New Moon "conspiracy." Lunar gravity is not 1/6th that of the Earth, but 64% and NASA is c

Smersh · May 6, 2008

As Phil Plait said, on his Bad Astronomy site somewhere, "When faced with such an avalanche of ridiculousness, it can be difficult to function coherently." Well, that's what I came across here, in a thread at the "Open Minds" forum. It is in a thread started by John Lear, [son of Lear Jet inventor Bill Lear.] The thread that John started is not actually about this Lunar gravity accusation, but about a book by one Paris Polter, [or was it Paris Hilton,] with some absurd theory that both Newton and Einstein were wrong, or something. I decided I didn't even want to go there [although, of course, anyone else here is welcome to join up at that forum, with far more knowledge than me of the maths and finer details, to debunk it if they wish.]When a subsequent post was made though, claiming that Lunar gravity is only .64% of Earth's, and suggesting a "coverup" by NASA, I thought "right, I'll have some of this," and posted the following: No I didn't know that. I have always understood it to be 1/6th Earth gravity. I first learned that from Sir Patrick Moore [who is probably the world's foremost expert on the Moon,] in books of his that I read over 40 years ago. Do you have any reliable links that back your claim that it is .64% of Earth? An answer was given, claiming that Wernher Von Braun got his figures wrong, in a Time magazine article in 1969, also that during the Apollo 8 mission, there was a discrepancy about the gravitational neutral point, between the Earth and the Moon. Further posts have been made, with a lot of maths given, that supposedly prove that Lunar gravity is .64% of Earth's, not 1/6th. Maths has never been my strongest point, so I would not be comfortable with trying to refute this by using maths, as certain as I am that it is all total nonsense. I have been searching and searching, trying to find out when it was first discovered that Lunar gravity is 1/6th of the Earth's. As I said, I'm sure Sir Patrick Moore stated it in some of his early books, which would have been before the Apollo missions even went there, but unfortunately, I cannot find any references. I feel that posting a link there to when the Lunar gravity figure was first discovered, or to some early astronomy books, giving the figure of 1/6th that of the Earth, should go some way to debunking this, without the need for complicated maths.If you read the thread that I have linked, the posts I'm referring to are all near the beginning, on the first page.Can anyone help please? Thanks!  [EDITED THREAD TITLE FROM .64% to 64%] <div class="Discussion_UserSignature"> <h1 style="margin:0pt;font-size:12px">----------------------------------------------------- </h1>Lady Nancy Astor: "Winston, if you were my husband, I'd poison your tea." Churchill: "Nancy, if you were my wife, I'd drink it."Website / forums </div>

derekmcd · May 6, 2008

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. <div class="Discussion_UserSignature"> <div> </div> <div>"If something's hard to do, then it's not worth doing." - Homer Simpson</div> </div>

Mee_n_Mac · May 6, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>Can anyone help please? Thanks!   Posted by Smersh</DIV>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 constant m = 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) mass r = the distance from the center of Object #2 to the center of Object #1 So 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=dHzVsLAhUCA Let 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 .....  <img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-foot-in-mouth.gif" border="0" alt="Foot in mouth" title="Foot in mouth" /> ) <div class="Discussion_UserSignature"> -----------------------------------------------------Ask not what your Forum Software can do do on you,Ask it to, please for the love of all that's Holy, STOP ! </div>

DrRocket · May 6, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>...at the "Open Minds" forum. ...  Posted by Smersh</DIV>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.   <div class="Discussion_UserSignature"> </div>

derekmcd · May 7, 2008

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.html The 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. <div class="Discussion_UserSignature"> <div> </div> <div>"If something's hard to do, then it's not worth doing." - Homer Simpson</div> </div>

Smersh · May 7, 2008

Thanks for all the posts so far, guys.  Replying to: <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. Posted by derekmcd</DIV>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.Replying to: <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 .....   ) Posted by mee_n_mac</DIV>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.Replying to: <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.  Posted by DrRocket</DIV> 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. 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.Replying to: <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. Posted by derekmcd</DIV>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.Oh btw, I just noticed that the author of the book challenging Newton and Einstein is Pari Spolter, not Paris ... <div class="Discussion_UserSignature"> <h1 style="margin:0pt;font-size:12px">----------------------------------------------------- </h1>Lady Nancy Astor: "Winston, if you were my husband, I'd poison your tea." Churchill: "Nancy, if you were my wife, I'd drink it."Website / forums </div>

derekmcd · May 7, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>so I felt that some attempt should be made to debunk it Posted by Smersh</DIV> Debunking it easy... getting them to listen is another story.  <div class="Discussion_UserSignature"> <div> </div> <div>"If something's hard to do, then it's not worth doing." - Homer Simpson</div> </div>

