So I was in the shower today...

[Torahteen]

You know he's right aga...


Don't shoot me. :lol:

[Agamemnus]

No it wouldn't decrease because F changes in the same proportion as m.

a = 1 / C

C... constant.

[SJ Zero]

Alright, SJ Zero, I have no freaking clue what your integration notations stand for, but integrate this, if you can:

Aga, I proved my own theory(and confirmed it with outside sources). If, considering your supposed math qualifications, you refuse to do the same, then I can't take you seriously in your protestations. Thus far, you've presented one critically flawed piece of mathematical evidence and little else.

Therefore, unless you have either numbers, or a corralary to the theory which actually helps this "cold light gas" theory get off the ground, act like an adult and accept that what you were thinking in the shower is incorrect. The physical world isn't going to change because you try to fire out a theory regardless of the logic behind it.

Now, I'd like to add a few things:

-We know the mass of the earth, more or less, based on the distance from the sun and the speed we're moving. This is something we've managed to do by knowing the same theory of gravitation we've been arguing over. If a large chunk of the earth were made out of light gass, then the density of the earth would be such that geologists would consistently find the mass of the earth to be much much less than expected.

-We know how sound travels through the earth. Seismographs taken around the world at the moment of an earthquake all show minute differences in how they register the 'sound' of an earthquake because the sound travels through the earth. The numbers given imply a solid outside, a molten semi-inside, and a hard core. A hollow earth would give spectacularly different results.

-Absolute Pressure remains relevant. The reason I keep talking about pressure is that there isn't going to be any 'super cold gas'(Which is an unscientific term as used here, by the way -- The terms super cooled and super heated refer to the temperature relative to phase change points. 1ata Steam at 100C is saturated steam. 1ata Steam at 110C is superheated steam. 1ata Steam at 90C is supercooled steam. Since there is no good reason to assume you meant the actual definition of the word, since you describe no phenomena to suggest a non-newtonian fluid, I'm assuming that you simply mean a relatively cold gas) that remains a gas unless the absolute pressure is almost vacuum. Simply put, there's no reason to believe that the center of our earth is a vacuum relative to outer space. It runs counter to the currently accepted theories about the creation of the planet. The forces would literally have to drag the matter in the middle of the earth outward with enough force to raise the surface to create this effect, and the only way that might happen is through alien intervention. No matter what you think gravity is capable of, the Law of Conservation of Energy more or less guarantees that Gravitational energy isn't going to raise and support the entire surface of the earth because of a bit of mass in the middle. Even if that were the case, the earth surely wouldn't accept this massive vacuum it could collapse upon.

Anyway, I'm done here. If you don't feel that this discussion is worth taking seriously, then I don't feel like having it with you. By all means, keep your unrealistic portrait of reality. Now everyone on this site knows what to make of your theories.

[Zack]

No it wouldn't decrease because F changes in the same proportion as m.

a = 1 / C

C... constant.

I don't understand what you are getting at (or missing).
As the mass of two objects increase, so does the gravitational attraction between them.
In a seperate law, F=ma, and by extension a=F/m, acceleration of an object=force applied/the mass of the object, as the mass increases, the rate of acceleration DECREASES. So as mass goes up, gravitational force does too, but we NEED more force to move a more massive object, therefore things of different masses fall at the same speed.

[Agamemnus]

SJ Zero, you're making the same ridiculous assertions about what I said as before. And, if you didn't realize, that was a definite integral that actually made sense, at least to me, which I wasn't sure how to solve. Just because your integration notation is incomprehensible doesn't mean you need to trash talk me... and also you had no sources to back up your result.

Zack, the rate of acceleration remains constant, and so does acceleration.

a is acceleration in "a = 1/C".

Let's say m1 = 5:

a = 1/C
da/dt = 0 (da/dt is the rate of acceleration over time)

Let's say m1 = 10:

a = 1/C (the same...)
da/dt = 0


As you're not yet quite as hopeless as SJ Zero, Zack, I'll dissect your statement:

>In a seperate law, F=ma, and by extension a=F/m, acceleration of an object=force applied/the mass of the object, as the mass increases, the rate of acceleration DECREASES.
Yes, that is true, assuming the force is held CONSTANT.

>So as mass goes up, gravitational force does too
Also true, assuming the acceleration DECREASES.

>but we NEED more force to move a more massive object
Yes, that is correct, assuming acceleration stays constant.

>therefore things of different masses fall at the same speed.
I assume you mean gravitational acceleration here. That would be true, if you somehow reduced something else proportionately-- for instance, if you reduced the mass of the other object, or increased the distance.

[Agamemnus]

In other news, I made an approximation of the gravity using my formulas and the result shows gravity first increasing and then sharply decreasing as the point of measurement reaches a depth at the center of the earth.

