Now we take another example of the law ofconservation of energy. Consider an object which initially has kinetic energyand is moving very fast, and which slides against the floor with friction. Itstops. At the start the kinetic energy is not zero, but at the end it iszero; there is work done by the forces, because whenever there is frictionthere is always a component of force in a direction opposite to that of themotion, and so energy is steadily lost. But now let us take a mass on the endof a pivot swinging in a vertical plane in a gravitational field with nofriction. What happens here is different, because when the mass is going up theforce is downward, and when it is coming down, the force is also downward. ThusF · ds has one sign going up and another signcoming down. At each corresponding point of the downward and upward paths thevalues of F · ds are exactly equal in size but ofopposite sign, so the net result of the integral will be zero for this case.Thus the kinetic energy with which the mass comes back to the bottom is thesame as it had when it left; that is the principle of the conservation ofenergy. (Note that when there are friction forces the conservation of energyseems at first sight to be invalid. We have to find another form of energy.It turns out, in fact, that heat is generated in an object when it rubs anotherwith friction, but at the moment we supposedly do not know that.)
现在，我们举另外一个能量守恒规律的例子。考虑一个对象，最初有动能，且移动很快，在地板上滑动，有摩擦力。它停下了。开始时，动能不是零，但最终是零；通过力，做了功，因为，任何时候只要有摩擦力，那么，在与运动方向相反的方向，就总有力的分量，于是，能量就会稳定地消耗。但是，现在，让我们看一个枢轴{？}，在万有引力场中，没有摩擦，在一个垂直的平面上摆动，其终端，有一个质量。这里发生的事情，有些不同，因为，当质量向上时，力是向下的，当质量向上时，力还是向下的。这样，F · ds 有一个符号向上，另一个向下。在向上的路径中的每一个点，与向下的路径中的每一个相应的点，这两个点的F · ds 的值，大小精确相等，但符号相反，于是，对于这个案例，积分的净值为零。这样，质量回到底部时所具有的动能，与它离开时所具有的动能，一样。（注意，当有摩擦力的时候，能量守恒，最初看上去，似乎无效。我们必须去找能量的其他形式。事实上，结果就是，当一个对象摩擦另一个对象时，在前者中，产生了热，但是，在这个时候，我们大概没有注意到这点。）

13–2 Work done by gravity
The next problem to be discussed is muchmore difficult than the above; it has to do with the case when the forces arenot constant, or simply vertical, as they were in the cases we have worked out.We want to consider a planet, for example, moving around the sun, or asatellite in the space around the earth.
下一个要讨论的问题，比上面的，要难的多；它必须处理这样的情况，当力不是常数，或不只是垂直的时候，我们前面处理的是这种简单情况。例如，我们考虑一个行星，绕日运动，或者一个卫星，在空间中绕地球运行。

We shall first consider the motion of an objectwhich starts at some point 1 and falls, say, directly toward the sun or towardthe earth (Fig. 13-2). Will there be a law of conservation of energy in thesecircumstances? The only difference is that in this case, the force is changingas we go along, it is not just a constant. As we know, the force is −GM/r2times the mass m, where m is the mass that moves. Now certainlywhen a body falls toward the earth, the kinetic energy increases as thedistance fallen increases, just as it does when we do not worry about thevariation of force with height. The question is whether it is possible to findanother formula for potential energy different from mgh, a differentfunction of distance away from the earth, so that conservation of energy willstill be true.
首先，我们将考虑的运动，是一个对象，从某个点1开始，且下落，比如直接落向太阳或地球（图13-2）。在这种情形中，会有能量守恒的规律吗？在此案例中，唯一的区别就是，随着我们向前走，力在变化，它不是一个常数。正如我们所知，力就是−GM/r2 乘以质量m，这里m就是那个运动中的质量。现在，当然，当一个物体落向地球时，动能随着落下距离的增加而增加，正如力随着高度的变化那样，我们无须为此担心。问题是，是否可能，为势能找到另一公式，不同于mgh的，一个远离地球的不同的函数，以致能量守恒，仍然成立。

