LRO- First images released

Originally posted by star in a jar
This site concerns the availability of raw LRO images.

Why should I trust a site from a guy I never knew before?

It's your money. These should be your images.

That's the funny part, it's not my money (I am Portuguese), but the images are also mine, now that is a good bargain.

Good photos, although not exactly as they should be, maybe next week.

For those that do not know it, the image made with several stripes is a colour image, if we join the right stripes together.

From Moongate

Another reputable source is the Encyclopaedia Britannica. This organization generally publishes information which is accepted by orthodox scientists. Therefore, their claim for the neutral point distance should be in close agreement with Wernher von Braun. In reference to Apollo 11, the Britannica stated the following in the 1973 printing within the topic "Space Exploration": Consideration of the actual dynamics of the Apollo trajectory will review the statements made above. The Apollo 11 spacecraft had been in Earth orbit at 118.5 mi. altitude, traveling at 17,427 mph. By firing the rocket motor at the exact moment when the spacecraft was precisely aligned along the proper trajectory, the velocity was increased to 24,200 mph. Because the Earth's gravitational pull continued to act upon the spacecraft during its two and three-quarters day (64 hr.) journey toward the Moon, the spacecraft velocity, with respect to the Earth, dwindled to 2,040 mph at a distance of 39,000 mi. from the Moon. At this point lunar gravitational attraction became greater than the Earth's and the spacecraft commenced accelerating as it swung toward and around the far side of the Moon, reaching a speed of 5,225 mph. By firing the spacecraft rocket propulsion system the velocity was reduced to 3,680 mph and the spacecraft entered an elliptical orbit about the Moon. Here the distance is 39,000 miles which is still close to the values given by Time and von Braun.

But the 1960 printing of the Encyclopaedia Britannica listed the neutral point distance as 19 Moon radii, or 20,520 miles, from the Moon. The distance discrepancy is between different printings of the same source. In We Reach the Moon, Wilford indicated that the spacecraft entered the lunar sphere of gravitational influence about 38,900 miles from the Moon.

In Footprints on the Moon written in 1969 by the Writers and Editors of the Associated Press, the neutral point is described as follows: Friday, Day Three of the mission, found Apollo 11 at the apex of that long gravitational hill between earth and the moon. At 1:12 p.m. EDT, the nose-to-nose spaceships passed the milestone where the moon's gravity becomes the more important influence. The astronauts were 214,000 miles from earth, only 38,000 miles from their rendezvous with the moon, leading their target like a hunter leads a duck.

There is no way to get around the discrepancy between the conventional, pre-Apollo distances of 20,000 to 25,000 miles, and the post-Apollo range of 38,000 to 43,495 miles. Even though the Earth to Moon distance varies between 221,463 and 252,710 miles, and spacecraft do not travel on a straight line between the Earth and Moon, this still does not explain the neutral point distance discrepancy. The logical conclusion is that the latest neutral point information reached the general public at about the time of the first Apollo lunar anding in 1969, even though it was determined as far back as 1959 from early lunar probes. Clearly, this discrepancy has not been pointed out to the public.

Is this enough to prove that pre-Apollo values for the neutral point have nothing to do with the reality of post-Apollo and moon probes flights?

[edit on 2/7/2009 by strNick]

reply to post by strNick


As I said, it is not a simple two body problem. Nor is it a static problem, everything was moving in relation to everything else. The Apollo craft was never exactly between the Moon and the Earth so the gravitational forces acting on it were not directly opposed to each other.

If the Moon wasn't moving and the spacecraft wasn't trying to catch it and the spacecraft was on straight line exactly between the Moon and the Earth then yes, the "neutral point" would have been at about 23,000 miles from the Moon. But it wasn't, it could not have been. As a result, the "neutral point" was farther away.

www.abovetopsecret.com...

www.apollo-hoax.me.uk...

[edit on 7/2/2009 by Phage]

reply to post by Phage

I see, you're ignoring the facts.

Then let me ask you: where this figure - 20 some thousand miles - came from? It's hypothetical (i.e. theory, based on assumption), right?

In the article (second link) the author uses this formula
and not the inverse-square law formula, which is without the mass.

At least, run a game based on Source engine (Half-Life 2 or Counter-Strike Source, for example) and set sv_gravity 100 (default value is 600) so it will be like on the Moon, then jump. Do the same with sv_gravity 384 (64%). Compare the results (and don't forget to compare them with the Apollo film).

[edit on 3/7/2009 by strNick]

Originally posted by VitalOverdose
At this point in time i don't think there is much that could provide solid evidence that the moon landing wasn't faked.

