- #211
heusdens
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Originally posted by Mr. Robin Parsons
Heusdens it doesn't matter how many times you tell me the wrong answer, it is still the wrong answer.
Take a 60 Metric tonne rock, place it on the moon, it weights in at 10 metric tonnes, place it at the Gravitational Absent Spatial Point, (GASP) (that is the Langerian point, between the Earth and the moon) and it weights NOTHING, but it still has gravitational energy and is attracted towards one, or the other, the Earth, or the Moon.
So heusdens, in Newtons proof, a very small part of the matter in the outer shell is what generates all of the Earth's gravity.
Now, the weighting of the Earth is done from the measure of the G force, since only a small percentage of the mass is actually generating that 'weight', (the rest, according to you, is 'self canceled(ling)) hence the true weight of the planet is factors larger then is the presently believed weight, and the density error problem also becomes factors larger, in accordance with the needed missing mass that you say 'self cancels', so your density error goes way up from the 5000 kg per m3, to the fact of, well, that shell game explanation I wrote up on the other page works out to about 20% of the outer shell as effective gravity generator, hence we need to multiply by 5 x's the density per m3, so you now need to prove that you have the ability to pressurize the mass to approx. 25,000 kg per m3!
Oooops, remember, the rock on the outer shell has been measured and proven, and tested, to be about 3000 kg per m3, so WOW, have you ever got a density problem NOW!
Also, (according to your promotion/theory) the Sun weights in at way more then we currently think, as does all of the rest of the Stars, the Galaxies, the universe's mass, and so on, and so on...
Sorry, (not really) but you are espousing the wrong answer as it might sound really nice, but it does not match the reality that has been measured and observed.
Oh, just in case you have missed how physics works, reality in it's presentation of the facts, ALWAYS Wins!
EDIT SP AND GRAMMER!
You have a great talent for misinterpreting someone else's ideas.
From where comes your idea that when standing outside of the earth, any of the gravitational interactions between some mass and that of the Earth would cancel out? Where was it stated? Nowhere! You made this up in your own mind.
I hold this dicussion on here with you no longer fruitfull, cause you keep misinterpreting me.
Here is something for you to figure out. Suppose you are standing in the middle of earth, at it's gravitational center.
Now your belief is that there is still a nett gravitational force.
It can be proven very easily however that right at the center, all forces of gravity of all the mass of earth, cancel out.
Now take any part of the Earth mass, and calculate this mass and it's distance to the center of gravity. That will enable you to calculate the force of gravity from that part of Earth's mass. Which is not zero. But at the exact opposite direction, we will find an equal mass at equal distance. This follows the fact that the density of Earth at any specific depth is more or less the same, and the Earth is a round spherical object. So we can deconstruct all of Earth's mass into small parts (as small as you want) and calculate the force of gravity from all parts. Since we always can find for every force vector an equal and exactly opposite directed force vector, this means that the resulting force of gravity exactly results in a force of zero.
Go calculate that yourself, and don't come back before you done that for every individual atom of earth!