Uncovering What Gives Gravity Its Power to Interact w/ Mass

[Odain]

Okay, yes, if I move an object (A) away from a massive object (B), I give the object (A) the potential energy needed to be pulled back down -- I fight against the force of gravity.
But what gives gravity the ability to fight against me for what seems like "free"?

For me, I was able to move that object away using my hands because of the food I had eaten which gave me the energy to move my arms, grab the object, and move it (Yes, I realize it goes a lot deeper than that, but let's just leave it there for the sake of argument and not getting out of control.)

What is giving gravity whateveritneeds to have an effect on the mass that I am moving?
How could it possible be affecting the object for free?

 

[TurtleMeister]

Welcome to PF Odain.

Try visualizing objects (A) and (B) with a spring connected between them. The spring represents gravity. I know it's a very simple analogy, but I think it may help you with your question. I'm not quite sure what you mean by "for free". There is nothing special about the force of gravity in this respect.

 

[Hetware]

Okay, yes, if I move an object (A) away from a massive object (B), I give the object (A) the potential energy needed to be pulled back down -- I fight against the force of gravity.
But what gives gravity the ability to fight against me for what seems like "free"?

For me, I was able to move that object away using my hands because of the food I had eaten which gave me the energy to move my arms, grab the object, and move it (Yes, I realize it goes a lot deeper than that, but let's just leave it there for the sake of argument and not getting out of control.)

What is giving gravity whateveritneeds to have an effect on the mass that I am moving?
How could it possible be affecting the object for free?

You are fighting against gravity. If you stop fighting against it, you will be less exhausted. Just lay down on the sofa and let the sofa support you.

The reason you perceive that gravity is fighting against you is because the human body is not a very rigid support. If you try to hold a bucket at arm's length in front of you, your arm will move up and down a bit. The upward motion requires energy. Hang the bucket on a hook fixed to a rigid body, and it will stay there without effort, because the rigid body doesn't move up and down.

 

[Chestermiller]

I think what you are asking is "how can gravity exert a force on a body if there is no contact involved?" This was part of Einstein's motivation for developing the General Theory of Relativity, which explains how gravity actually works.

Body in Free fall:
Newton: When you are in free fall, there is a gravitational force acting on you and you are accelerating.

Einstein: When you are in free fall, there is no gravitational force acting on you, and you are not accelerating (relative to curved spacetime)

Stationary body:

Newton: When you are stationary relative to a large massive body, there is a gravitational force acting on you, an upward contact force acting on you, and you are not accelerating.

Einstein: When you are stationary relative to a large massive body, there is no gravitational force acting on you, but there is an upward contact force acting on you, and you are accelerating (relative to curved spacetime).

To learn more about this, you need to study General Relativity.

 

[Odain]

...too much to put in a quote...

This might be what I'm looking for... I know about general relativity from a layman perspective, but not much more. I didn't think it had what I was looking for, but in the way you explained it, it might.
I'll definitely take a deeper look at it after my classes.
We've just stared special relativity, but I don't think we're going to get into general relativity at all.

Thanks for the replies from you other guys too, and the welcoming. I'm thinking I just didn't describe my question well enough. Chester seems to be closest to what I'm asking.

 

[Hetware]

I think what you are asking is "how can gravity exert a force on a body if there is no contact involved?" This was part of Einstein's motivation for developing the General Theory of Relativity, which explains how gravity actually works.

Body in Free fall:
Newton: When you are in free fall, there is a gravitational force acting on you and you are accelerating.

Einstein: When you are in free fall, there is no gravitational force acting on you, and you are not accelerating (relative to curved spacetime)

Stationary body:

Newton: When you are stationary relative to a large massive body, there is a gravitational force acting on you, an upward contact force acting on you, and you are not accelerating.

Einstein: When you are stationary relative to a large massive body, there is no gravitational force acting on you, but there is an upward contact force acting on you, and you are accelerating (relative to curved spacetime).

To learn more about this, you need to study General Relativity.

