Meaning of Gravitational Field Outside Uniform Sphere

AI Thread Summary
The discussion centers on the gravitational field outside a uniform sphere of mass M, which is equivalent to that of a point mass M located at the center of the sphere. A participant initially misunderstands the concept by relating it to the gravitational field at the center, where the radius r equals zero, leading to confusion about infinite values. Clarifications reveal that the comparison is only valid for points outside the sphere, where the gravitational field behaves consistently with that of a point mass. The key takeaway is that for any point outside the sphere, the gravitational field can be calculated using the formula g = G(M/r^2), and this holds true regardless of the distance as long as it exceeds the sphere's radius. Ultimately, the participants reach a consensus on the correct interpretation of the gravitational field's behavior outside a uniform sphere.
songoku
Messages
2,467
Reaction score
382

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


g=G\frac{M}{r^2}


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
 
Physics news on Phys.org
songoku said:
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?
 
Hi Doc Al
Doc Al said:
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 :

g=G\frac{M}{R^2}

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
 
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.
 
Last edited:
songoku said:
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 :

g=G\frac{M}{R^2}

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.
 
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 !
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top