
Fig. 13-6. A thin spherical shell of massor charge.图3-6 一个薄的球形的质量壳或电壳。
We now demonstrate the correctness of thismiracle. In order to do so,however, we shall consider a thin uniform hollow shellinstead of the whole earth. Let the total mass of the shell be m, and let uscalculate the potential energy of a particle of mass m’ a distance R away fromthe center of the sphere (Fig. 13-6) and show that the potential energy is thesame as it would be if the mass m were a point at the center. (The potentialenergy is easier to work with than is the field because we do not have to worryabout angles, we merely add the potential energies of all the pieces of mass.)If we call x the distance of a certain plane section from the center, then allthe mass that is in a slice dx is at the same distance r from P, and thepotential energy due to this ring is −Gm’ dm/r. How much mass is in thesmall slice dx? An amount
(13.17.1)
where μ = m/4πa2 is the surface density of mass onthe spherical shell. (It is a general rule that the area of a zone of a sphereis proportional to its axial width.)Therefore the potential energy due to dm is
(13.17.2)
But we see that
r2 = y2 + (R − x)2 = y2 +x2 + R2 − 2Rx
= a2 + R2 − 2Rx.
Thus
2r dr = −2Rdx
Or
dx/r= −dr/R.
Therefore,
dW = Gm’2πaμ dr / R ,
and so
(13.18)
Thus, for a thin spherical shell, the potentialenergy of a mass m’, external to the shell, is the same as though the mass ofthe shell were concentrated at its center. The earth can be imagined as a seriesof spherical shells, each one of which contributes an energy which depends onlyon its mass and the distance from its center to the particle; adding them alltogether we get the total mass, and therefore the earth acts as though all thematerial were at the center!
我们现在演示这个奇迹的正确性。然而,为了这么做,我们将考虑一个薄的、均匀的空壳,而不是整个地球。设壳的总质量为m,另有质量为m’的粒子,距球心的距离为R(图13-6),让我们计算其势能,并指出,这个势能,与质量m全集中在球心时的势能,是一样的。(势能比场好处理,因为我们不用担心角度,只需把所有小的质量块的势能,加在一起就行。)如果我们称x,为某个平面到中心的距离,那么,在切片dx上的所有质量,到P的距离,都为r,可归于这个环的势能就是−Gm’ dm/r。小的切片dx上,有多少质量呢?一个量:

(13.17.1)
这里,是球壳表面的质量密度。(一个球形地带的面积,正比于它的轴线的宽度,这是一条普遍规则。)因此,归于dm的势能就是:

(13.17.2)
但我们可以看到
r2 = y2 + (R − x)2 = y2 +x2 + R2 − 2Rx
= a2 + R2 − 2Rx.
这样
2r dr = −2Rdx
或者
dx/r= −dr/R.
因此
dW = Gm’2πaμ dr / R ,
于是

(13.18)
这样,对于一个薄的球形壳,一个处于壳外的质量m’,其势能,与壳的质量全集中在中心时的势能,是一样的。地球可以被想象为一系列球形壳,每一个都贡献一份能量,此能量,只依赖于该壳的质量,及其中心与粒子的距离;把这些质量全加起来,我们得到总质量,因此,地球的表现,就与其所有材料都在中心,是一样的。