JonClarke · May 7, 2008

if lunar gravity was approximately 2/3s earth's, rather than 1/6, the CSM complex would not be able to enter lunar orbit, or leave again.  It would not have enough propellant.  the LM would have not had enough propellant or a powerful enoughy engine to land.  or, if by some miracle it did, to take off again.  the same applies to all the unmanned spacecraft that entered lunar orbit or soft landed.  It would require a conspiracy not just by the US but also the USSR, Japan, ESA, and China.And why would there be a conspiracy?  If the lunar graviational field was different to what was expected this would be a major discovery and celebrated as such.Like all the lunar conspiracy stuff it is just nuts. Jon <div class="Discussion_UserSignature"> Whether we become a multi-planet species with unlimited horizons, or are forever confined to Earth will be decided in the twenty-first century amid the vast plains, rugged canyons and lofty mountains of Mars Arthur Clarke </div>

derekmcd · May 7, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>It would require a conspiracy not just by the US but also the USSR, Japan, ESA, and China.And why would there be a conspiracy? Posted by jonclarke</DIV> So you're saying it's a global conspiracy... cool!!!  Even better. <img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-laughing.gif" border="0" alt="Laughing" title="Laughing" /> <div class="Discussion_UserSignature"> <div> </div> <div>"If something's hard to do, then it's not worth doing." - Homer Simpson</div> </div>

3488 · May 7, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>if lunar gravity was approximately 2/3s earth's, rather than 1/6, the CSM complex would not be able to enter lunar orbit, or leave again.  It would not have enough propellant.  the LM would have not had enough propellant or a powerful enoughy engine to land.  or, if by some miracle it did, to take off again.  the same applies to all the unmanned spacecraft that entered lunar orbit or soft landed.  It would require a conspiracy not just by the US but also the USSR, Japan, ESA, and China.And why would there be a conspiracy?  If the lunar graviational field was different to what was expected this would be a major discovery and celebrated as such.Like all the lunar conspiracy stuff it is just nuts. Jon Posted by jonclarke</DIV>I totally agree Jon. These people have too much time on their hands & not enough to do in their lives.Certainly it is time for <img src="http://sitelife.space.com/ver1.0/Content/images/store/9/12/992ea599-283d-42aa-8cbe-430f070ab29e.Medium.gif" alt="" />, possibly <img src="http://sitelife.space.com/ver1.0/Content/images/store/2/11/d29092ef-c3db-4b99-8bad-6098c7324de0.Medium.gif" alt="" /> & maybe even <img src="http://sitelife.space.com/ver1.0/Content/images/store/3/4/a332d8d2-68ad-462c-b4d4-c0bc59839ad6.Medium.gif" alt="" /> .This kind of woo woo nonsense does make me very cross & goes to show the poor state of science teaching these days.Andrew Brown. <div class="Discussion_UserSignature"> "I suddenly noticed an anomaly to the left of Io, just off the rim of that world. It was extremely large with respect to the overall size of Io and crescent shaped. It seemed unbelievable that something that big had not been visible before". Linda Morabito on discovering that the Jupiter moon Io was volcanically active. Friday 9th March 1979.http://www.launchphotography.com/ http://anthmartian.googlepages.com/thisislandearthhttp://web.me.com/meridianijournal </div>

