Gravitational field in a hollow sphere

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The gravitational field inside a uniform hollow sphere is zero due to the symmetrical distribution of mass. According to Gauss's law for gravity, any point inside the sphere experiences equal gravitational forces from opposing mass portions, which cancel each other out. This cancellation occurs because the gravitational force is inversely proportional to the square of the distance from the mass, leading to equal but opposite forces at any point inside the sphere. Visualizing this with cones extending from a point inside the sphere helps clarify that the forces from the surface areas do not result in a net gravitational pull. Therefore, the total gravitational force at any point within the hollow sphere is zero.
maddys_daddy
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Can someone please give me a qualitative justification for the gravitational field inside a uniform hollow sphere being zero? I'm having a lot of trouble understanding this. Prof. said (in class) not to worry about the higher order polynomials involved, just be concerned with it "qualitatively" at this point. I'm lost.
Thanks,
m_d
 
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maddys_daddy said:
Can someone please give me a qualitative justification for the gravitational field inside a uniform hollow sphere being zero? I'm having a lot of trouble understanding this. Prof. said (in class) not to worry about the higher order polynomials involved, just be concerned with it "qualitatively" at this point. I'm lost.
Thanks,
m_d

You did not indicate what level of physics you already have. So, I have no idea if what I will say here makes any sense to you.

There is a Gauss's law equivalent for gravitational field. So if you have seen gauss's law applied to electrostatic, you should understand the identical principle applied to gravitational field. So apply that.

http://scienceworld.wolfram.com/physics/GausssLaw.html

Zz.
 
Qualitatively, think of it like this: At any given point inside the sphere, there is x amount of mass to the right in the shape of a bowl, and 1-x amount of mass to the left in the shape of a shperical shell missing a bowl shaped cap. These two complementary portions have centers of mass at let's say rx and r1-x. Then, qualitatively, you can imagine that rx < r1-x by just such an amount that:

x/rx2 = (1-x)/r1-x2.

It would probably help to draw a picture.

Qualitatively, you're supposed to realize that Fgrav is larger for larger amounts of mass and smaller at larger distances away from the mass. So, qualitatively, you can imagine that the effect of more mass (1-x) is canceled by the effect of further away (r1-x).
 

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turin,
Thanks, I think I can understand that. However, when I click on the link to your .gif, I get a "you need to login to view this" message--well, I am! Oh, well, I'll take that up with the admins.
ZapperZ, FWIW, I'm in first year college physics for engineers (PHYS141 @ UoArizona). Next semester is my Electricity and Magnetism. So I have no idea what Gauss' law is, but I guess I'll find out.
Thanks for the help!
 
From any point inside the sphere, imagine a cone extending to a portion of the surface. Now extend that cone back to the opposite side of the sphere. If your point is not exactly at the center, your two cones will not intersect equal areas (and, so, not equal masses) because area is proportional to the square of the distance. HOWEVER, since gravitational force is inversely proportional to the square of the distance, the gravitational force from each of those portions of the surface will be the same. Since they are in opposite directions, the two equal but opposite forces cancel. The total gravitational force is 0.

(The same thing is true of magnetic force as well.)
 
I like HallsofIvy's explanation better than mine. It more clearly incorporates the inverse square dependence.
 
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