It looks like you're using an Ad Blocker.

Please white-list or disable AboveTopSecret.com in your ad-blocking tool.

Thank you.

 

Some features of ATS will be disabled while you continue to use an ad-blocker.

 

An End To The Moon Conspiracy!

page: 122
29
<< 119  120  121    123  124  125 >>

log in

join
share:

posted on Nov, 28 2007 @ 10:06 AM
link   
This is NASA's technology in 2007

youtube.com...



Bad day


Pete Worden, a Lunar Lander Challenge judge – and director of NASA's Ames Research Center, told SPACE.com that the engine blew up, with the rocket's engine chamber tossing out pieces onto the pad.


"It's over for them for this X Prize Cup," Worden said. But he added: "I do think they are getting there...it's a robust design. I think they'll make it. Once again, it proves that rocket science is hard."


"Number one is not to have any injuries or fatalities when you're dealing with rockets. So from a safety perspective, everything went by the books," said Brett Alexander, Executive Director of Space Prizes and the X Prize Cup.



Mod Edit: Big Quote – Please Review This Link.



[edit on 3-12-2007 by Jbird]




posted on Nov, 28 2007 @ 03:36 PM
link   
That's not NASA's technology; that's a small aerospace start-up's technology.


jra

posted on Nov, 28 2007 @ 06:11 PM
link   
reply to post by jra-2
 


jra-2 eh? You couldn't have come up with something slightly more original?

And like cdrn pointed out, this doesn't have anything to do with NASA. It was made by Armadillo Aerospace which has been completely self funded and the people who work there do it in there free time, generally about twice a week or so. And as of 2006 they had only spend $2 million on there project. I'd say they're doing really well considering. They've have many successful flights as well as failures, that's just the way it goes.



posted on Nov, 30 2007 @ 04:29 AM
link   
Lunar Lander Challenge is open also to all big space industries.

But no one is able to build a spacecraft that can land going backwards.

For this reason we see only Armadillo failures.

Otherwise, we would see also Grumman failures, Boeing failures and so

on.






posted on Dec, 3 2007 @ 12:42 PM
link   
Conclusion: Americans never went to the Moon.


anon... .nasa-global.edgesuite.net/anon.nasa-global/CEV/Orion.mov&_id=79493&_title=To the Moon and Beyond&_tnimage=156322main_Orion_lunar_orbit_100.jpg" target="_blank" class="postlink" rel="nofollow">www.nasa.gov...://anon.nasa-global.edgesuite.net/anon.nasa-global/CEV/Orion.mov|anon... .nasa-global.edgesuite.net/anon.nasa-global/CEV/Orion.mov&_id=79493&_title=To the Moon and Beyond&_tnimage=156322main_Orion_lunar_orbit_100.jpg


All already seen. BUT THEY ARE NOT ABLE TO LAND THIS WAY ON THE MOON.

How is that to go to the Moon they will use a biggest rocket and to come back they will use a very little one with solar panels?
Electric rocket? How does it work?

However they have learnt better to do 3D animation movies with 3D Studio Max.

BUT THEY ARE ONLY AND EVER ANIMATED CARTOONS, SCIENCE FICTION.





[edit on 3-12-2007 by jra-2]



posted on Dec, 3 2007 @ 01:06 PM
link   
reply to post by jra-2
 


jra-2 appears to be nothing less, nothing more than a 'troll'.

Just review the history of the posts, it should be evident....



posted on Dec, 3 2007 @ 02:24 PM
link   
reply to post by jra-2
 


welcome back skeptic-friend (sarcasm). Please present your ideas in a understandable and respectful manor and I'm sure people will want to debate you instead of ban you AGAIN. Please seriously consider what I'm saying. Thank you.



posted on Dec, 3 2007 @ 02:42 PM
link   
I am still trying to figure out how a planet (the Moon) which is about one sixth the size of our planet (Earth) can have .64g (or 64 percent the gravity of our planet, Earth).

Please, can anyone tell us (me) how this is possible?

Is Kepler wrong? Is Galileo wrong? Is Copernicus wrong???

Just asking.....



posted on Dec, 3 2007 @ 03:08 PM
link   

Originally posted by weedwhacker
I am still trying to figure out how a planet (the Moon) which is about one sixth the size of our planet (Earth) can have .64g (or 64 percent the gravity of our planet, Earth).

