Gravitational potential at center of Universe

[Johnny5]

This question arose in post 17 of http://www.scienceforums.net/forums/showthread.php?t=11414from

 

It was off topic so I'm moving it here.

 

Off topic:

Originally Posted by Johnny5

Isn't the total gravitational field at the center of the universe necessarily zero? Maybe he could distinguish it that way.

 

At the center of universe the gravitational potential should be at maximum' date=' the sum of gravity forces is zero.[/quote']

 

All I really meant was that if you put something at the center of the universe, the net gravitational force on it should be zero, and thus it will remain at rest there.

 

I believe it was Sir Isaac Newton who proved this.

 

In order to prove that at the center of the universe the gravitational potential should be a maximum, you have to use gravitational field theory.

 

If and when Latex is working, perhaps I will check what you say by doing that.

 

First you have define gravitational potential, analagous to electrical potential, and go from there. Mathematically speaking, its really easy.

 

You even come up with the idea of objects which gravitationally repel each other. Mach's idea.

 

Just postulate negative inertial mass.

[J.C.MacSwell]

If you had a closed system, and chose an inertial frame based on the center of mass being at rest, then the center of mass would stay at rest. The gravitational net force at this point would not necessarily be zero, nor would it necessarily stay zero if it was.

[Johnny5]

If you had a closed system, and chose an inertial frame based on the center of mass being at rest, then the center of mass would stay at rest. The gravitational net force at this point would not necessarily be zero, nor would it necessarily stay zero if it was.

 

Can we discuss this further?

 

Introduce the mathematical tools necessary to discuss it.

[Guest Geneticist]

It depends on your reference frame. If you set it at the center of the earth, then it will be zero, since potential gravitational energy is m*g*h. Now, if the reference frame was placed on the surface of the earth, then it would be different.

[Johnny5]

It depends on your reference frame. If you set it at the center of the earth, then it will be zero, since potential gravitational energy is m*g*h. Now, if the reference frame was placed on the surface of the earth, then it would be different.

 

mgh is not an exact formula.

 

This is why i want to use gravitational field theory for everything.

[J.C.MacSwell]

Can we discuss this further?

 

Introduce the mathematical tools necessary to discuss it.

 

Picture yourself in space at the center of a system that includes you, a solar mass 200 million miles away to your right and a 2 solar mass 100 million miles away to your left. You are at the center of mass.

 

Which way would you be pulled by gravity?

[Johnny5]

Picture yourself in space at the center of a system that includes you' date=' a solar mass 200 million miles away to your right and a 2 solar mass 100 million miles away to your left. You are at the center of mass.

 

Which way would you be pulled by gravity?[/quote']

 

Both directions, yet equally and so i would remain at rest in the CM frame.

[J.C.MacSwell]

Both directions, yet equally and so i would remain at rest in the CM frame.

 

Check again.

 

Center of mass is linear and gravitation is inverse squared.

 

Hint: You are closer to the bigger mass!

[Johnny5]

Check again.

 

Center of mass is linear and gravitation is inverse squared

 

I didnt do any calculation i just sort of blurted out the answer.

 

Let me see...

 

You say i have a mass of 2M on my left and a mass of M on my right.

 

but the mass of 2M is twice as far away from me as the M mass.

 

Let me use the mathematics, its the only way for me to be certain of the answer.

 

denote my mass by m

 

The gravitational force on me to the left is given by:

 

F = (2M)(m)/R^2

 

and the gravitational force on me to the right is given by:

 

F = (M)(m)/r^2

 

and the distance from me to the mass on my right is 200 million miles

and the distance from me to the mass on my left is 100 million miles.

 

therefore

 

R=2r

[CanadaAotS]

Thats really easy...

you'd be pulled to the left... really hard to, 8x more force then to the right

[CanadaAotS]

now if you meant the other way around (1 SM to the right at 100 mil miles, and 2 SM to the left at 200 mil miles) then itd be 2x more force to the right.

[J.C.MacSwell]

now if you meant the other way around (1 SM to the right at 100 mil miles, and 2 SM to the left at 200 mil miles) then itd be 2x more force to the right.

 

Then you would not be at the center of mass. Your first answer (8x) is correct.

