If energy is represented by Acceleration=A, then

am=F ams=W (ams)/t=P

Acceleration is always present when motion is present, acceleration is the energy.

Since gravity translates into acceleration, and acceleration turns to Force with mass, and force turns to work with distance, and work to power with time, doesn't it stand to reason that acceleration never goes away and that therefore acceleration represents energy, therefore gravity is energy because acceleration is energy?

Gravity acts on all different masses in a peculiar fashion. If you measure the amount of force it takes to accelerate a 100 lbs ball to 10 mph on an even surface and the amount of force needed to accelerate a 10 lbs ball to 10 mph on an even surface, the larger mass would require more force.

But gravity doesn't exert force in the way you and I do, it doesn't respond to mass at all except in one way, Gravity accelerates mass. When gravity accelerates the 10 lbs ball at 32.2 ft/s^2 and when gravity accelerates the 100 lbs ball at 32.2 ft/s^2, two different forces are generated simultaneously.

I suggest this is because gravity behaves the way an electro-magnetic field behaves, where momentum is ignored during acceleration.

Gravity ignores mass when accelerating, so what is gravity?

Gravity appears to act equally on some part of all the molecules in an object, creating a density effect. The energy of gravity gives mass its weight and that is why weight is measured like a force. I am starting to think all energy comes from life in some way form or fashion, but mass gets weight from energy vectors. Energy and force behave the same way accept that pure energy is expressed in acceleration where as force is expressed as a product of mass and energy.

Maybe gravity accelerates all molecules equally, thus generating unequal forces (varying energy) but constant acceleration. Like the more molecules, the more points of access gravity has.

Guide on translating forces:

Acceleration to force: multiply the acceleration by mass. F=ma Force to work: multiply the Force by distance. W=Fs (s=distance) Work to power: divide by time. P=W/t

Power to Work: Pt=W Work to Force: F=W/s (s=distance) Force to Acceleration: F/M=A

Power to force multiply the power by the time and divide by the distance. Acceleration to Power: (AMs)/t=P (s=distance t=time) Power to Acceleration: Pt/(Ms)=A (s=distance t=time)

P(t/s)=W/s = F = MA = M (v^2/r)
[s=distance, t=time, P=power, W=work, F=force, M=mass, A=acceleration, V=velocity, r=radius]


closed as off-topic by CuriousOne, user36790, David Z Oct 28 '15 at 9:31

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From another comment:

I wonder why the object with greater mass does not accelerate slower

Maybe it will help, to think of it like this:

If you (theoretically) split up the object in many, many (let's say infinitely many) pieces of equal mass, then gravity can pull equally in each of those pieces. They will all fall with equal acceleration $g$.

  • If you put those pieces together to be one object, then of course the total gravitational pull $F$ on this object is the sum of all those pulls.
  • The total mass $m$ is the sum of all the small masses.
  • But each piece of mass is still only pulled downwards with the acceleration $g$ - that is not summed up.

Maybe this helps.

  • $\begingroup$ It's like mass interacts with an unseen force at the molecular level. The unseen force that generates this energy is called gravity. What is the source of the energy then do you think? Matter itself? Would that mean that matter has an attraction energy field? $\endgroup$ – d w Oct 28 '15 at 0:27
  • $\begingroup$ Yes, mass is indeed being pulled at the molecular level. And we call that force the gravitational force or the weight. And exactly as CuriousOne mentioned in a comment above, understanding the reason that the gravitational force exists (that means, understanding why mass attracts other mass) is not at all easy. This is still a major challenge of physics, and as far as I know, noone actually understands this yet fully. So it is understandable that you have this question, but I am afraid I cannot give you the reason for gravity's existence - I can only help you understand how it works. $\endgroup$ – Steeven Oct 28 '15 at 0:33
  • 1
    $\begingroup$ @dw, force fields don't generate energy, they store it. Imagine yourself lifting a heavy stone up to the top of a building. You are doing work on the stone (i.e., exerting energy). We know that energy is conserved, so where did it go? It is stored in the gravitational attraction between the Earth and the stone. When energy is stored in a field like that, we call it potential energy. If you drop the stone off the roof, you get the stored energy back as kinetic energy (at least, up until the point where the stone smacks into the sidewalk.) $\endgroup$ – Solomon Slow Oct 28 '15 at 3:21
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    $\begingroup$ all force is energy, that remains a fact But this is untrue - this is not a fact. Again, the units are completely different - and while a force is a vector, energy is a quantity. They are two different types of proporties. There are several differences. Force is not equal to energy! There is no more to it - but force can of course be included in a formula to calculate energy of some sort. How can you have displacement without force? Think of a spaceship drifting in space. It moves (at constant speed) but no forces are exerted on it. $\endgroup$ – Steeven Oct 28 '15 at 14:37
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    $\begingroup$ @dw About your equations. This is true: $$P=Fv=W/t$$ ...but it is not equal to $F$. This is true (the last equal sign just for circular motion): $$F=ma=m v^2/r$$ ...but it is not equal to energy $E$. About the units: Force is in Newton's $[N]$ which can be rewritten as: $$[N]=[kg\cdot m/s^2]$$ Energy is in Joules $[J]$, which can be rewritten as: $$[J]=[N\cdot m]=[kg\cdot m^2/s^2]$$ Power $P$ is in Watts $[W]$, which can be rewritten as: $$[W]=[J/ s]=[kg\cdot m^2/s^3]$$ Does this help clearing out the relationships? $\endgroup$ – Steeven Oct 30 '15 at 11:19

