Simple Nature - Light and Matter

the apple. A kilogram of dirt a few feet under his garden in Englandwould interact much more strongly with the apple than a kilogramof molten rock deep under Australia, thousands of miles away. Also,we know that the earth has some parts that are more dense, andsome parts that are less dense. The solid crust, on which we live,is considerably less dense than the molten rock on which it floats.By all rights, the computation of the total gravitational energy ofthe apple should be a horrendous mess. Surprisingly, it turns out tobe fairly simple in the end. First, we note that although the earthdoesn’t have the same density throughout, it does have sphericalsymmetry: if we imagine dividing it up into thin concentric shells,the density of each shell is uniform.Second, it turns out that a uniform spherical shell interacts withexternal masses as if all its mass were concentrated at its center.The shell theorem: The gravitational energy of a uniform sphericalshell of mass M interacting with a pointlike mass m outside itequals −GMm/s, where s is the center-to-center distance. If massm is inside the shell, then the energy is constant, i.e., the shell’sinterior gravitational field is zero.Proof: Let b be the radius of the shell, h its thickness, and ρits density. Its volume is then V =(area)(thickness)=4πb 2 h, andits mass is M = ρV = 4πρb 2 h. The strategy is to divide theshell up into rings as shown in figure j, with each ring extendingfrom θ to θ + dθ. Since the ring is infinitesimally skinny, its entiremass lies at the same distance, r, from mass m. The width ofsuch a ring is found by the definition of radian measure to be w =b dθ, and its mass is dM = (ρ)(circumference)(thickness)(width)=(ρ)(2πb sin θ)(h)(b dθ)=2πρb 2 h sin θ dθ. The gravitational energy ofthe ring interacting with mass m is thereforeGm dMdU = −r= −2πGρb 2 hmsin θ dθrIntegrating both sides, we find the total gravitational energy of theshell:∫ πU = −2πGρb 2 sin θ dθhm0 rThe integral has a mixture of the variables r and θ, which are relatedby the law of cosines,r 2 = b 2 + s 2 − 2bs cos θ ,and to evaluate the integral, we need to get everything in terms of eitherr and dr or θ and dθ. The relationship between the differentialsis found by differentiating the law of cosines,2r dr = 2bs sin θ dθ ,.j / A spherical shell of massM interacts with a pointlike massm.Section 2.3 Gravitational Phenomena 103