Smersh · May 7, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>I totally agree Jon. These people have too much time on their hands & not enough to do in their lives.Certainly it is time for , possibly  & maybe even  .This kind of woo woo nonsense does make me very cross & goes to show the poor state of science teaching these days.Andrew Brown. Posted by 3488</DIV>ROFLMAO !!! Cheers Andrew, I love those "Dead Horse" emoticons of yours. Keep em coming! <img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-wink.gif" border="0" alt="Wink" title="Wink" />Replying to: <DIV CLASS='Discussion_PostQuote'>if lunar gravity was approximately 2/3s earth's, rather than 1/6, the CSM complex would not be able to enter lunar orbit, or leave again.  It would not have enough propellant.  the LM would have not had enough propellant or a powerful enoughy engine to land.  or, if by some miracle it did, to take off again.  the same applies to all the unmanned spacecraft that entered lunar orbit or soft landed.  It would require a conspiracy not just by the US but also the USSR, Japan, ESA, and China.And why would there be a conspiracy?  If the lunar graviational field was different to what was expected this would be a major discovery and celebrated as such.Like all the lunar conspiracy stuff it is just nuts. Jon Posted by jonclarke</DIV>Great points Jon. Thanks!Btw guys, I hope nobody has any objections if I quote some of what you are all saying over there, when I do eventually post. If anyone does object to that, I'll try to rephrase it, but would like to give credit where it's due. I guess I could put a link over there to this thread here too, so they can read all your points.  <div class="Discussion_UserSignature"> <h1 style="margin:0pt;font-size:12px">----------------------------------------------------- </h1>Lady Nancy Astor: "Winston, if you were my husband, I'd poison your tea." Churchill: "Nancy, if you were my wife, I'd drink it."Website / forums </div>

TheOscarMP · May 7, 2008

I am a little confused.  Are they saying that the moon's gravity is 64% that of earth or .64%.  there is a big difference at 64% they are saying the moon's gravity is nearly 4X that of the accepted valueat .64% they are saying the moon's gravity is almost negligible (some ~26X less than the accepted value) in either case...these guys are nutters <div class="Discussion_UserSignature"> Screw you guys...I'm going home: Eric CartmanSILENCE...I KILL YOU!!!!: Achmed, the dead terrorist </div>

CalliArcale · May 7, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>I am a little confused.  Are they saying that the moon's gravity is 64% that of earth or .64%.  there is a big difference at 64% they are saying the moon's gravity is nearly 4X that of the accepted valueat .64% they are saying the moon's gravity is almost negligible (some ~26X less than the accepted value) in either case...these guys are nutters Posted by TheOscarMP</DIV>Actually, if they have fallen prey to the common (among the innumerate) mistake of calling 64% ".64%", that might go some way towards explaining why they've gotten their math so dreadfully wrong.  <div class="Discussion_UserSignature"> "People assume that time is a strict progression of cause to effect, but actually from a non-linear, non-subjective viewpoint it's more like a big ball of wibbly wobbly . . . timey wimey . . . stuff." -- The Tenth Doctor, "Blink" </div>

Smersh · May 7, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>I am a little confused.  Are they saying that the moon's gravity is 64% that of earth or .64%.  there is a big difference at 64% they are saying the moon's gravity is nearly 4X that of the accepted valueat .64% they are saying the moon's gravity is almost negligible (some ~26X less than the accepted value) in either case...these guys are nutters Posted by TheOscarMP</DIV>Thanks, I did notice that in the thread they put .64%, in fact I did mention it in one of my posts in this thread. I'm pretty sure they mean 64%, in fact in the OM thread 64% was mentioned once or twice.One thing being queried in the OM thread is why the Apollo astronauts only managed "small hops" so I'm pretty sure they do mean 64%.However, they should have put the correct percentage, [as they see it,] I agree. Replying to: <DIV CLASS='Discussion_PostQuote'>Actually, if they have fallen prey to the common (among the innumerate) mistake of calling 64% ".64%", that might go some way towards explaining why they've gotten their math so dreadfully wrong.  Posted by CalliArcale</DIV>Good point Calli, thanks. I think I may mention that over there when I post. (EDIT) I am certain in my own mind they mean 64%, so I have changed the thread title here from .64% to 64%. I still intend pointing it out over there though.  <div class="Discussion_UserSignature"> <h1 style="margin:0pt;font-size:12px">----------------------------------------------------- </h1>Lady Nancy Astor: "Winston, if you were my husband, I'd poison your tea." Churchill: "Nancy, if you were my wife, I'd drink it."Website / forums </div>