This one-dimensional case is easy to treatbecause we know that the change in the kinetic energy is equal to the integral,from one end of the motion to the other, of −GMm/r2 times thedisplacement dr:
(13.11)
There are no cosines needed for this casebecause the force and the displacement are in the same direction. It is easy tointegrate dr/r2; the result is −1/r, so Eq. (13.11)becomes
(13.12)
Thus we have a different formula forpotential energy. Equation (13.12) tells us that the quantity (mv2/2−GMm/r) calculated at point 1, at point 2, or at any other place, has aconstant value.
这种一维的案例，容易处理，因为我们知道，动能的变化，就等于−GMm/r2 乘以位移dr的积分，从运动的一端到另一端。
(13.11)
对于这种情况，不需要cos，因为力和位移，在同一个方向。积分dr/r2很容易；结果是−1/r，于是方程（13.11）就变为：
(13.12)
这样，对于势能，我们就有了一个不同的方程。方程(13.12)告诉我们，量(mv2/2− GMm/r)，无论是在点1、点2，还是在任何其他地方计算，恒为常数。

We now have the formula for the potentialenergy in a gravitational field for vertical motion. Now we have an interestingproblem. Can we make perpetual motion in a gravitational field? Thegravitational field varies; in different places it is in different directionsand has different strengths. Could we do something like this, using a fixed,frictionless track: start at some point and lift an object out to some otherpoint, then move it around an arc to a third point, then lower it a certaindistance, then move it in at a certain slope and pull it out some other way, sothat when we bring it back to the starting point, a certain amount of workhas been doneby the gravitational force, and the kinetic energy of the object is increased?Can we design the curve so that it comes back moving a little bit faster thanit did before, so that it goes around and around and around, and gives usperpetual motion? Since perpetual motion is impossible, we ought to find outthat this is also impossible. We ought to discover the following proposition:sincethere is no friction the object should come back with neither higher nor lowervelocity—it should be able to keep going around and around any closed path. Statedin another way, the total work done in going around a complete cycle shouldbe zero for gravity forces, because if it is not zero we can get energy outby going around. (If the work turns out to be less than zero, so that we getless speed when we go around one way, then we merely go around the other way, becausethe forces, of course, depend only upon the position, not upon the direction;if one way is plus, the other way would be minus, so unless it is zero we willget perpetual motion by going around either way.)
对于万有引力场中的垂直运动，我们现在就有了其势能的公式。现在，我们有一个非常有趣的问题。在万有引力场中，我们能造成永动吗？万有引力场在变；位置不同，则场的方向不同，强度不同。我们如下能做这种事情吗，用一个固定的、无摩擦的轨道：在某点开始，把一个对象抬到另一个点，然后，让它绕着一个弧，运动到第三个点，然后，把它降低一段距离，然后，让它在一个坡道上移动，且把它往外拉一点{？}，这样，当我们把它带回起点时，万有引力场，就做了一定数量的功，且对象的动能也增加了？我们能设计出这样的曲线吗：让它回来时，比出发时要快一点，这样，它就能来回不断地走，给我们提供永动？由于永动是不可能的，我们应该找出，这也是不可能的。我们应该发现下面的命题：由于没有摩擦，所以对象回来时，矢速既不可能更高，也不可能更低—在任何闭合的路径中，它应该能够不断地走。换一种方式陈述就是，对于一个完全循环的路径，绕着它转圈走，重力所做的功，应该是零，因为如果不是零，那么我们就可以通过转圈走，得到能量。（如果最后的功，比零小，那么就是，当我们按某一个方向转着走一圈时，速度变小了，那么，我们反着方向走，当然是因为，力只依赖于位置，而不依赖于方向；如果一个方向是加，另一方向就应该是减，所以，除非它是零，否则，我们就可以通过选择另一方向转圈，获得永动。）