[Johnny5]

Let me make sure I copied everything down correctly, here is your exact question again:

 

Originally Posted by J.C.MacSwell

Picture yourself in space at the center of a system that includes you, a solar mass 200 million miles away to your right and a 2 solar mass 100 million miles away to your left. You are at the center of mass.

 

Which way would you be pulled by gravity?

 

I will use a diagram this time, to represent the given information:

 

 

M..........X....................m

 

X represents my position, M_x will represent my mass.

The distance from me to m is twice the distance from me to M.

 

M=2m

 

F_L = (2m)(M_x)/R²

 

F_R =(m)(M_x)/(2R)²

 

F_L = 4(2m)(M_x)/4R²

F_L = 4(2m)(M_x)/(2R)²

F_L = 8(m)(M_x)/(2R)²

 

F_L = 8F_R

 

The answer is above is correct.

 

The gravitational force to the left is 8 times greater than the gravitational force to the right.

 

I see your point. Center of mass is linear, and gravitation is inverse r^2.

 

I could just have known that from the formulas for both.

 

Let me comare things.

 

Here is the formula for the center of mass of an N body system.

 

MR = S m_ir_i

 

The indice ranges from i=1 to i=N, in the summation. M is the total system mass, and mi is the mass of the ith body. R is the position vector of the center of mass in some arbitrary reference frame, and ri is the position vector of the ith body.

 

So:

 

M = S m_i

 

Let us choose our frame to be the CM frame, then in particular, R=0... the zero vector.

 

So, in the center of mass frame, the following statement is true:

 

M0 = S m_ir_i

 

Hence:

0 = S m_ir_i

 

Now, the system is a closed three body system hence the following statement is true in the CM frame:

 

0 = m₁r₁+m₂r₂+m₃r₃

 

Using the original symbols for the three masses we have:

 

0 = Mr₁+M_xr₂+mr₃

 

Now, my position happens to be at the center of mass of the system, that is, r2=0, but this should be inferrable.

 

Let me not a-priori assume that my position is the center of mass of the system, but let me formulate true statements from a frame in which the center of mass of Mx is the origin. Therefore, for sure r2=0, and it should be inferrable that this frame happens to be the CM frame of the system.

 

 

Since r2=0 it follows that:

 

0 = Mr₁+mr₃

 

0 = 2mr₁+mr₃

 

0 = 2r₁+r₃

 

-2r₁=r₃

 

The previous statement was stipulated to be true, hence the center of mass of Mx is the CM of the frame.

[Zeth]

There is no one point center of the universe. Relativity shows this to be the case since every observer in constant motion can claim that he is stationary and every object on average is moving away from him. Due to supersymmetry there is also no net force on the largest imaginable scale.

[Spyman]

All I really meant was that if you put something at the center of the universe, the net gravitational force on it should be zero, and thus it will remain at rest there.

Hey, You could have told me, I almost missed this thread.

 

I realized that You ment something like that but since they talked about measuring time dilation, I wanted to point out that time dilation will grow when moving closer to the center.

 

Which Janus explained to me in this thread: http://www.scienceforums.net/forums/showthread.php?t=10761

 

For the record: I didn't mean that You where off topic, it was my comment to You, that was off topic.

 

If you had a closed system, and chose an inertial frame based on the center of mass being at rest, then the center of mass would stay at rest. The gravitational net force at this point would not necessarily be zero, nor would it necessarily stay zero if it was.

Would this also apply to a roughly uniform spherical system (3D) where the initial set had the mass center at the center of the system ? (And the maximum gravitational potential.)

 

I tried to start a discussion in this thread: http://www.scienceforums.net/forums/showthread.php?t=10511

(But the interest was low.)

 

There is no one point center of the universe. Relativity shows this to be the case since every observer in constant motion can claim that he is stationary and every object on average is moving away from him. Due to supersymmetry there is also no net force on the largest imaginable scale.

Yeah, You can always claim that You are the one at rest and the Universe is moving.

 

But I guess it would be a hard thing to prove, that the Universe is moving...

 

Lets say we could take a complete picture of the Universe in 3D, in the picture nothing would be moving, would the picture have a center ?