Mee_n_Mac · May 7, 2008

Replying to: <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. Posted by derekmcd</DIV>EDIT : by OP I mean the poster over @ Open Minds that believes the Moon's surface gravity is indeed 64% of the Earths.I'm not sure exactly what the OP is objecting to.  I suspect he isn't (sure) either. He either thinks that Newton was wrong (true though not in the sense the OP implies) and gravity doesn't follow the stated laws or he thinks the equations are correct but the mass of the Moon is not what we think it is.  If it's the first case then why does the OP use the inverse square law to (attempt) to compute the relative surface gravities from the (wrong) neutral gravity point ?  It would seem he believes in Newton when doing said calculation so therefore (ha !) he must think the Moon masses more than (the rest of us think) it does. If that's the case then derekmcd's point above is a good one.  You can calculate the Moon's mass from it's observed period and it's orbital radius.  I note the OP gives a good approximation to the true radius so unless he measures a vastly different persiod from the ~ 1 month (27.3 days) the rest of us see, I can't see where the OP can back up his 64% claim.  Clearly the OP is not understanding what von Braun meant (or WvB had what's known in the technical world as a brain fart).  Clearly he's not understanding what ever Apollo 8 "data" he has or thinks he has. On the latter topic ....Just a quick looksee on the WWW to see what NASA claims as the neutral gravity point (for Apollo 8) didn't yeild me much other than this article from Astronautix which says :"Following the second video presentation, the crew neared a new stage in manned space flight - travel to a place where the pull of earth's gravity was less than that of another body. At 3:29 in the afternoon on Monday, 23 December, that historic crossing was made. At that point, the spacecraft was 326,400 kilometers from the earth and 62,600 from the moon, and its velocity had slowed to 1,218 meters per second. Gradually, as the ship moved farther into the moon's gravitational field, it picked up speed."Note that the number there (which has probably been misintepreted in woo-woo land) is 62,600 km (39,000 mi) is larger than the aformentioned ~24,000 mi.  At first it might appear that the woo-woos are onto something until you remember that ;a) the spacecraft isn't travelling in a straight line and b) that's it's headed to where the Moon will be when it (the spacecraft) gets there. The Moon is many hours behind that point in it's orbit (it will have to travel another 28,000+ miles) and so is further away from the spacecraft than you might think at first glance. When the spacecraft is 24,000 miles from the Moon's orbit, I think it's still ~39,000 miles from the Moon itself.   <div class="Discussion_UserSignature"> -----------------------------------------------------Ask not what your Forum Software can do do on you,Ask it to, please for the love of all that's Holy, STOP ! </div>

qso1 · May 7, 2008

Proving once again, there is never a shortage of conspiracy theories. <div class="Discussion_UserSignature"> My borrowed quote for the time being:There are three kinds of people in life. Those who make it happen, those who watch it happen...and those who do not know what happened. </div>

derekmcd · May 7, 2008

"Note that the number there (which has probably been misintepreted in woo-woo land) is 62,600 km (39,000 mi) is larger than the aformentioned ~24,000 mi."  The 24,000 mi distance is the misinterpreted number.  It's derived by neglecting the gravitational force of the 2nd object.  The moon actually lessens the gravitational pull of the earth.  What ever distance you plot using the inverse square law from earth would actually be less when the gravitational of the moon is factored in.  Hence the discrepancy. "b) that's it's headed to where the Moon will be when it (the spacecraft) gets there. The Moon is many hours behind that point in it's orbit (it will have to travel another 28,000+ miles) and so is further away from the spacecraft than you might think at first glance."  I'm only guessing here, but given that the moon's orbital speed is 1023 km/s and you cited a quote with the spacecraft's speed at 1218 km/s, I would assume the spacecraft is precisely in between (matching the moons orbital speed with a little extra to approach the moon), or maybe slightly behind allowing for Wherner's 43,495 quote to be accurate.  This, if anything, only confirms the 39,000 miles distance from the moon being the average neutral point and just another tool for confirming 1/6th the gravity. <div class="Discussion_UserSignature"> <div> </div> <div>"If something's hard to do, then it's not worth doing." - Homer Simpson</div> </div>