Id love to see NASA try another moonshot using the same equipment as they did in the Apollo missions. With no radiation shielding for the suits or the ship and the computing power of a Commodore64.

That's all you got?? Radiation shielding and low-memory computers?

I'm sorry, but those criticisms have been thoroughly debunked by everyone a long time ago. Try again.


back on topic...
As LRO, it's good to hear that everything seems to be in working order.

reply to post by strNick


Is that formula correct. Say close to the Moon of mass M and
space craft of mass m and avoiding the Earth mass E for now.

Force = F = mass of craft X gravity g = GMm/rxr

Thus g = GM/rxr

The force of Moon mass M may have the Earth mass E included
by some formulation but closeness counts as in horse shoes.

ED: The Moon Missions where in a time of the German free energy
ufo scientists control as their counterparts needed a war to make
more money as in Germany so we got Nam in a big way after the
JFK assassination. It was Nam and the Moon cause free energy was
not considered.




[edit on 7/3/2009 by TeslaandLyne]

reply to post by strNick

No.
It is you who are ignoring the facts. Or do not understand them.

Fact: The formula is the same. Your source uses the general inverse square law, mine uses the inverse square law applied specifically to gravity. The formula used by your source includes mass as a component of "G". G (surface gravity, usually expressed as "g") is directly related to M (mass).

Fact: Your sources use the formula incorrectly. Your sources assume that at 43,000 miles the spacecraft was on a direct line between the Earth and the Moon. It was not.

I thought G was the gravitational constant used in the force between
two masses:

www.google.com...

reply to post by TeslaandLyne


It is. But G is a constant, it is not a variable or a force. The guys saying the Moon has .6 Earth's gravity are using it incorrectly. They used it as the force of gravity which is supposed to be "g".

[edit on 7/3/2009 by Phage]

Originally posted by strNick
....At least, run a game based on Source engine (Half-Life 2 or Counter-Strike Source, for example) and set sv_gravity 100 (default value is 600) so it will be like on the Moon, then jump. Do the same with sv_gravity 384 (64%). Compare the results (and don't forget to compare them with the Apollo film)....

We are talking about the gravity of the moon acting upon an object on or nearits surface. This is the Gravity Due To Acceleration we are talking about. We aren't talking about the gravitational attraction between the Earth and the Moon.

Why should we consider the distance between the Earth and Moon when trying to figure out the force of gravity on an object or body near the surface of the Moon? That makes no sense. We don't take the Sun's distance to the Earth into consideration when we figure out the Acceleration due to Gravity on the Earth, do we?? When we talk about the gravity on Mars (which is a little more that 1/3 Earth) we don't care how far Mars is from the Earth --

-- so why do you care now how far the Earth is to the Moon?


Here's how you calculate the force of gravity on a body (Accelaration due to Gravity) on the surface of the Moon and Earth.


Acceleration due to gravity (Agrav) = GxM / R^2

Calculate the acceleration due to gravity on the Moon (Agrav[moon]):

- The Gravitational Constant (G) is 6.67 x 10^-11 meters^3 per kg^-1 per second^-2
- Moon's mass (M) is 7.35 × 10^22 kg.
- The Moon’s radius (R) is 1.74 × 10^6 meters


Agrav [moon] = GxM / R^2

= (6.67 × 10^-11) x (7.35 × 10^22) / (1.74 × 10^6)^2

Acceleration due to Gravity [moon] = 1.62 meters/sec^2


Calculate the acceleration due to gravity on the Earth (Agrav[earth]):

- The Gravitational Constant (G) is 6.67 x 10^-11 meters^3 per kg^-1 per second^-2
- Earth's mass (M) is 5.97 × 10^24 kg.
- Earth’s radius (R) is 6.37 × 10^6 meters

Agrav [earth] = GxM / R^2

= (6.67 × 10^-11) x (5.97 × 10^24) / (6.37 × 10^6)^2

Acceleration due to Gravity [earth] = 9.8 meters/sec^2


Therefore, the difference in Acceleration due to gravity of the Moon and Earth =:

Agrav[moon] / Agrav[earth] =
1.62 / 9.81 =
0.16503 or 1/6

therefore the gravity acting on a body on or near the surface of the Moon is 1/6 the gravity acting on a body on or near the surface of the Earth. The distance of the earth to the Moon is not practically relevant in this calculation


EDIT TO ADD:
Here's a link to a picture of the calculations that might be easier to read (it's hard to show math on this board):

files.abovetopsecret.com...