If I am in an elevator in deep space with a giant pulling the cable at g=9.8 m/s^2, I will feel as if I am on Earth. The force acting on me will be my mass times g. The distance I travel will be 1/2 g t^2. Work is force times distance, so my kinetic energy has changed.

Why is my kinetic energy constant on Earth?

 

[K^2]

Spring example is by far the best. When you are doing work against gravity, you are storing energy in gravitational field. It's kind of like storing energy in a spring by stretching or compressing it. You get same thing with electrostatic fields.

It's a bit more complicated in General Relativity, but you still, effectively, store energy in space-time curvature. So the spring analogy does apply loosely.

 

[HallsofIvy]

If I am in an elevator in deep space with a giant pulling the cable at g=9.8 m/s^2, I will feel as if I am on Earth. The force acting on me will be my mass times g. The distance I travel will be 1/2 g t^2. Work is force times distance, so my kinetic energy has changed.

Why is my kinetic energy constant on Earth?

"kinetic energy" relative to what?

 

[russ_watters]

Okay, yes, if I move an object (A) away from a massive object (B), I give the object (A) the potential energy needed to be pulled back down -- I fight against the force of gravity.
But what gives gravity the ability to fight against me for what seems like "free"?...

What is giving gravity whateveritneeds to have an effect on the mass that I am moving?
How could it possible be affecting the object for free?

Forget gravity and try the same thing with a spring. Conservation of energy is the reason and nothing happens for "free". You seemed to have discarded the potential energy you gave to the object by lifting it, when you shouldn't have.

 

[Chestermiller]

If I am in an elevator in deep space with a giant pulling the cable at g=9.8 m/s^2, I will feel as if I am on Earth. The force acting on me will be my mass times g. The distance I travel will be 1/2 g t^2. Work is force times distance, so my kinetic energy has changed.

Why is my kinetic energy constant on Earth?

These are very interesting and perceptive questions.

If you are in the elevator frame of reference, your kinetic energy has not changed. To you, you feel a force acting on you, but you are not moving.

Why is your kinetic energy constant on Earth? I interpret this to mean, if there is no gravitational force acting on you but you feel an upward contact force acting on you imposed by the earth, how come you are not starting to move upward?

Imagine that you are out in space again, and you have a ball attached to a string. You swing the ball around in a circle, and the string has tension on it. There is a radial acceleration, but no radial velocity component. The velocity is in the circumferential direction, and the acceleration is the result in the velocity vector changing direction, but not magnitude. The kinetic energy is constant.

Have you ever been on that arcade ride called the Roundup. You go into a cylindrical room and the room begins to rotate around the cylinder axis. When the angular velocity reaches a certain value, the angular velocity is held constant, and the floor drops out of the room. You are pinned against the cylinder wall, and the wall is exerting a radially inward force on you. But you are not moving radially, and, from your perspective, your kinetic energy is zero. As far as you are concerned, there is radial gravity present.

Now consider the case where you are standing on the surface of the earth. According to Einstein, there is no gravitational force acting on you, but the surface of the Earth is exerting a radially upward force on you, and you are accelerating upward. The situation here is very similar to the case of the Roundup. In the Roundup, the directions involved are the circumferential and radial directions. The acceleration on the Roundup is in the radial direction, and the velocity is in the circumferential direction. In the theory of relativity, there are 4 dimensions, and, for a guy standing on the surface of the earth, the two directions involved are the radial direction and the time direction. The velocity is in the time direction (and is very large, equal to the speed of light), and the acceleration is in the radial direction. As 3D beings, we cannot see the velocity in the time direction, just as, on the Roundup, you cannot see the velocity in the circumferential direction. The basic thing that is happening is that the direction of the velocity vector in the time direction is changing. This results in the radial acceleration and radial force that you feel. I don't know how to explain it any more precisely without using the general relativity math involved.

 

[Hetware]

These are very interesting and perceptive questions.

