The True Nature of Gravity

Note: the original posts were removed, because they became unnecessary for this thread after I worked out a solution. Special thanks to Lozmaster for helping out with the preliminaries.

I wanted to derive an equation that can model objects falling from over long distances, where the acceleration due to gravity is not assumed to be constant. In other words, the difference in a(t) is not negligible. The reason I wanted to solve this equation is that I was taught motion due to gravity is parabolic, and I've known it isn't due to Newton's law. At first I tried finding the solution by looking online and in textbooks, but I couldn't find anything. Eventually, I gave up and attempted the problem myself.




















Making the following substitutions makes the equation cleaner.







Starting here, I use a different notation so that I can use the separation of variables method to integrate the function.







Substituting u as an operator allows for use of trigonometric integrals.











For the following substitutions, draw a triangle.

















I integrated from 0 to t. I used this notation, because it is impossible for me to find θs that correspond to r(0) and r(t). I wanted to find the functions that correspond to these substitutions before plugging back 0 and t.



Notice that the constant C on the left side is due to the r(0) term. Notice that the definite integral on the right side yields no second term.





I substituted back from the triangle I drew earlier.







I substituted sqrt® back in the place of the u operator.



Now that r is back, I can determine the constant term, which is basically just plugging in 0 into the function r. I noted this as r(0). I noted r(t) as just r, because r(t) is already implied by r.









I changed the variables so that they are less annoying to deal with. I chose h for initial height, r(0), and v for initial velocity, r'(0).







I'm going to simplify the asinh term first:



Plugging back in:

















This is just about the most simplified form I could get the above equation to. I like this form because when r=h, it's immediately apparent that t=0 and the ln term must subtract from the first part of t®.

Alternatively, the equation can be written as the following:



Where:



The equation gives the exact trajectory of the falling object when graphed. Albeit, I was unable to write r in terms of t, but it's easy to graph the function and just change the axes.

At a glance, the ln term will approximately cancel out when the difference between r and h is negligible. Without the ln term, the motion is exactly parabolic.

When the equation is graphed, it appears that the motion is parabolic over small distances, but over large distances, the time it actually takes for an object to fall through increasing acceleration is about half that of the time it actually takes for an object to fall through constant acceleration.


Someone please check my work. Much scrutiny is appreciated.

I think your physics is all messed up.

Gravitational acceleration is a constant. G is well defined as 6.67428 blah blah blah. Gravitational acceleration is always the same, but the force is ever increasing. So the force is dependent on both the mass of the Earth and the mass of the falling body and the square of the distance between them along with the G constant, not gravity.

EDIT: I didn't see your edits. Hold on.

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Doesn't gravity equal the amount of mass calculated against the collective force of the atomic binding of all contained matter therein to an infinitesimal point at which matter is broken down into pure energy... or is that the Black Hole theory?

In the equation you're using, G is a constant:

http://scienceworld.wolfram.com/physics/...stant.html

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F/M = Gm/r^2 = a

This makes sense when M is the reference point, because the acceleration due to gravity is infinitesimally small at r=inf and huge near r=0. This prompts me to integrate for the total acceleration from r2 to r1. Thus, I get the equation:

a® = Gm (1/r1 - 1/r2)

Ok, you've derived the equation for a change in gravitational potential. Now what? Since you want to find out the *true* shape of the displacement time curve for an object moving under gravity, i'd think integrating from your start point with respect to time would be better. (Gm/r^2 = a)
Then you'd need to express a or r as function(s) of time. I'd go for r, since you can just go with int(a)dt= velocity, int(v)dt= displacement
You would need to use a equation of motion of some sort to express r as some function of t, however since all the equations of motion that spring to my mind at the moment will give you a bunch of other variables that are also functions of t, I'm not entirely sure how well this approach would work.

Uhm. No.

F=ma

Your mass doesn't change when moving from Earth to space. Thus, experiencing weightlessness in space is evidence of the decreased acceleration due to gravity.

The weightlessness felt in space is because of the orbit around the earth. When in orbit, one is always accelerating towards the earth at a rate equal to about g, but a little less because of the increased distance.

This seems kind of needless. By the time that you get to falling distances high enough that you would actually see a difference in acceleration, an object would have reached terminal velocity.

Also, it is true that all objects have the same acceleration, regardless of mass. A more massive object is being pulled to the earth with more force, but since it has more mass it also takes more to move it. The equation F=ma proves it.

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For lack of a better source, this is visual proof that the acceleration due to gravity changes with r:



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

Also remember that we calculate gravity from the center of the earth to us, so the radius at the surface is actually about 6440 km, while at a reasonable distance away, say in orbit around the earth, is about 6760 km.

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As for acceleration, F=GmM/r^2 proves that the acceleration is dependent on mass. Negligible does not mean nonexistent, as I highlighted in the first post.

Yes- the mass of the sun/planet/satellite/whatever being orbitted. Not the object that is orbiting. The orbiting particles mass is irrelevant. You said it yourself, F=ma gives you the equation a=GM/r^2 where M is the mass of the other astronomical body. Not the one you are currently trying to find the acceleration due to gravity of.

You entirely missed the question.

It's kind of pointless to answer a question when there are misconceptions in getting there.

Also, does M equal the earth's mass, or does m?

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Thus, if this is true, then objects move in parabolic motion, because the integral of a constant acceleration gives a linear velocity, for which the integral gives a parabolic position.

Looking at Newtonian Gravitation, if you want to integrate with respect to time t, then you are stating that the Force F is some function where F(t) = GMm/r(t)^2.
Given that F(t) is a force, it could also be appropriately written as ma(t), where acceleration is a function due to time, which is appropriate in gravitation. Thusly, after dividing out the constant mass, and integrating with respect to time, we have
v(t) = integral of GM/r(t)^2 dt
z(t) = integral of v(t) dt

Which shows that for two objects, the acceleration is a function with respect to the distance they are away from each other, assuming they are in a state where movement is possible, and not truly constant, and that velocity is non-linear.

Acceleration is considered constant due to the fact that the change in the distance, r, is so insignificantly small that it need not be considered in most practical uses of the equation.