranging to find the ratio of the distances squared ;dm^2/de^2 = Mm/Me Taking the square root of each side gives me the ratio of the distances ;dm/de = sqrt(Mm/Me)So IF we accept that the Earth's mass is ~81.3 times that of the Moon then the ratio of distances is about 9. That is the distance from the Earth to the equi-gravity point is 9 times the distance from the Moon to that point. If their centers are 239,000 miles apart then it's 23,900 miles from the Moon's center (~22,820 mi from Moon surface) and/or 215,100 mi from the Earth's center (~211,140 mi from surface). Note that I haven't had to use surface gravities to figure this out but let's now talk about them.Fge = (G*m*Me)/re^2 (force of gravity on me on Earth's surface)Fgm = (G*m*Mm)/rm^2 (force of gravity on me on Moon's surface)So what's the ratio of the Moon to the Earth ? Again the G*m terms drop out and we have ;(Mm/rm^2) / (Me/re^2) So with Mm/Me about 1/81.3 and rm (Moon's radius) = 1080 mi and Earths = 3960 we get
1/81.3) * (3960^2/1080^2) = 0.165 or the approx 1/6th that everyone uses.So does the OM poster accept the relative masses of the Earth to Moon to be ~81 ?I have no idea what von Braun really said or what Apollo 8 measured. I find it intriguing that NASA is somehow hiding the Apollo 8 data and yet somehow the OP found it. I wonder what later Apollo missions measured ? Why didn't the OP mention them ? Hmmmmm !?!?!Lasty look at the video of the famous hammer and feather drop.
http://www.youtube.com/watch?v=dHzVsLAhUCALet say the distance to the ground is about 5 feet. On Earth the drop time (neglecting air resistance) would be about 0.56 secs. With 64 % of Earth's gravity it would be ~0.7 secs and with the number above, ~1.37 secs. Which do you think is closer ?(No doubt NASA doctored the video ..... ) <br /> Posted by mee_n_mac</DIV></p><p>Thanks for the maths. <img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-wink.gif" border="0" alt="Wink" title="Wink" /> Good point as well, I think about the other Apollo missions and the gravitational neutral point. Also the hammer and feather footage, I forgot about that one. As you suggest though, the standard answer to any of this is quite likely to be that NASA doctored it, so possibly hitting a brick wall there.</p><p>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I took a quick look at the article that you found and at the "Open Minds" forum in general. There seems to be no point in debunking this idiocy. All I found was conspiracy theories, UFO sightings, etc. -- lunatic central. There is no chance that logic and rationality will have the slightest impact on the illogical and irrational. This one is particularly crazy. Why on earth would anyone want to cover up the "true" gravitational force of the moon ? Debunking this junk with a calculation would be like trying to teach calculus in kindergarten. <br /> Posted by DrRocket</DIV> </p><p>Point taken. The fact still remains though that this pseudo science has all been posted, and that forum does have a large, rapidly growing membership [a lot of ex "Above Top Secret" people etc] and many guests reading the posts, so I felt that some attempt should be made to debunk it, otherwise a lot of [young people especially,] are going to get the wrong ideas in their heads, which does not help in the progress of science. </p><p>Regarding the UFO threads, I won't go too much into that here, but I do have an interest in the subject, as I did see something once, [and only once] at very close quarters, that I could not explain, hence my reason for joining there. I may start a thread here about my sighting sometime, but haven't got around to it yet.</p><p>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Your way works Mee-n-mac, but it can be a bit much to digest. I was going to simplify it. To determine the moon's gravity, all you need to know (easier said than done) is the moon's mass and radius, and Earth's mass, radius, and gravitational force. All of which can be measured quite precisely here on earth.The moon's mass is about 1/100th of the Earth and the radius^2 is about 1/16th that of Earth.1/100 x 16 = .16 or about 1/6. 9.8m/s^2 / .16 = 1.56 (rounded 1.6)I used approximations, but the result is close enough.As for Wernher's statement he was correct. The 23,900, as shown on his graphic representation, was a point at which their paths crossed when charted on a graph neglecting the others gravity (which is not the true neutral point or L1) as seen here:
http://www.apollo-hoax.co.uk/neutral.htmlThe distance from center to center is about 239,000 miles. The Lagrange point (L1... nearly identical to the neutral point) is, on average 200,000 miles from Earth's center. Thus, 39,000 miles from the Moon's center, which is about 1/6 (coincidence??). The quote of 43,495 miles might be derived from the apogee (furthest distance since the orbit of the moon is elliptical) and his 200,000 miles was, likely, just an approximation (as seen by him saying "some 200,000") for the layman.So he basically started out with faulty information and used faulty math to give a wrong answer.Ask him to use km next time... i hate converting. <br /> Posted by derekmcd</DIV></p><p>Thanks! I am reading everyone's posts carefully. I do intend making another post there at some stage, of some sort. I'm also hoping that calikid posts again there, as he is also questioning the views, but possiby is experiencing the same problems as me, in how to go about tackling it.</p><p>Oh btw, I just noticed that the author of the book challenging Newton and Einstein is Pari Spolter, not Paris ...</p> <div class="Discussion_UserSignature"> <h1 style="margin:0pt;font-size:12px">----------------------------------------------------- </h1><p><font color="#800000"><em>Lady Nancy Astor: "Winston, if you were my husband, I'd poison your tea."<br />Churchill: "Nancy, if you were my wife, I'd drink it."</em></font></p><p><font color="#0000ff"><strong>
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