If you are in the elevator frame of reference, your kinetic energy has not changed. To you, you feel a force acting on you, but you are not moving.

Why is your kinetic energy constant on Earth? I interpret this to mean, if there is no gravitational force acting on you but you feel an upward contact force acting on you imposed by the earth, how come you are not starting to move upward?

Imagine that you are out in space again, and you have a ball attached to a string. You swing the ball around in a circle, and the string has tension on it. There is a radial acceleration, but no radial velocity component. The velocity is in the circumferential direction, and the acceleration is the result in the velocity vector changing direction, but not magnitude. The kinetic energy is constant.

Have you ever been on that arcade ride called the Roundup. You go into a cylindrical room and the room begins to rotate around the cylinder axis. When the angular velocity reaches a certain value, the angular velocity is held constant, and the floor drops out of the room. You are pinned against the cylinder wall, and the wall is exerting a radially inward force on you. But you are not moving radially, and, from your perspective, your kinetic energy is zero. As far as you are concerned, there is radial gravity present.

Now consider the case where you are standing on the surface of the earth. According to Einstein, there is no gravitational force acting on you, but the surface of the Earth is exerting a radially upward force on you, and you are accelerating upward. The situation here is very similar to the case of the Roundup. In the Roundup, the directions involved are the circumferential and radial directions. The acceleration on the Roundup is in the radial direction, and the velocity is in the circumferential direction. In the theory of relativity, there are 4 dimensions, and, for a guy standing on the surface of the earth, the two directions involved are the radial direction and the time direction. The velocity is in the time direction (and is very large, equal to the speed of light), and the acceleration is in the radial direction. As 3D beings, we cannot see the velocity in the time direction, just as, on the Roundup, you cannot see the velocity in the circumferential direction. The basic thing that is happening is that the direction of the velocity vector in the time direction is changing. This results in the radial acceleration and radial force that you feel. I don't know how to explain it any more precisely without using the general relativity math involved.

Are you familiar with the Ehrenfest Paradox?

 

[Chestermiller]

Are you familiar with the Ehrenfest Paradox?

Sorry, no.

Chet

 

[Hetware]

Sorry, no.

Chet

I believe this is completely legal, or I would not post it. See page 61:

http://www.combat-diaries.co.uk/diary29/Link%2014%20Einstein.PDF

I don't pretend to comprehend all of what is there.

 

[Chestermiller]

I believe this is completely legal, or I would not post it. See page 61:

http://www.combat-diaries.co.uk/diary29/Link%2014%20Einstein.PDF

I don't pretend to comprehend all of what is there.

Thanks Hetware. I will look it over. I'm not sure if my studies are far enough along to understand all the details.

Chet

 

[Chestermiller]

Hi again Hetware,

I didn't realize the link you sent was for Einstein's book The Meaning of Relativity (I believe it was edited by Brian Greene). I thought the link was about the Ehrenfest Paradox. I have a personal copy of Einstein's book, and have read it. Thanks.

Chet

 

[Hetware]

Hi again Hetware,

I didn't realize the link you sent was for Einstein's book The Meaning of Relativity (I believe it was edited by Brian Greene). I thought the link was about the Ehrenfest Paradox. I have a personal copy of Einstein's book, and have read it. Thanks.

Chet

The discussion in the page I reference pertained to the so-called Ehrenfest Paradox, but not by that name.

I'm not sure anybody fully understand the subtelties of Einstein's The Meaning of Relativity.

"In its original formulation as presented by Paul Ehrenfest 1909 in the Physikalische Zeitschrift,[1] it discusses an ideally rigid cylinder that is made to rotate about its axis of symmetry. The radius R as seen in the laboratory frame is always perpendicular to its motion and should therefore be equal to its value R0 when stationary. However, the circumference (2πR) should appear Lorentz-contracted to a smaller value than at rest, by the usual factor γ. This leads to the contradiction that R=R0 and R<R0.

"
http://en.wikipedia.org/wiki/Ehrenfest_paradox