Mee_n_Mac · May 7, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>"Note that the number there (which has probably been misintepreted in woo-woo land) is 62,600 km (39,000 mi) is larger than the aformentioned ~24,000 mi."  The 24,000 mi distance is the misinterpreted number.  It's derived by neglecting the gravitational force of the 2nd object.  The moon actually lessens the gravitational pull of the earth.  What ever distance you plot using the inverse square law from earth would actually be less when the gravitational of the moon is factored in.  Hence the discrepancy. {snip} Posted by derekmcd</DIV> OK, I did some more digging and as they say, a picture is worth 1000 words (more like 33,742 of mine).  Look at the picture of a typical translunar trajectory in this thread on the very topic we've been discussing.  Then recall what was said above :"Following the second video presentation, the crew neared a new stage in manned space flight - travel to a place where the pull of earth's gravity was less than that of another body. At 3:29 in the afternoon on Monday, 23 December, that historic crossing was made. At that point, the spacecraft was 326,400 kilometers from the earth and 62,600 from the moon, and its velocity had slowed to 1,218 meters per second. Gradually, as the ship moved farther into the moon's gravitational field, it picked up speed."Read jayutah's (from the above link) explanation carefully.  I interpret the words italicized above pretty directly, the distances quoted are where the spacecraft stopped slowing and start gaining speed due to the relative influences of the Earth's and Moon's gravity.  Given the flight path is not headed directly towards the Moon at that particular moment what is happening is that the component of the Moon's gravitational force (think vector) that lies along the flight path (at that moment) is larger than the component of the Earth's gravity vector lying on, but opposite in sign/direction, the same flight path. Because the angle of the flight path relative to a direct path to the Earth may be different than the angle of the flight path relative to a direct path to the Moon, the total forces (due to gravity) need not be equal in magnitude to have the spacecraft gain speed.  The Earth may exerting a larger total force on the spacecraft than is the Moon, but the component lying on the flightpath is smaller than that of the Moon.  The quoted material above might be intepreted to mean the spacecraft crossed the equigravsphere but that isn't what I gather from the numbers given above. The point described by those numbers does not lie on the equigravsphere.  If I had the Apollo 8 flight trajectory and relative positions of Earth and Moon along it's way, I could determine if I'm on the right path but I don't and I'm too lazy to try to find them to answer a woo-woo's question (though I have gained new insight as a result). @ dereckmcd = Can you explain the red curve in the link you previously gave ? It isn't Fe - Fm so what is it ? <div class="Discussion_UserSignature"> -----------------------------------------------------Ask not what your Forum Software can do do on you,Ask it to, please for the love of all that's Holy, STOP ! </div>

qso1 · May 7, 2008

Chances are, the calculations being done will be in vain. Anyone who believes this lunar gravity stuff probably already believes we never made it to the moon. <div class="Discussion_UserSignature"> My borrowed quote for the time being:There are three kinds of people in life. Those who make it happen, those who watch it happen...and those who do not know what happened. </div>

Mee_n_Mac · May 7, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>Chances are, the calculations being done will be in vain. Anyone who believes this lunar gravity stuff probably already believes we never made it to the moon. Posted by qso1</DIV> If the calcs were being done to convince the OP over at Open Minds I'd agree.  I think that's the general point behind the whole "Moongate conspiracy", that the Moon's real gravity is so high that the Moon landings were either faked (because we couldn't get back off if landed there) or somehow altered so the world wouldn't know what the Moon's gravity "truely" is.  Why NASA and the rest of the scientific world would go to such lengths to hide this bit of info, that I'm sure the majority of the remainder of the world doesn't care about, is beyond my understanding.  For me I'm interested in what the true intepretation of the data presented in the Astronautix article I linked to is. It isn't because the Moon's gravity is 50% or 64% of the Earth's. <div class="Discussion_UserSignature"> -----------------------------------------------------Ask not what your Forum Software can do do on you,Ask it to, please for the love of all that's Holy, STOP ! </div>

Kalstang · May 8, 2008

Derogatory and sarcastic comments or insinuations about mental capacity......are not welcome and offenders will be notified.

One of the rules over there...I think of this were really implemented here then half the people here would be banned

lol <div class="Discussion_UserSignature"> I have an answer for everything...you may not like the answer or it may not satisfy your curiosity..but it will still be an answer. "Imagination is more important then Knowledge" ~Albert Einstien~ Guns dont kill people. People kill people.Solar System </div>

Smersh · May 8, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>

Derogatory and sarcastic comments or insinuations about mental capacity......are not welcome and offenders will be notified.