Please, can anyone tell us (me) how this is possible?

Is Kepler wrong? Is Galileo wrong? Is Copernicus wrong???

Just asking.....


We don't know anything about gravitational forces that link together the various heavenly bodies.

We don't know gravity of the Moon because nobody have gone there and nobody have measured it.

We do know instead that the studies to build a rocket that can land on the Moon or on another planet are at this point:

youtube.com...



Try to balance a coke can on your finger. Difficult? Impossible?



[edit on 3-12-2007 by jra-2]



posted on Dec, 3 2007 @ 04:11 PM
link   
reply to post by weedwhacker
 


Your information is a bit off...

The Moon is about 1/4 the diameter of the Earth, and has about 1/80 the mass.
The gravity of the Moon is .165 g (or 1/6 Earth's gravity), NOT .64 g

When calculating gravity, the mass of the object AND its radius must be considered.


EDIT: fixed transposed digits

[edit on 12/3/2007 by Soylent Green Is People]


jra

posted on Dec, 3 2007 @ 06:57 PM
link   

Originally posted by jra-2
Try to balance a coke can on your finger. Difficult? Impossible?


Sure that's hard to do, but that has nothing to do with a VTOL rocket. It's a horrible comparison really. A coke can has its mass uniformly distributed throughout the entire can, where as the LM, for example, has most of it's mass on the bottom half and its center of mass is close to the point of thrust, so it's quite balanced, unlike a coke can on top of ones finger.



posted on Dec, 3 2007 @ 08:33 PM
link   
Originally posted by weedwhacker



I am still trying to figure out how a planet (the Moon) which is about one sixth the size of our planet (Earth) can have .64g (or 64 percent the gravity of our planet, Earth).

Please, can anyone tell us (me) how this is possible?

Is Kepler wrong? Is Galileo wrong? Is Copernicus wrong???

Just asking.....


No weedwhacker none of those guys are wrong.

What we are doing weedwhacker is figuring the moons gravity relative to earth using the Bullialdus/Newton inverse-square law and the neutral point verified by von Braun, Michael Collins (Apollo 11) and Gene Cernan, Apollo 17..

The inverse-square law (Bullialdus/Newton) is any physical law stating that some physical quantity or strength is inversely proportional to the square of the distance between them, specifically, the gravitational attraction between two massive objects, in additional to being directly proportional to the product of their masses, is inversely proportional to the square of the distance between them.

Therefore using the following values:

Re = radius of the Earth = 3,960 miles
Rm = radius of the Moon = 1,080 miles
X = distance from the Earth’s center to the neutral
Point = 200,000 miles
Y = Distance from the Moon’s center to the neutral point = 43,495 miles
Ge = Earth’s surface gravity
Gm = Moons surface gravity

Since the forces of attraction of the Earth and the Moon are equal at the neutral point, the inverse-square law yields:

Ge (Re²/X²) = Gm(Rm²/Y²)

Gm/Ge = Re²Y²/Rm²X²

= (3,960)2 (43,495)2/(1,080)2 (200,000)2

= .64

Therefore, Gm = .64 Ge

So the gravity on the moon is approximately .64 that of earths gravity or almost two thirds. Now we understand why the Apollo astronauts were making those pitiful 18 inch hops on the moon. It should also be obvious why they tired so quickly.

If the moon’s gravity was in fact, one-sixth that of earth or approximately 16.66% we could work the problem in reverse and come out with a neutral point from the moon of about 24,000 miles. There is no evidence that the neutral point is that close to the moon.

That the gravity on the Moon is one sixth that of earths is one of the biggest con jobs in the history of mankind.

Thanks for the post weedwhacker.




[edit on 3-12-2007 by johnlear]


jra

posted on Dec, 3 2007 @ 08:50 PM
link   

Originally posted by johnlear
So the gravity on the moon is approximately .64 that of earths gravity or almost two thirds. Now we understand why the Apollo astronauts were making those pitiful 18 inch hops on the moon. It should also be obvious why they tired so quickly.