Fig. 13-3. A closed path in a gravitationalfield. 图13-3 万有引力场中的一条闭合路径。
Is the work really zero? Let us try todemonstrate that it is. First we shall explain more or less why it is zero, andthen we shall examine it a little better mathematically. Suppose that we use asimple path such as that shown in Fig. 13-3, in which a small mass is carriedfrom point 1 to point 2, and then is made to go around a circle to 3, back to4, then to 5, 6, 7, and 8, and finally back to 1. All of the lines are eitherpurely radial or circular, with M as the center. How much work is donein carrying m around this path? Between points 1 and 2, it is GMm timesthe difference of 1/r between these two points:
（13-W12）
From 2 to 3 the force is exactly at rightangles to the curve, so that W23 =0.
The work from 3 to 4 is
（13-W34）
In the same fashion, we find that W45= 0, W56 = GMm(1/r6 − 1/r5),W67 = 0,W78 = GMm(1/r8− 1/r7), and W81 = 0. Thus
（13-W）
But we note that r2 = r3, r4= r5, r6 = r7, and r8 = r1. Therefore W =0.
功真的为零？让我们来演示一下，它确实如此。首先，我们先简单解释一下，为什么它是零，然后，我们将用数学方法，更仔细地考察。假设我们用一个简单的路径，如图13-3，在其中，一个小的质量，被从点1带到点2，然后，绕着圈走，从3到8，最后回到1。所有的线，要么是纯粹径向的，要么是圆弧的，以M为中心。让m走一圈，能做多少功呢？在点1和2之间，功是GMm 乘以这两点间的1/r 的微分：
（13-W12）
从2到3，力与曲线垂直，所以W23 =0.。
从3到4的功是：
（13-W34）
以同样的方式，我们得到 W45 = 0, W56 = GMm(1/r6− 1/r5), W67 = 0,W78 = GMm(1/r8− 1/r7), 及W81 = 0。这样 （13-W）
但我们注意到r2 = r3, r4 = r5, r6 = r7, and r8= r1。因此W = 0。

Fig. 13-4. A “smooth” closed path, showinga magnified segment of it approximated by a series of radial andcircumferential steps, and an enlarged view of one step.图13-4 一个“光滑的”闭合路径，其放大的片段，显示出，通过一系列径向的和环绕的台阶来近似，及一个台阶的增强视图。

Of course we may wonder whether this is tootrivial a curve. What if we use a real curve? Let us try it on a realcurve. First of all, we might like to assert that a real curve could always beimitated sufficiently well by a series of sawtooth jiggles like those of Fig. 13-4,and that therefore, etc., Q.E.D., but without a little analysis, it is notobvious at first that the work done going around even a small triangle is zero.Let us magnify one of the triangles, as shown in Fig. 13-4. Is the work done ingoing from a to b and b to c on a triangle the sameas the work done in going directly from a to c? Suppose that theforce is acting in a certain direction; let us take the triangle such that theside bc is in this direction, just as an example. We also suppose thatthe triangle is so small that the force is essentially constant over the entiretriangle. What is the work done in going from a to c? It is
（13-Wac）
since the force is constant. Now let uscalculate the work done in going around the other two sides of the triangle. Onthe vertical side ab the force is perpendicular to ds, sothat here the work is zero. On the horizontal side bc,
（13-Wbc）
Thus we see that the work done in goingalong the sides of a small triangle is the same as that done going on a slant,because s cos _ is equal to x. We have proved previouslythat the answer is zero for any path composed of a series of notches like thoseof Fig. 13-3, and also that we do the same work if we cut across the cornersinstead of going along the notches (so long as the notches are fine enough, andwe can always make them very fine); therefore, the work done in going aroundany path in a gravitational field is zero.
当然，我们可以琢磨，这条曲线，太琐碎了。如果我们用一条真实的曲线，又会如何呢？让我们取一条真实的曲线看看。首先，我们可能喜欢断言，一条真实的曲线，总是可以通过一系列锯齿状的跳动，而被充分地模仿，如图13-4，因此，证明完毕，但是，首先，没有一点分析的话，甚至绕着一个小的三角形，所做的功都是零这一点，并不明显。让我们放大一个三角形看看，如图13-4所示。在三角形上，从a到b再从b到c所做的功，与从a直接到c所做的功，是否相同？假设力的作用方向，是一定的；让我们这样取三角形，边bc就是在力的方向，正如例子所示。我们还假设，三角形是如此之小，以至于在整个三角形上，力是常数。从a到c所做的功是什么呢？就是：
（13-Wac）
由于力是常数。现在，我们计算，沿着三角形的另外两条边所做的功。在垂直边ab上，力垂直于ds，所以这里的功是零。在水平边bc上：
（13-Wbc）
这样，我们看到，沿着三角形的各边所做的功，与沿着斜边所做的功，相同，因为，s cosθ就等于x。我们前面已经证明了，对于任何包含着若干缺口的路径，如图13-3所示那样，答案都是零，并且，如果我们抄近路，而不是沿着缺口走的话（只要缺口足够光滑，而我们总是可以做到此点），所做功也一样；因此，在一个万有引力场中，绕着任何路径所做的功，都是零。