One of the rules over there...I think of this were really implemented here then half the people here would be banned

lol Posted by kalstang</DIV>Wait till I make my next post over there and put a link to this thread. Then everyone at each forum can read all the comments in each others' threads. Maybe both threads will end up getting locked ... <img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-surprised.gif" border="0" alt="Surprised" title="Surprised" />However, there is a similar rule here already about that, re ad-hominems, etc. as you pointed out. <div class="Discussion_UserSignature"> <h1 style="margin:0pt;font-size:12px">----------------------------------------------------- </h1>Lady Nancy Astor: "Winston, if you were my husband, I'd poison your tea." Churchill: "Nancy, if you were my wife, I'd drink it."Website / forums </div>

Mee_n_Mac · May 8, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>Wait till I make my next post over there and put a link to this thread. Then everyone at each forum can read all the comments in each others' threads. Maybe both threads will end up getting locked ... However, I was under the impression that there was a similar rule here already about that, re ad-hominems, etc.  Posted by Smersh</DIV> I'd ask whether "they" think the Moon masses more than generally thought or if they don't believe in Newton's law re: gravity ? If it's the former how do they explain the Moon's orbit (see derekmcd's post). If it's the latter why is it (Newtonian physics) used (by the OP) to produce the 64% number from the supposed equigrav point ? <div class="Discussion_UserSignature"> -----------------------------------------------------Ask not what your Forum Software can do do on you,Ask it to, please for the love of all that's Holy, STOP ! </div>

derekmcd · May 8, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>@ dereckmcd = Can you explain the red curve in the link you previously gave ? It isn't Fe - Fm so what is it ? Posted by mee_n_mac</DIV> Refer to link for graph: http://www.apollo-hoax.co.uk/neutral.html Well, the green line represents the plot points as if the craft took off from earth and there was no other influence of gravity other than the source being earth.  The blue line is just opposite.  The point were they cross is really an arbitrary point as it doesn't include influence from both masses.  As you can see with the green line, the point where it begins to diverge is where the influence of the moon is becoming significant enough to noticeably influence the spacecraft.I think the red line is more representative to the Lagrange point (L1) which, although similar, is not the same as the gravity neutral point.  Trying to figure out the L1 point for the moon/earth system is a bit beyond me, but every google search has the L1 ranging from 38-39,000 miles.  I came up this trying to figure out the gravity neutral point:My math skill are pretty limited, but...F = G M s/e^2 = G m s/x^2 G = Gravitational constant m= mass of moon M=Mass of earth s = mass of spacecraft e= distance between L1 and earth x= distance between L1 and Luna d= distance between earth and moon centers of mass e + x = d drop G and s M/e^2 = m/x^2 e/x = sqrt (M/m) =  sqrt (5.98 x 10^24)/(7.36 x 10^22) e/x = sqrt (81.25) e/x = 9.014 e = 9.014x   --- />   9.014x + x = d x = d/9.014 x = 384,400km/9.014 =  42,644.77km (or about 26500mi) from the center of the moon.Close enough to the 'gravity neutral point' as represented by the green and blue lines.  The only problem I have with this neutral point is that it doesn't reflect the real world concerning object accelerating around a barycenter, spacecraft trajectories, libration points and 'spheres' of influence.  The gravity neutral point might work for 2 static object and will represent their mass properly, but that about all it's good for.  I was going to take a stab at the Lagrange formula to show it's close to what the astronauts and Wherner von Braun claim, but that's a bit more complicated than I'm prepared to deal with.My scorecard has von Braun - 1... Hoaxers - 0.  Of course, that's not surpising.<img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-laughing.gif" border="0" alt="Laughing" title="Laughing" />   <div class="Discussion_UserSignature"> <div> </div> <div>"If something's hard to do, then it's not worth doing." - Homer Simpson</div> </div>

Smersh · May 8, 2008

Replying to: <DIV CLASS='Discussion_PostQuote'>I'd ask whether "they" think the Moon masses more than generally thought or if they don't believe in Newton's law re: gravity ? If it's the former how do they explain the Moon's orbit (see derekmcd's post). If it's the latter why is it (Newtonian physics) used (by the OP) to produce the 64% number from the supposed equigrav point ? Posted by Mee_n_Mac</DIV>Cheers Mee_n_Mac. I'll quote that over there if you don't mind, along with Derekmcd's, Calli's and several other points. I think I'll post pretty soon there but will take me a while to construct it. Could be a big post. I won't quote everything, though, I can always make subsequent posts anyway, after others have posted there. Maybe I'll just post in "bite-size chunks."Thanks again for everyone's help here. <img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-wink.gif" border="0" alt="Wink" title="Wink" /> <div class="Discussion_UserSignature"> <h1 style="margin:0pt;font-size:12px">----------------------------------------------------- </h1>Lady Nancy Astor: "Winston, if you were my husband, I'd poison your tea." Churchill: "Nancy, if you were my wife, I'd drink it."Website / forums </div>

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