How do you explain the footage of some astronauts jumping 4 to 5 feet in height? They didn't do it often however, due to there center of mass being more behind them. With a heavy life support system on ones back. One tends to tip over backwards when jumping up high, as one astronaut found out, when he landed on his PLSS. He was lucky nothing happened to it. It's not worth the risk to jump as high as you can.

As for them getting tired. Those suits are rather stiff, it takes some effort to move around in those things. And when you're doing an EVA for several hours stright, you're going to get tired in that thing.



posted on Dec, 3 2007 @ 09:01 PM
link   

Originally posted by jra
How do you explain the footage of some astronauts jumping 4 to 5 feet in height?


Come now JRA you know better than that.... you are going to have to show us this footage... Link please....

I saw that in the training videos but I would really appreciate you backing this up

And I am truly sorry your evil twin got out


[edit on 3-12-2007 by zorgon]



posted on Dec, 3 2007 @ 09:06 PM
link   

Originally posted by jra
As for them getting tired. Those suits are rather stiff, it takes some effort to move around in those things. And when you're doing an EVA for several hours stright, you're going to get tired in that thing.


Well considering how heavy those suits are supposed to be they seem to get around in them alright in 1 G during training...

And that Nevada desert gets about as hot as the Moon






[edit on 3-12-2007 by zorgon]



posted on Dec, 3 2007 @ 10:20 PM
link   
reply to post by johnlear
 


johnlear, I understand the inverse square law of gravity, and I see your math...my real question involved the MASS of the two bodies in question, not just their radii. Can you re-work the equation?

In other words, how dense is the Moon, relative to Earth? If the Moon is a construct, as I've seen mentioned here on ATS forums, then one would have to assume it is not nearly as massive as a .64g surface gravity would indicate...I mean, I've seen some people claim the Moon is hollow...how does this jive?

Thanks for your indulgence

ps...since on the subject, what's the gravity on Mars?



posted on Dec, 3 2007 @ 10:28 PM
link   

Originally posted by weedwhacker
In other words, how dense is the Moon, relative to Earth?


Good question... seems no one knows for sure... its all just based on calculations not actual measurements. So far as I know no one ever weighed the moon
I hear tell that they added an iron core to the Earth to make it 'more' dense' to 'fudge' their figures...

But thats just a nasty rumor...





ps...since on the subject, what's the gravity on Mars?


come come now surely you can google that yourself?

[edit on 3-12-2007 by zorgon]



posted on Dec, 4 2007 @ 09:53 AM
link   
reply to post by zorgon
 


Yes, of course I can Google or Wiki MARS gravity...and when I Google or WIKI the Moon's gravity I get a different answer than .64g. So, my question was simply, is there any other data we're not being told about Mars' mass and its resulting gravity?

We know that the acceleration due to gravity on Earth is 9.81m/sec/sec

IF the Moon had a mass that is 1/6 that of Earth, then acceleration would be about 1.63m/sec/sec. Of course, at 64% on the Moon, the Mg would be about 6.23m/sec/sec. Perhaps we can watch some of the historic video from Apollo EVAs, especially the clearer vids from Apollo 15 thru 17, and someone can calculate the rate at wich objects are observed to fall, in light of an understanding of how objects fall in a gravitational field. Anyone with the computer/science savvy game for the challenge?



posted on Dec, 4 2007 @ 12:07 PM
link   
reply to post by johnlear
 


John --

What does the DISTANCE a celestial body is from the Earth have anything to do with How many g's that celestial body exerts on a person or object near the surface of that body? The only reason the Earth needs to be considered at all is because the answer is a RATIO of the gravity on Earth (thus the MASS and the RADIUS of the Earth needs to be considered, but not the DISTANCE of the Earth to that celestial body.)

Using this logic means, for example, that we would then need to figure the distance the Earth is from an extrasolar planet (for instance Gliese 581c) when calculating the ratio of its gravity to the Earth's. That would not be logical -- the distance of that extrasolar planet from the Earth would be irrelevant.

So you're saying that if I wanted to know the force of gravity that I would feel on Gliese 581c (expressed in g's), I would need to consider the distance that the planet is from the Earth?? That makes no sense.

Furthermore, your calculations ignore mass completely -- so are you trying to say that the mass of a celestial body is irrelevant when discussing the force of gravity one feels while on taht body? This also makes no sense.

