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> Spherical mass can be replaced with a point mass I'm not a physicist, but that doesn't sound right. One example is that if you are inside of the sphere, at l
by asdfasgasdgasdg 6y ago
> Spherical mass can be replaced with a point mass
I'm not a physicist, but that doesn't sound right. One example is that if you are inside of the sphere, at least some of the mass will be pulling you away from the point at its center. I think what you mean is that a point mass closely approximates a spherical mass, but the degree to which that is true becomes less and less the closer you get to the center. I don't think you have successfully contradicted what the person you responded to said.
- klodolph 6y agoIn Newtonian physics, a sphere and point mass are exactly interchangeable as long as you are not inside the sphere. If you are outside the sphere, the equivalence is exact, regardless of distance. Proving this is a classic problem in undergraduate physics.
- tedunangst 6y agoWe are inside the atmosphere at least...
- thewakalix 6y agoAnd that brings fluid dynamics into the picture, which produces discrepancies orders of magnitude greater than the parabolic/elliptical/etc. distinction!
- asdfasgasdgasdg 6y agoThanks! Shame I never took undergraduate physics but now that you've rung my bell I think we may have discussed this in high school. What an unintuitive result that being even a meter under ground breaks what is, up to that point, a fine model.
- klodolph 6y agoThe way the result works is that if you are a meter underground, it is equivalent to standing on top of a sphere with the top meter of mass removed. Or equivalently, if you are inside a shell (sphere outside, hollow sphere inside) then the gravitational effect is zero. This can be explained by analogy with light which also follows the inverse square law--changing your position inside a hollow sphere does not change the fact that you see the sphere all around you. Same reason that there is no electrical field inside a hollow conductive object.
- scotty79 6y agoCheck out gauss law for electromagnetism and gravity. It says that total flux through closed surface is proportional to strength of field sources inside the surface. Flux from outside sources cancels out. You can use this law to see what's the gravity fields as you move under ground and in other very symmetrical cases.
- ChrisLomont 6y ago>What an unintuitive result that being even a meter under ground breaks what is, up to that point, a fine model. It doesn't break it at all. The meter above you can be treated as a hollow shell, which surprisingly has zero net pull, and the solid sphere below can be treated as a point mass just as before. Just remember these two facts, each provable with a simple integral calculation, usually done in high school physics or freshman college physics: a uniform sphere has the same gravitational pull on an object as a point mass at it's center, and the net gravitational pull on an object inside a spherical shell is zero. This all works under perfect spheres, uniform (at the spherical shell level at least) density... There are other cases it works, but this simple case is the basic idea.
- hugh-avherald 6y agohttps://en.wikipedia.org/wiki/Shell_theorem https://en.wikipedia.org/wiki/Shell_theorem
- sudosysgen 6y agoThe intuition we came up with when we had to solve this issue in undergrad physics was interesting. For each point of any given distance, you can find a disk of points at the plane of the same distance whose gravitational pull will be equivalent to a single point at their center. You do this for all distances x, and you will find an equivalent rod going from the attracted objected to the center of the sphere, of a non-uniform but symmetrical density. We find that the density is linearly correlated to the area of the corresponding section, and we then take the closer half of the rod, multiply the density by 1/r^2, and find that it is constant! For the far half of the rod, we cannot do this, so instead we divide it into two halves of equal pull, then divide those to halves and so on, and find that it this reduces to the equivalent of a point. Now that we have two points in-line with the object, we find the point with the same total mass that exerts and equivalent force, and lo and behold it is at the center. I'm sure there is a much simpler way to intuit it, though :)
- pdonis 6y ago> if you are inside of the sphere, at least some of the mass will be pulling you away from the point at its center If the mass is spherically symmetric, this will not be the case; all of the Newtonian forces from the masses further away from the center than you are will cancel out. This is called the "shell theorem", and it turns out to hold even in General Relativity.
- willis936 6y agoI don’t think this is correct. I think the shell theorem says that the gravitational forces cancel if you are on the inside of a hollow sphere and all mass is on the surface. A perfect sphere of uniform density would not meet the shell theorem assumptions.
- srtjstjsj 6y agoYes, it would. A solid ball (filled sphere) is just a union of many shells (hollow spheres), so the theorem still applies. en.wikipedia.org/wiki/Shell_theorem
- deleted 6y ago[deleted]
- pdonis 6y ago> I think the shell theorem says that the gravitational forces cancel if you are on the inside of a hollow sphere and all mass is on the surface. No, it's stronger than that. It says that any spherically symmetric distribution of matter outside a certain radius exerts no "gravitational force" on anything inside that radius, whether there is matter inside that radius or not.
- willis936 6y agoWhat if my distance to one point on the shell was zero? Would I not feel infinite acceleration towards the massive point on the shell that I was infinitely close to?