This is a very remarkable result. It tellsus something we did not previously know about planetary motion. It tells usthat when a planet moves around the sun (without any other objects around, noother forces) it moves in such a manner that the square of the speed at anypoint minus some constants divided by the radius at that point is always thesame at every point on the orbit. For example, the closer the planet is to thesun, the faster it is going, but by how much? By the following amount: ifinstead of letting the planet go around the sun, we were to change thedirection (but not the magnitude) of its velocity and make it move radially,and then we let it fall from some special radius to the radius of interest,thenew speed would be the same as the speed it had in the actual orbit, because thisis just another example of a complicated path. So long as we come back to thesame distance, the kinetic energy will be the same. So, whether the motion is thereal, undisturbed one, or is changed in direction by channels, by frictionless constraints,the kinetic energy with which the planet arrives at a point will be the same.
这个结果，非常值得注意。有些关于行星运动的事情，我们以前不知道，现在它告诉了我们。它告诉我们，当行星绕日运动时（没有任何其他对象在旁边，没有其他的力），其移动方式为，在任何点，速度的平方，减去某些常数除以该点的半径，对于轨道上的任一点，都一样。例如，行星离日越近，走的越快，但是多少呢？通过下面的量：不是让行星绕日运行，我们要改变其矢速的方向（而不是大小），让它在径向运动，然后，我们让它从某些特殊的半径，下落到感兴趣{？}的半径，新的速度，应该与它在实际轨道中的速度一样，因为这不过是另外一个复杂路径的例子。只要我们回到同样的距离，动能就是一样的。所以，无论运动是真实的、不受干扰的，还是方向，已经被轨道{？}、被无摩擦的常数，给改变了，行星到达一个点时所具有的动能，都一样。

Thus, when we make a numerical analysis ofthe motion of the planet in its orbit, as we did earlier, we can check whetheror not we are making appreciable errors by calculating this constant quantity,the energy, at every step, and it should not change. For the orbit of Table 9-2the energy does change,* it changes by some 1.5 percent from thebeginning to the end. Why? Either because for the numerical method we usefinite intervals, or else because we made a slight mistake somewhere inarithmetic.
* The energy per unit mass is (v2x+v2y)/2− 1/r in the units of Table 9-2
这样，对于行星在其轨道中的运动，当我们对此做一个数值分析时，正如我们早前所做那样，我们可以检查，我们是否通过计算这个常数量，即在每一步的能量，而在犯可察觉到的错，该量本不应该变化。对于表9-2的轨道，能量并不变化*，它从始至终，大约改变了1.5%。 为什么？要么，因为对于数值方法，我们使用了有限的间隔，要么，我们在算法的某处，犯了小错。
*在表9-2的单位中，每单位质量的能量是 (v2x+v2y)/2− 1/r 。

13-3 Summation of energy 13-3 能量的总结
Now we go on to the more generalconsideration of what happens when there are large numbers of objects. Supposewe have the complicated problem of many objects, which we label i = 1, 2, 3, .. . , all exerting gravitational pulls on each other. What happens then? Weshall prove that if we add the kinetic energies of all the particles, and addto this the sum, over all pairs of particles, of their mutual gravitationalpotential energy, −GMm/rij , the total is a constant:
现在，我们继续考虑更普遍的情况，当有大量对象时，会发生什么。假设我们有复杂问题，其中包含很多对象，它们被标为i = 1, 2, 3, . . . ,所有存在的万有引力，都相互吸引。那么，会发生什么呢？我们将证明，如果我们把所有粒子的动能，都加起来，并且，每对粒子之间，都有万有引力的势能−GMm/rij，我们把所有粒子对之间的这个势能，都加到前面的动能总和中，那么，总体是一个常数：
(13.14)

