Gravitational Force

1. Nov 18, 2014

Eric Hentschel

Gravity has such a long range and such a weak magnitude . Might it be due to the fact that it's acting on the three-dimensional vehicle as a mode exerting its force this way gives limitations, its inability to conserve energy due to friction resistance among others. Also is gravity applied evenly over the surface of the party in mass being attracted or is there the correlation between the point closest to the center of the large mass is proportionally transferring gravitational magnitude throughout the object. So is the force exerted on the center particle of the small mass?

2. Nov 18, 2014

Bandersnatch

Hi Eric,

Not sure I understand you questions. All forces act in three dimensions. How's gravity different there?

The second seems to be asking whether gravity varies over extended bodies - the answer is yes it does. The difference in strength and direction is what gives rise to tidal forces, but the body needs to be relatively near the centre of the gravitational field for the effects to become noticeable.

3. Nov 18, 2014

Eric Hentschel

Sorry, I can see how the wordings confusing. Does gravity act only the center of the small mass or the entirety of it. could it be concentrated solely on the center and the rest of the mass follow suit. But either way I see gravity losing a lot of its magnitude due to three-dimensional resistance is that are experienced by the smaller mass is this commonly which attributed to that loss.

4. Nov 18, 2014

A.T.

The highest tidal forces aren't necessary near the center of the mass. It depends on the mass distribution.

5. Nov 18, 2014

A.T.

The entirety of it. The center of mass is used to simplify calculations, when the gravity gradient is negligible.

6. Nov 18, 2014

Eric Hentschel

with relevance to quantum gravity they led fail to find the correlation between gravitys range and magnitude in relation to the other three forces. How have physicist take into account that gravitational forces one has to be exerted on three-dimensional objects which will experience reresistance

7. Nov 18, 2014

Staff: Mentor

I'm not sure what kind of "resistance" you mean, but if you mean inertia, general relativity addresses that by not modeling gravity as a force at all. Instead it arises from the curvature of spacetime caused by the presence of mass and energy.

8. Nov 18, 2014

Eric Hentschel

Yes , I am familiar with that but my question is essentially can we calculate the rate that the magnitude decrease due variables which are limited to the objects of 3 dimensions such as gravity of a nearby mass pulling it on an angle losing some of the force. Think of it in terms of hypnosis for a moment, When I become the hypnotized by a "bigger" hypnotist , I am told to walk towards him. Whether I stumble doesn't mean that the hypnosis was any weaker for I had the SAME desire to arrive there. Is it possible to think of gravity as the initial need to move and everything else more along the lines of stumbling?

9. Nov 18, 2014

DaveC426913

It sounds like you're asking of the inertia of a massive body has an effect on the force of gravity.

It does not. It certainly affects the acceleration of the massive body, but it does not reduce the force of gravity.

10. Nov 18, 2014

Staff: Mentor

How would gravity "pull on it at an angle"? Can you describe a specific example of this?

Once again, in relativity, gravity is thought of as arising from the curvature of spacetime. It has nothing to do with an object's "need to move". An object moving solely under the influence of gravity is in free fall, feeling no force; the trajectory it takes is determined only by the curvature of spacetime, and doesn't depend on any particular property of the object.

11. Nov 18, 2014

Staff: Mentor

In Newtonian physics, yes, this is true. If we're talking about how gravity is modeled in relativity (and this is, after all, the relativity forum), then gravity is not a force, and the inertia of a body moving under gravity does not affect the acceleration of the body at all; as per my previous post, an object moving solely under the influence of gravity is in free fall, and therefore has zero proper acceleration, regardless of its inertia. The body's inertia only comes into play when a (non-gravitational) force is exerted on it--if, for example, it fires rockets. Any such force will cause a nonzero proper acceleration, and how much proper acceleration will be felt in response to a given (non-gravitational) force is certainly affected by the body's inertia, yes. But gravity doesn't cause any proper acceleration.

12. Nov 18, 2014

Eric Hentschel

I included the word angle to indicate its force was taking the object of the straight trajectory! Instead of trying to convey my ideas of the subject at this juncture, I will just ask you if there is a curvature in space time how can the object not be experiencing any energy. The object is moving whether falling or not it moves from a point of reference it does possess kinetic energy and there in lies a force, however subtle you interpret it. I look forward to your replies!

13. Nov 18, 2014

Staff: Mentor

The quantity that you're calling "kinetic energy", the thing that is $(mv^2)/2$ in classical mechanics, is a function of only three of the four components ($\sqrt{x_1^2+x_2^2+x_3^2}$ - you'll notice that $x_0$ doesn't appear in that expression) of the velocity vector (also called "four-velocity" - google for it) of an object in four-dimensional spacetime.

It is clear that I can always find a coordinate system in which any three of the four components of a vector are zero, and therefore that I can find a coordinate system in which the kinetic energy is zero. In classical terms, this comes down to saying that the kinetic energy of a 10 kg anti-tank shell moving at 1000 m/sec is $5\times{10}^6$ Joules as far as the tank it strikes is concerned - but an insect riding along with the shell could reasonably claim that the kinetic energy of the shell is zero.

Thus, there isn't any particular kinetic energy necessarily associated with an object moving through spacetime. Choose where you're standing and how fast you're moving relative to that object, and you can arrange to make its kinetic energy whatever you please.

14. Nov 18, 2014

Staff: Mentor

I don't understand what "experiencing energy" means.

Nugatory already responded to the part about possessing kinetic energy; as he said, that's coordinate dependent, so we can always find a frame in which it is zero.

Also, I don't understand why possessing kinetic energy requires a force. As I have repeated several times now, an object moving solely under the influence of gravity is in free fall and feels no force. Yet it has kinetic energy, at least in some coordinates. Its kinetic energy will even be changing in some coordinates. But it feels no force.

15. Nov 19, 2014

Eric Hentschel

Its an equal an opposite reaction and passing kinetic energy to the object even changing and using the edge of the mass's field as where the object began its trajectory is what I am using to reference as a starting point.

16. Nov 19, 2014

Staff: Mentor

This only applies in relativity if an actual force (one that can be felt) is being exerted. So I still don't see how an object in free fall that possesses kinetic energy is experiencing a force.

I'm not sure I understand what this means, but it seems to mean that you are defining "motion" as being relative to the gravitating mass, and defining "force" as "causing coordinate acceleration relative to the gravitating mass". To me, this just means you are giving an excellent illustration of why GR does not define "force" this way, as I have already described several times.