I think you are wondering why heavy and light things accelerate at the same rate when in free-fall (let's ignore air resistance for simplicity).

You probably notice that the force on the heavy object is greater than the force on the light object. This is correct.

You then wonder why the heavy object does not accelerate more, since the force is greater.

The answer is inertia. The heavy object feels a greater force, but it is harder to accelerate because it has more mass. The two effects exactly cancel out (inertial mass == gravitational mass) and so acceleration in a gravitational field is independent of mass.

One way to think about it is to imagine the object splitting up as it falls. Would you really expect the pieces to accelerate less as they separated? [remember, we're doing this in a vaccum].

  • $\begingroup$ "You then wonder why the heavy object does not accelerate more, since the force is greater." No, I wonder why the object with greater mass does not accelerate slower. Why do large masses get more force put on them by gravity? $\endgroup$ – d w Oct 27 '15 at 15:12
  • $\begingroup$ Inertia is not the answer, inertia is what is being ignored by gravity. All masses accelerate at the same rate. That is an infinitely different amount of forces generated by gravity instantaneously. I want to know why inertia is ignored. Seems like a field to me. $\endgroup$ – d w Oct 27 '15 at 15:13
  • $\begingroup$ @dw Why do large masses get more force put on them by gravity? Imagine a 1lb brick - it gets a force $F$. Now bring up another 1lb brick - it also gets a force $F$. Now glue the two bricks together to make a 2lb brick. What is now the force? (Answer: $2F$). $\endgroup$ – Oscar Bravo Oct 28 '15 at 8:42
  • $\begingroup$ I under stand if you double the mass, you get double the weight. I was wondering why when you get double the mass, why you get the same acceleration with double the force. Force and acceleration force are the same and not the same since they are on opposite sides of the equation. They represent the same idea, accept force is a product of mass and acceleration. The question was answered, gravity acts on molecules, not mass. That is why weight is a good indicator of mass. $\endgroup$ – d w Oct 28 '15 at 13:33
  • $\begingroup$ I want to correct the last sentence of the last statement saying, "gravity acts on molecules, not mass" by saying, "gravity acts on individual molecules of a mass." $\endgroup$ – d w Oct 28 '15 at 15:38

Gravity does not ignore the mass because the mass of the balls actually provides some of the gravitational pull. Their masses are do small however that they have little effect on the force of gravity. Mass is the fundamental property that governs the strength of interaction between the earth and the balls.

  • $\begingroup$ The difference in force generated by gravity in a 1 lbs ball vs the force generated by gravity in a 100,000 ton ball is due to the increased gravity from the ball? Any math to go with that? $\endgroup$ – d w Oct 27 '15 at 7:36
  • $\begingroup$ $F = {Gm_1m_2 \over r^2}$... There you go. $\endgroup$ – Oscar Bravo Oct 27 '15 at 14:44
  • $\begingroup$ Let me be more clear. There is math that shows why earth has gravity and it is based on mass. The question is, since the difference between a 1 pound brick and a 2 pound brick is not going to create a difference in the acceleration due to mass, but the force is measurably different. That is what you have to figure out to answer this question. $\endgroup$ – d w Oct 27 '15 at 15:25
  • $\begingroup$ @dw: The force on a 2lb brick is twice the force on a 1lb brick. But... the 2lb brick is twice as hard to move (i.e., it has twice as much inertia). Hence the acceleration is the same. I think you are mixing up the chain of events: Presence of mass in a gravitational field -> Force -> Acceleration. $\endgroup$ – Oscar Bravo Oct 28 '15 at 8:38
  • $\begingroup$ @OwenBoyle Twice the mass means twice as much energy to accelerate. Why is gravity putting out twice as much energy to force the mass down as the mass right next to it? Why doesn't the smaller mass accelerate faster since it requires less energy to move? $\endgroup$ – d w Oct 28 '15 at 13:38

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