How do we prove it? We differentiate eachside with respect to time and get zero. When we differentiate miv2i/2 , we find derivatives of the velocity that are the forces, just as inEq. (13.5). We replace these forces by the law of force that we know fromNewton’s law of gravity and then we notice that what is left is minus the timederivative of
我们如何证明它呢？我们对两边求时间的导数，得到零。当我们对miv2i /2求导时，我们发现，矢速的导数，是力，正如公式（13.5）所示。从牛顿的重力规律中，我们知道了力的规律，我们用这些规律，把这些力替换掉，然后，我们注意到，剩下的就是值 的负的对时间的导数。
The time derivative of the kinetic energyis
动能的时间导数是：
(13.15)
The time derivative of the potential energyis
势能的时间导数是：
（13.15.1）
But
但是
（13.15.2）
so that
于是
（13.15.3）
since rij =−rji, while rij = rji.Thus
由于当 rij = rji 时，rij= −rji,这样
(13.16)
Now we must note carefully what and means.In Eq.(13.15), means that i takes on all values i =1, 2, 3, . . . in turn, and for each value of i ,the index j takes on allvalues except i. Thus if i = 3, j takes on the values 1, 2,4, . . .
现在，我们必须仔细注意， 和 意味着什么。在方程（13.15）中，意味着i依次取所有的值i = 1, 2, 3, . . .，对每一个i的值，下标j取除i之外的所有值。这样，如果i=3，j取的值就是1, 2,4, . . .。

In Eq. (13.16), on the other hand, means
that given values of i and j occur only once. Thus the particle pair 1 and 3contributes only one term to the sum. To keep track of this, we might agree tolet i range over all values 1, 2,3, . . . , and for each i let j range onlyover values greater than i. Thus if i = 3, j could only have values 4, 5, 6, .. . 另一方面，在方程（13.16）中，意味着i和j的被给予的值，只出现一次。这样，1和3这一对粒子组合，对于总和，只贡献一个项。要跟踪这个，我们可以让i取所有1, 2,3, . . .的值，对于每个i，让j只取比i大的值。这样，如果i=3，j就只能取4, 5, 6, . . 。But we notice that for each i, j value there are two contributionsto the sum, one involving vi, and the other vj , and thatthese terms have the same appearance as those of Eq. (13.15), where all valuesof i and j (except i = j) are included in the sum.但是，我们注意到，对于每个i，j值，对总和有两个贡献，一个包含vi，另一个包含vj ，且这些项，与方程（13.15）中的那些项，外观一样，在那里，所有i和j的值（除i=j外），都被包含在总和中。Therefore, by matching the terms one by one, we see that Eqs. (13.16)and (13.15) are precisely the same, but of opposite sign, so that the timederivative of the kinetic plus potential energy is indeed zero. 这样，通过逐个比较各项，我们看到方程（13.16）和（13.15）精确地相同，只是符号相反，所以，动能加势能的时间导数，确实就是零。Thus we see that, for many objects, the kinetic energy is the sum ofthe contributions from each individual object, and that the potential energy isalso simple, it being also just a sum of contributions, the energies between allthe pairs.这样，我们看到，对于很多对象，动能就是每个独立对象的贡献的总和，而势能也简单，它是所有每一对对象组合对势能的贡献的总和。 We can understand why it should be the energy of every pair this way:Suppose that we want to find the total amount of work that must be done tobring the objects to certain distances from each other. We may do this in severalsteps, bringing them in from infinity where there is no force, one by one. 我们可以理解，为什么它是每一对组合的能量这种方式：假设我们想找出，为了把对象，带到相互间有一定的距离，需要做的总的功。我们可以按这样的步骤做，把它们从没有力的无限远处，一个一个地带进来。First we bring in number one, which requires no work, since no otherobjects are yet present to exert force on it. Next we bring in number two,which does take some work, namely W12 = −Gm1m2/r12.Now, and this is an important point, suppose we bring in the next object toposition three.首先，我们带进第一个，这不需要功，因为，其他对象，都不在场，对它没有作用力。然后，我们带进第二个，这需要一些功，即W12 = −Gm1m2/r12 。现在，到了关键点，假设我们把下一个对象，带到位置3。At any moment the forceon number 3 can be written as the sum of two forces—the force exerted by number1 and that exerted by number 2. Therefore the work done is the sum of the worksdone by each, because if F3 can be resolved into the sum of twoforces, 在任何时候，作用于第三个的力，都可写作两个力的和，即第一个和第二个作用的力。因此，所做的功，就是每个所做的功的和，因为，如果F3可以被分解成两个力的和：
F3 = F13 + F23,
then the work is
那么功就是：
(13.F3ds)
That is, the work done is the sum of thework done against the first force and the second force, as if each actedindependently. Proceeding in this way, we see that the total work required toassemble the given configuration of objects is precisely the value given in Eq.(13.14) as the potential energy. It is because gravity obeys the principle ofsuperposition of forces that we can write the potential energy as a sum overeach pair of particles.
也就是说，所做的功，就是对第一个力和第二个力的功的总和，如果每个都是独立起作用的话。以这种方式处理，我们看到，对于被给予的对象配置，集成它们所需要的总的功，正是方程（13.14）所给的值。这是因为，重力遵循力的叠加原理，这样，我们就可以把势能写作：每一对粒子间势能的总和。