[edit on 7/3/2009 by Soylent Green Is People]

reply to post by Soylent Green Is People

Just for fun, and to complicate things a bit, here's another thing to consider.

When the Apollo craft was 200,000 miles from Earth the Sun was exerting 1.5 times the gravitational influence of the Earth upon it. The Sun was exerting 5 times the gravitational influence of the Moon.

Orbital mechanics make it a very complex problem. Since everything was orbiting the Sun it may not be relevant but since the Apollo craft was changing its orbit, it may be. Not really something you can calculate on a cocktail napkin.


[edit on 7/3/2009 by Phage]

reply to post by Phage



That's another great point... but thanks for complicating it!...


Seriously though, the point I'm trying to make is quite simple, and I don't want it getting lost...
....what I'm asking is why are some people putting the Earth into the equation at all when trying to figure out what kind of gravity an astronaut feels on the Moon.

The Earth has no practical relevance in the matter whatsoever.

I'm agreeing with Teslaandlyne (for once ) in saying that the equation for figuring out how much gravity an astronaut will feel on the Moon is

GM / R^2

Not

GMm / R^2


The only mass to consider is the Moon's mass, and the only distance to consider is the distance from the center of the Moon to its surface (its radius). Leave the Earth out of it.



[edit on 7/3/2009 by Soylent Green Is People]

reply to post by Soylent Green Is People


I'm not sure what a video game on the surface of the Moon has to do with anything but they are using the supposed "neutral point turnover" of the Apollo missions (which is influenced by Earth) to prove that the Moon's gravity is stronger than claimed.

Their calculations are too simplistic (see above).

[edit on 7/3/2009 by Phage]

reply to post by Phage


That does bring up the multi bodied system.
Well I was going to outline an approach and just avoid numbers
and then thought find some numbers and then figured NASA or
the web may be lacking in these for ready access and I figured
no numbers and then I remembered a whole bunch of numbers
at one place in a discussion about the Moon in 1919 and wondered
how good those numbers were from today's.


Disregard the topic, just watch the numbers roll by:

The moon is a nearly spherical body, of a radius of about 1,081.5 miles, from which I calculate its volume to be approximately 5,300,216,300 cubic miles. Since its mean density is 3.27, one cubic foot of material composing it weighs close to 205 pounds. Accordingly, the total weight of the satelite is about 79,969,000,000, 000,000,000,000 and its mass 2,483,500,000,000,000,000 terrestrial short tons. Assuming that the moon does physically rotate upon its axis, it performs one revolution in 27 days 7 hours 43 minutes and 11 seconds, or 2,360,591 seconds. If, in conformity with mathematical principles, we imagine the entire mass concentrated at a distance from the centre equal to two-fifths of the radius, then the calculated rotational velocity is 3.04 feet per second, at which the globe would contain 11,474,000,000,000,000,000 short foot tons of energy, sufficient to run 1,000,000, 000 horsepower for a period of 1,323 years. Now, I say that there is not enough energy in the moon to run a delicate watch.

www.rastko.rs...
Yeah the units and number size, I just caved in to wondering how
any are close to our present understanding.

reply to post by yeti101



Ooops.Posted on wrong one.

Sorry.


[edit on 6/7/09 by gallifreyan medic]

Back on topic....

The official website from the people operating the LRO's Camera...
lroc.sese.asu.edu...
....also has images -- the images are zoomable and the raw image "TIF" files are downloadable [go to the web page and click on "images"]

Originally posted by Phage
Just for fun, and to complicate things a bit, here's another thing to consider.

Apollo 14



www.nasa.gov...



[edit on 17-7-2009 by zorgon]

So NASA says its a spacecraft and all the little Skeptic run and create a page that says "See" Its not a rock, its proof..." and now they tell me "Oh yeah but this" Yuppers its a SPACECRAFT"



But when Zorgon says "Its an artifact on Eros, and Zorgon's picture is much better resolution than NASA's, those same skeptics say "Its just a rock"



Funny how the Lemmings operate isn't it?

reply to post by zorgon

It's impossible to tell from either image what each is. Both could be rocks. The difference is that we know there are vehicles on the Moon. The difference is that we know exactly where they are. The difference is that the Eagle is exactly where it's supposed to be so we know it is not a rock.

Originally posted by Phage
It's impossible to tell from either image what each is. Both could be rocks.

Quite true... but when I say that in the Apollo Hardware thread it gets deleted as off topic

I hear my Medieval group calling. Since Mike got the axe its gonna be boring here anyway.

Your right one can extrapolate that IF it was expected it must be... Score one point for your team But I was always in the camp that they went, just faked what they saw. No reason to change that view yet