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Author Topic: Would length contraction increase surface gravitational field strength?  (Read 4660 times)

Offline jeffreyH

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It's all in the question apart from the fact that we are talking about the direction of motion or the direction of the gravitational field when considering the contraction. Meaning that the assumed increase will be in the same direction as the motion or the field.


 

Offline PmbPhy

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Quote from: jeffreyH
It's all in the question apart from the fact that we are talking about the direction of motion or the direction of the gravitational field when considering the contraction. Meaning that the assumed increase will be in the same direction as the motion or the field.
I don't understand this question. Length contraction of what? Why do you think it should? Since mass is the source of gravity what mass are you talking about?
 

Offline jeffreyH

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Consider this Pete. Take a flat 1mm thick disk with the same mass as the earth. Place an object perpendicular to the plane of the disk at the centre of gravity and at a distance that is equal to the radius of the earth. Now what is the value of g coincident with this object? Is it 9.80665 m/s^2?
 

Offline PmbPhy

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Quote from: jeffreyH
Consider this Pete. Take a flat 1mm thick disk with the same mass as the earth. Place an object perpendicular to the plane of the disk at the centre of gravity and at a distance that is equal to the radius of the earth. Now what is the value of g coincident with this object? Is it 9.80665 m/s^2?
I don't know. I'd have to calculate it. Would you like me to do this or do you know calculus well enough to do it yourself? If I do it I can do it up very nice and save me work in a PDF file and upload it to my website for you to download if you'd like. Or do you simply want the answer?
 

Offline jeffreyH

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Quote from: jeffreyH
Consider this Pete. Take a flat 1mm thick disk with the same mass as the earth. Place an object perpendicular to the plane of the disk at the centre of gravity and at a distance that is equal to the radius of the earth. Now what is the value of g coincident with this object? Is it 9.80665 m/s^2?
I don't know. I'd have to calculate it. Would you like me to do this or do you know calculus well enough to do it yourself? If I do it I can do it up very nice and save me work in a PDF file and upload it to my website for you to download if you'd like. Or do you simply want the answer?

I would actually be very interested in the calculus. I'm sure others would be interested too. Thanks for the offer.
 

Offline PmbPhy

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Quote from: jeffreyH
I would actually be very interested in the calculus. I'm sure others would be interested too. Thanks for the offer.
I forgot to mention that this problem is fairly intractable because in general it can't be done in closed form. At least not easily that I'm aware of. However if the disk is in the xy-plain and the center of the disk is at the origin of the coordinate system then I can find an expression for the gravitational field along the z-axis fairly easy so that's what I'll do. When I have the chance that is.
 

Offline PmbPhy

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Quote from: jeffreyH
I would actually be very interested in the calculus. I'm sure others would be interested too. Thanks for the offer.
Dear Jeff,

I've decided to work on this tomorrow. For now I'll say this. As I said above, to my knowledge this can't be done in closed form in general so I'll find the field along an axis normal the plane of the disk passing through the center. Since the solution will have the same form if the disk had a uniform charge on it then the electric field will have the same functional form as the gravitational field. The functional form of the disk is

3bd4551699eae0f5e1fd0fb30cb976c2.gif

So when I post the one for the gravitational field expect it to have the same form.
 

Offline jeffreyH

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I have saved the equation and I'll work through it. I will have questions! It is an interesting exercise. I look forward to your formula for the gravitational field.
 

Offline PmbPhy

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« Last Edit: 03/12/2014 03:15:05 by PmbPhy »
 

Offline jeffreyH

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Thanks Pete that helped enormously.
 

Offline PmbPhy

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Thanks Pete that helped enormously.
You're welcome. Do you want me to work out the math myself or can it be found in those files?
 

Offline jeffreyH

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I'm going to try it but it won't be for a while until I have sorted out some other things.
 

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