Your calculations have more to do with the effects of the Earth's Gravity on the Moon itself (at least I think that was the mistake you made). Your calculation has nothing to do with the force of gravity one would feel while on a celestial body (such as the Moon). Again, I don't understand how the distance from the Earth to to the Moon (or any body in question) has anything to do with the calculation.


Here is the ACTUAL formula for calculating any body's "g", or the ratio of its acceleration due to gravity compared to the Earth's:

Fo = the force of gravity due to acceleration of an Object (in this case, the Moon)
Fe = the force of gravity due to acceleration of the Earth
**see Note 1 below why I used "F" in the place that you (John) used "G"**
Mo = the Mass of the object (in this case the Moon) = 7.348 X 10^22 kg
Ro = the Radius of the object (in this case the Moon) = 1,738,000 m
Me = the Mass of the Earth = 5.974 X 10^24 kg
Re = the Radius of the Earth = 6,356,800 m

Fo / Fe = the ratio of the gravitaional force of an object compared to the Earth's gravity (this is commonly called "g" (lower case "g"), ie the gravitational force on the Earth is earth is 1 g)

Here's the calculation:

Fo / Fe = (Mo / Ro²) / (Me / Re²)
Fo / Fe = (7.348 X 10^22 kg / 1,738,000 m^2) / (5.974 X 10^24 kg / 6,356,800 m^2)

*** Fo / Fe = 0.164 or about 1/6 ***


ANOTHER WAY TO LOOK AT IT:

Let's calculate the ACTUAL acceleration due to gravity on the Earth and the Moon separately.

a = acceleration due to gravity
G = the Gravitational Constant = 6.6726x10^-11 N*m² / kg² (this is expressed as a "Big G" as opposed to a lower case "g") Gravitaional Constant Definition
N = 1 "Newton" = 1 kg*m / s²

Here's the calculation of "a" for the Earth:

a = (G * Me) / Re²
a = (6.6726x10^-11 N*m² / kg² * 5.974x10^24 kg) / 6,356,800 m²

a = 9.86 m/s²


Here's the calculation of "a" for the Moon:

a = (G * Mo) / Ro²
a = (6.6726x10^-11 N*m² / kg² * 7.348x10^22 kg) / 1,738,000 m²

a = 1.62 m/s²


...therefore
the ratio of the Moons gravity to Earth's gravity (g) =

*** 1.62 / 9.86 = 0.164 or about 1/6 ***

You get the same answer using either calculation method.



Notes:

1. I used "F" for the force of gravity instead of the "G" that you used, John. I did not want to confuse the force of gravity (usually designated as lower case "g") with the concept of the "Gravitational Constant" (which is usually desiganated as an uppercase "G"). Suffice it to say that my "F" is the same as your "G".

2. The Gravitational Constant "G" (uppercase "G") does not appear in the formula for my first calculation (Fo /Fe) for clarity purposes -- because it would appear in both the Numerator and the Denominator, and thus it would cancel out of the equation completely. You could put it in the formula, but you would end up with the same answer.

3. MAYBE in on some small level, the distance from the Earth to the Moon DOES matter, since the Earth pulls on the Moon (and vice-versa), thus changing the force of the Moon's gravity somewhat (like the Moon pulls on the Earth, thus creating the tides), but this is negligible when considering the NOMINAL gravity due to acceleration.


EDIT: formatting, clarification, and typos



[edit on 12/4/2007 by Soylent Green Is People]



posted on Dec, 4 2007 @ 02:59 PM
link   
Nobody know the real gravity of the Moon because nobody went there and measured it.

However gravity of the Moon has not to be so little if it can lift Oceans water.

It is instead a big truth that neither Northrop Grumman (Lunar Module builders) nor Lockheed Martin Corporation (Shuttle builders) have technology to build a rocket that can be used on the Moon to land there going backwards.

Actually LUNAR LANDER CHALLENGE is open to all big space companies.

The real American RETROROCKETS technology is this:

youtube.com...





Excuse me for repeating things but I'm waiting for someone that tell me:
"You are right".

[edit on 4-12-2007 by jra-2]



new topics

top topics



 
29
<< 119  120  121    123  124  125 >>

log in

join