Gravitational field

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  • #1
songoku
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Homework Statement



I found this statement from my book :
For points outside a uniform sphere of mass M, the gravitational fields is the same as that of a point mass M at the center of the sphere.
My question : what is the meaning of it?

Homework Equations


[tex]g=G\frac{M}{r^2}[/tex]


The Attempt at a Solution


I don't think it will be the same.
At the center of the sphere, r = 0 and g will be infinite ??
And for point outside a uniform sphere, for a certain value of r, the point will have certain value of g, so how can they be the same?

Thanks
 

Answers and Replies

  • #2
Doc Al
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At the center of the sphere, r = 0 and g will be infinite ??
They are talking about outside a uniform sphere of mass M. What's that got to do with r = 0? (At r=0, M=0 also. So the division will be undefined.)
And for point outside a uniform sphere, for a certain value of r, the point will have certain value of g, so how can they be the same?
Why don't you figure it out and see? What's the value of g immediately above the surface of that uniform sphere of mass M and radius R. What's the g value at a distance R from a point mass M?
 
  • #3
songoku
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Hi Doc Al
They are talking about outside a uniform sphere of mass M. What's that got to do with r = 0? (At r=0, M=0 also. So the division will be undefined.)
I interpret the statement is about comparison between gravitational field of point outside the sphere and the point at center of the sphere, that's why I tried to find g at r = 0. Maybe I am wrong but I keep thinking like that based on the question I read. Am I wong?

Why don't you figure it out and see? What's the value of g immediately above the surface of that uniform sphere of mass M and radius R. What's the g value at a distance R from a point mass M?

The value of immediately above the surface of that uniform sphere of mass M and radius R :

[tex]g=G\frac{M}{R^2}[/tex]

The g value at a distance R from a point mass M is the same as above.

But outside the sphere can be the point located 2R from the mass M, so the value of g will be different.

Thanks
 
  • #4
Gear300
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Usually in physics we deal with point particles. What they're saying is that if you're taking into account a gravitational field from a body of mass, such as Earth's, you're taking into account a bunch of point masses...so the question is what is Earth's gravitational field? For uniform spheres of mass, it turns out that it is the same as a point particle (except in cases when you're inside the sphere). The Earth isn't a uniform sphere, but this perspective in general does help.
 
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  • #5
Doc Al
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I interpret the statement is about comparison between gravitational field of point outside the sphere and the point at center of the sphere, that's why I tried to find g at r = 0. Maybe I am wrong but I keep thinking like that based on the question I read. Am I wong?
Yes, I'd say you are interpreting it incorrectly. They are comparing the field at a distance r from the center of a uniform sphere of mass M (as long as r > radius of the sphere) with the field at a distance r from a point mass M. The field is the same for both. At no point are you comparing anything at r = 0.


The value of immediately above the surface of that uniform sphere of mass M and radius R :

[tex]g=G\frac{M}{R^2}[/tex]

The g value at a distance R from a point mass M is the same as above.

But outside the sphere can be the point located 2R from the mass M, so the value of g will be different.
Exactly.
 
  • #6
songoku
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Hi Doc Al and Gear300

Ahh, I see what you mean, Doc Al. And now I can also see what the statement really means.

Thanks a lot !
 

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