# The wierdness of gravity

1. Feb 20, 2012

### phillip1882

so i would like to start off with what is known and work from there.
equation for falling objects
d = 1/2 *a*t^2
equation for solar orbits
d^3 /t^2 = k
or if you prefer,
v^2 *d = k
equation for galactic orbits
v = d/k

so; for the first observation, i would like to note, neither force nor mass seems to be a factor.
for the second observation, i would like to note that the first and third equations flow naturally from the definitions of acceleration and velocity.
so, whatever is causing objects to be attracted must a) not involve mass, and b) seamlessly tie together these three equations. i don't know what that would be, but maybe we should start here in order to work forward.

2. Feb 20, 2012

### Jorriss

These equations are seamlessly tied together!

F=dp/dt + F(gravity)=G(m1)(m2)/(r^2) => classical gravity (Also, google Kepler's laws of planetary motion)

Also, it's not fair to say nothing depends on mass because you do not see 'm'. The proportionality constants may depend on mass.

3. Feb 20, 2012

### Ivan Seeking

Staff Emeritus
From the posting guidelines for this forum

4. Feb 20, 2012

### phillip1882

My apologies, I thought the skepticism forum to be the best place for that post since it's outside mainstream physics. I should have read the forum rules there before posting.
I hope it now resides in the correct forum.

to Joriss; I would like to point out that Newton starts breaking down at the galactic level.
a full 90% of matter is "missing". We've tried to detect particles smaller than neutrinos that might make up this matter but to no avail. We know about where it should be , based on the gravitational lensing effect, but the matter itself seems to be beyond our current instrumentation. I'm not necessarily saying Newton is wrong. He may very well be right. He didn't really predict what gravity would do at the galactic level. Maybe there is dark matter; or maybe Modified Newtonian Dynamics is the way to go. Or perhaps there is a third yet uncharted path. I don't know. I was just exploring some thoughts.

5. Feb 21, 2012

### Jorriss

Sure, you can get into GR (which I know nothing of) but you posted some very basic classical mechanics formulae then wanted to discuss bringing them together. Those formulae aren't even right if you want to talk GR/quantum.

I probably don't really know what you were asking.

6. Feb 21, 2012

### elfmotat

Well there's clearly a force (in the context of Newton). How else do you explain the acceleration?

The gravitational acceleration experienced by an object is (locally) independent of the properties of that object (i.e. mass, shape, etc.). This is one way of stating the equivalence principle.

http://en.wikipedia.org/wiki/Equivalence_principle

I'd like to add that classical gravity is very well understood.

7. Feb 21, 2012

### zoobyshoe

I'm confused by this. The distance a mass covers during an acceleration certainly seems dependent on the force applied, since the acceleration is dependent on it. a = F/m

8. Feb 21, 2012

### gmax137

Then what's the value for "a" for a falling object? I'm assuming you're thinking, "here on earth it is 9.8 m/sec^2". Where does this value come from? You should be able to calculate it from Newton's law of gravitation.

9. Feb 21, 2012

### nasu

I am not sure what are you trying to prove. However your starting point is flawed, at least regarding the second equation.
I suppose you mean Kepler's third law. Here d is the major axis (or semi-axis, depends on what is K) of the orbit. It is not distance traveled, as in your first equation.
This is a constant for a given orbit so using v=d/t, as in your second form of the equation, is nonsense. It would also imply a constant speed around the orbit which is not the case for elliptic orbits.
Also does not make sense to compare this with the free fall equation. Labeling different quantities with the same letter does not make them similar.

10. Feb 21, 2012

### phillip1882

to nasu:
right, d in both equations is not ment to be distance traveled, but distance from orbiting body.
the third equation, v = d/k is for Galactic orbits, not the solar obits. (again d here is not distance traveled but distance from galactic center.)
Also I am well aware that labeling different quantites with the same letter does not make them similar. however, we know that all three equations should be similar in some unifying fashion. (in other words; all three equations should obey the same base law of gravity.)
to gmax:
i'd like to point out that Galileo knew the constant a to be roughly 10 meters per second^2 before Newton developed the theory of gravity. How? He simply observed falling obects, and calculated the rate.
I am well aware F = m*a.
however, gravity seems to behave in a very odd fashion at the galactic level with Newton's equation.

11. Feb 21, 2012

### gmax137

phillip, I will be more blunt. You said that whatever causes objects to be attracted does not involve mass. Several posters pointed out that your "a" in equation 1 involves both force and mass. I invited you to show us what "a" means. The answer is, "a" = GM/r^2, where G is newton's gravitational constant, M is the mass of the earth, and r is the radius of earth at the surface (where your equation 1 obtains).

I don't understand the remainder of your posts either.

12. Feb 21, 2012

### Integral

Staff Emeritus
In response to the OP.

The equations you quote are not fundamental. They are derived, under the assumptions for which they are true mass is not needed. What you are doing is looking at a cake and claiming that it does not have eggs and flour as ingredients because you cannot see them.

I would recommend that before making claims about physical theories that you actually research them a little.

13. Feb 21, 2012

### nasu

In the equation
d = 1/2 *a*t^2
"d" is distance traveled. This equation is not actually specific for gravity.
It is valid for motion with constant acceleration so it applies approximately for free fall in proximity of the planetary surface.
It does not apply to orbital motion of planets.

In the Kepler's law d is an orbital parameter and not the distance from the "orbiting body". It is a fixed quantity for a given orbit. The distance of the orbiting body from the central body is a variable quantity.

Well, your statements seems to indicate that you are not.
Regarding the statement that the equations should "obey" some law, is not very clear what you mean. The first equation is valid for motion under any constant force so there is really no need for it to "obey" the law of gravity any more than other laws.

14. Feb 21, 2012

### phillip1882

You gentlemen are very nit-picky. I realize the details are important, and I wasn't as clear on them as I should be, but no one has yet to even address my primary concern.
let me state it again. THERE IS NO KNOWN EQUATION FOR GRAVITY WHICH UNIFIES THE THREE GIVEN EQUATIONS.

Again, Newton falls short at the galactic level unless you assume 90% of matter is undetectable. A very precarious state in my opinion.

15. Feb 21, 2012

### Staff: Mentor

Both that and the other concern you listed are wrong: the link between the three is Newton's law of gravity (or for higher accuracy, GR).
Precarious or not, there isn't anything else to go on. Given the choice of two assumptions (there is a different law of gravity acting on the galactic level or there is more mass we can't see), you should pick the simpler of them.
I'll be blunt too...

We're not being nit-picky: What you did seems purposely intended to obfuscate the falseness of your claims. It makes you look like a crackpot.

16. Feb 21, 2012

### genericusrnme

'so, whatever is causing objects to be attracted must a) not involve mass'

This is like me saying the repulsion between a + charge and a + charge isn't due to the electric charge because they cancel on both sides...

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