Code:

defint A-Z
screen 9
DIM C as double, F1 as double, F2 as double, intF1 as double, intF2 as double, F as double
DIM bigR as double, r as double, x as double
dim g as double, pi as double


G = 6.67300 * 10^-11 'm^3 kg^-1 s^-2
pi = 3.14159265
r = 63.78 'hundreds of kilometers
m1 = 5.98 * 10^24 'kg
C = G*pi*r*m1
'C = 10: r = 1000

for n = 0 to r

F1 = 0: for x = 0 to n
bigR = (r^2 - x^2)^.5
if bigR <> 0 then
intF1 = 0
for b = 0 to bigR
if n^2+b^2 <> 0 then
intF1 = intF1 + b/(n^2+b^2)
end if
F1 = F1 + intF1
next b
end if
next x

F2 = 0: for x = 0 to 2*r-n
bigR = (r^2 - x^2)^.5
if bigR <> 0 then
intF2 = 0
for b = 0 to bigR
if (2*r-n)^2+b^2 <> 0 then
intF2 = intF2 + b/((2*r-n)^2+b^2)
end if
F2 = F2 + intF2
next b
end if
next x
F = F1 - F2

if n > 1 then
'print F2;F;F1
'line ((n-1)*2+50, 300-log(abs(lastF))*10)-(n*2+50, 300-log(abs(F))*10), 15
line ((n-1)*2+50, 300-abs(lastF)/5)-(n*2+50, 300-abs(F)/5), 1
line ((n-1)*2+50, 300-abs(lastF1)/5)-(n*2+50, 300-abs(F1)/5), 2
line ((n-1)*2+50, 300-abs(lastF2)/5)-(n*2+50, 300-abs(F2)/5), 4
line ((n-1)*2+50, 300)-(n*2+50, 300), 15
end if
lastF2 = F2
lastF1 = F1
lastF = F
next n
'print abs(F)

sleep
system

[Zack]

SJ Zero, you're making the same ridiculous assertions about what I said as before. And, if you didn't realize, that was a definite integral that actually made sense, at least to me, which I wasn't sure how to solve. Just because your integration notation is incomprehensible doesn't mean you need to trash talk me... and also you had no sources to back up your result.

Zack, the rate of acceleration remains constant, and so does acceleration.

a is acceleration in "a = 1/C".

Let's say m1 = 5:

a = 1/C
da/dt = 0 (da/dt is the rate of acceleration over time)

Let's say m1 = 10:

a = 1/C (the same...)
da/dt = 0


As you're not yet quite as hopeless as SJ Zero, Zack, I'll dissect your statement:

>In a seperate law, F=ma, and by extension a=F/m, acceleration of an object=force applied/the mass of the object, as the mass increases, the rate of acceleration DECREASES.
Yes, that is true, assuming the force is held CONSTANT.

>So as mass goes up, gravitational force does too
Also true, assuming the acceleration DECREASES.

>but we NEED more force to move a more massive object
Yes, that is correct, assuming acceleration stays constant.

>therefore things of different masses fall at the same speed.
I assume you mean gravitational acceleration here. That would be true, if you somehow reduced something else proportionately-- for instance, if you reduced the mass of the other object, or increased the distance.

Maybe I'm being misleading. Suppose we have two objects, one which is 5 kg and the other which is 10kg. To accelerate the 5kg object to 10 m/s, we need how much force? F=ma, F=5kg*10m/s, F=50 N. We need 50 Newtons of force. To accelerate a 10kg object to 10 m/s, we need double the force, because the mass doubled. We need 100 N. Suppose that we are getting our force from gravity. Like any other force, we need double to accelerate the 10kg object. Luckily, the force of gravity increases whenever the mass goes up. So 10kg is being pulled on more than 5kg. But we need more force to pull the 10kg object. So, the conclusion is that the 5kg and 10kg object will be moved at the same rate.

[Agamemnus]

>Suppose that we are getting our force from gravity. Like any other force, we need double to accelerate the 10kg object.
OK.

>Luckily, the force of gravity increases whenever the mass goes up. So 10kg is being pulled on more than 5kg.
Yes...

>But we need more force to pull the 10kg object.
We need more force to push a 10kg object, not pull it...

>So, the conclusion is that the 5kg and 10kg object will be moved at the same rate.
Now you're talking about velocity.. speed. v = d/t. So you're saying that the masses will pull towards each other at the same speed regardless of the mass, which is completely different than "as the mass increases, the rate of acceleration DECREASES"...

Your assertion is: the rate at which one object approaches the other object due to gravity is independent of the first object's mass.

Code:

v = d/t, a = v/t

F1 = m1*a1

F = G*m1*m2/d^2

Now suppose that F = F1, then you would be correct, I suppose.
F = F1 = m1*a1 = G*m1*m2/d^2 , so: v1 = t * G*m2/d^2

Can you say that F1 = F? Obviously if you do this you will be confusing the frame of reference. If F1 = F, then F2 = F, and F1=F2:
F1 = F2
m1*a1 = m2*a2

But we know that a1 = a2 (and v1=v2), and to say that m1=m2 is ridiculous. Ifthe point of reference was simply the object in question (e.g. F1=F2=F), NOT the other object:

1) The first object's velocity towards its own position is not dependant on the mass of the first object, but rather the second object's mass.

2) The second object's velocity towards its own position is not dependant on the second object's mass, but rather the first object's mass.