4 ms·

Nope. It's not immediately obvious though. Let the radius of the Earth be R. Suppose that you are somewhere between the center of the Earth and the Earth's sur

by jules 15y ago

Nope. It's not immediately obvious though. Let the radius of the Earth be R. Suppose that you are somewhere between the center of the Earth and the Earth's surface, say at x meters from the center. Then you can divide the Earth into two pieces: one smaller ball of radius x below you, and a shell with inner radius x and outer radius R "above you". If you are inside or at the inner surface of a spherical shell, then the gravity from all the atoms in the shell is pulling at you in all directions. Now it so happens that all these little pulls exactly cancel each other out. This can be shown with Gauss' law. [Does anybody know an intuitive proof of this?] So effectively only the inner ball of radius x counts. Now the question is whether this ball has bigger or smaller gravity than standing on the Earth's surface. It's not completely clear that the gravity is smaller because the density of the smaller ball could be different. If we assume that the density of the Earth is roughly uniform however, then it is clear that the gravity inside the Earth is lower than on its surface. The precise condition for the gravity being lower at height x is that x/R < a/b where a is the average density of the of the Earth and b is the average density of the inner ball.

iwwr 15y ago
For reference, Earth's interior gravitational acceleration: http://en.wikipedia.org/wiki/File:EarthGravityPREM.jpg http://en.wikipedia.org/wiki/File:EarthGravityPREM.jpg As you see, gravity gets stronger up to a maximum in the outer core, because the core is much more dense. But the issue here is not about spherical Earths at all. Earth deviates significantly from a spheroid and even an ellipsoid shape-wise and gravity-wise. Just saying "location X lies closer to the center of the earth and therefore experiences less gravity" is somewhat irrelevant here.
VladRussian 15y ago
>If you are inside or at the inner surface of a spherical shell, then the gravity from all the atoms in the shell is pulling at you in all directions. Now it so happens that all these little pulls exactly cancel each other out. This can be shown with Gauss' law. [Does anybody know an intuitive proof of this?] imagine rays through you position that form narrow cones to the closest and the farthest point of the shell - if you're 2 times closer to one side than to another, they'd cut 4 times less (proportional to distance squared) of the shell on the closer side than on the farthest. The gravitational pull (reverse of the distance squared) per unit mass is 4 times stronger from the pieces on the closer side. Thus 4 times stronger pull per unit mass from 4 times less mass - the same. (for complete mathematical correctness now make the cones infinitesimally narrow, slice the shell into infinitesimally thin shells and integrate over the 2 angles [pie x pie] to cover the surface of the shell and over the thickness of the shell to cover all the layers - each such infinitesimal would be 0 as described above and integrating 0 would produce 0 in this case. For a general Gauss law it is based on the same distance squared proportionality of the surface element size and reverse distance squared proportionality of the field gradient value (ie. it works only for such fields) )

by jules 15y ago

iwwr 15y ago

For reference, Earth's interior gravitational acceleration: http://en.wikipedia.org/wiki/File:EarthGravityPREM.jpg http://en.wikipedia.org/wiki/File:EarthGravityPREM.jpg As you see, gravity gets stronger up to a maximum in the outer core, because the core is much more dense. But the issue here is not about spherical Earths at all. Earth deviates significantly from a spheroid and even an ellipsoid shape-wise and gravity-wise. Just saying "location X lies closer to the center of the earth and therefore experiences less gravity" is somewhat irrelevant here.

VladRussian 15y ago

>If you are inside or at the inner surface of a spherical shell, then the gravity from all the atoms in the shell is pulling at you in all directions. Now it so happens that all these little pulls exactly cancel each other out. This can be shown with Gauss' law. [Does anybody know an intuitive proof of this?] imagine rays through you position that form narrow cones to the closest and the farthest point of the shell - if you're 2 times closer to one side than to another, they'd cut 4 times less (proportional to distance squared) of the shell on the closer side than on the farthest. The gravitational pull (reverse of the distance squared) per unit mass is 4 times stronger from the pieces on the closer side. Thus 4 times stronger pull per unit mass from 4 times less mass - the same. (for complete mathematical correctness now make the cones infinitesimally narrow, slice the shell into infinitesimally thin shells and integrate over the 2 angles [pie x pie] to cover the surface of the shell and over the thickness of the shell to cover all the layers - each such infinitesimal would be 0 as described above and integrating 0 would produce 0 in this case. For a general Gauss law it is based on the same distance squared proportionality of the surface element size and reverse distance squared proportionality of the field gradient value (ie. it works only for such fields) )