13-4 Gravitational field of largeobjects 13-4 大的对象的万有引力场
Now we shall calculate the fields which aremet in a few physical circumstances involving distributions of mass. We havenot so far considered distributions of mass, only particles, so it isinteresting to calculate the forces when they are produced by more than justone particle. First we shall find the gravitational force on a mass that isproduced by a plane sheet of material, infinite in extent.The force on a unitmass at a given point P, produced by this sheet of material (Fig. 13-5), willof course be directed toward the sheet. Let the distance of the point from thesheet be a, and let the amount of mass per unit area of this huge sheet be μ. Weshall suppose μ to be constant; it is a uniform sheet of material. Now, whatsmall field dC is produced by the mass dm lying between ρand ρ+dρ from the point O of the sheetnearest point P? Answer: dC = −G(dmr/r3). But this field is directedalong r, and we know that only the x-component of it will remain when we addall the little vector dC’s to produce C. The x-component of dC is
（13.dcx）
Now all masses dm which are at the same distancer from P will yield the same dCx, so we may at once write for dm thetotal mass in the ring between ρand ρ+ dρ, namely dm = μ2πρdρ (2πρdρ is the area of a ring ofradius ρ and width dρ, if dρ<<ρ). Thus
（13.dcx2）
在有些物理情形中，我们遇到的场，会牵扯到质量的分布，现在，我们将计算这些场。至今，我们只考虑了粒子，而未考虑质量的分布，所以，当力不是由一个粒子、而是由更多粒子产生时，计算它们，也很有趣。首先，我们将找出，由一个材料平板，在一个质量上，所产生的万有引力，平板的扩展是无限的。在一个单位质量上，在给定点P，由这个材料平板，所产生的力（图13-5），当然，被指向此平板。让从点到板的距离为a，这个巨大平板的单位面积质量为μ。我们假定μ是常数；它是质量均匀的材料板。现在，O是板上距点P最近的一点，ρ为板上距点O的距离，在ρ和ρ+dρ之间的质量为dm，那么，由dm所产生的小的场dC，是什么？但是，这个场方向，沿着r，且我们知道，当我们把所有小的矢量dC们，都加起来，以产生C时，只有其x分量，会被保留。dC的x分量就是：
（13.dcx）
现在，与P的距离都为r的所有质量dm，将产生同样的dCx ，于是，我们可以立即为dm写出，在环ρ和ρ+ dρ之间的总质量，即dm = μ2πρdρ (如果dρ<<ρ，那么2πρdρ，就是半径为ρ，宽度为dρ的环的面积)。这样：
（ 13.dcx2）

Fig. 13-5. The gravitational field C at amass point produced by an infinite plane sheet of matter.图13-5由一个无限的材料平板，在一个质量上的一点，所产生的万有引力场C。

Then, since r2 = ρ2 + a2,ρdρ= r dr. Therefore,
(13.17)
Thus the force is independent of distance a!Why? Have we made a mistake?One might think that the farther away we go, theweaker the force would be. But no! If we are close, most of the matter ispulling at an unfavorable angle; if we are far away, more of the matter issituated more favorably to exert a pull toward the plane. At any distance, thematter which is most effective lies in a certain cone. When we are farther awaythe force is smaller by the inverse square, but in the same cone, in the sameangle, there is much more matter, larger by just the square of the distance! Thisanalysis can be made rigorous by just noticing that the differentialcontribution in any given cone is in fact independent of the distance, becauseof the reciprocal variation of the strength of the force from a given mass, andthe amount of mass included in the cone, with changing distance.The force isnot really constant of course, because when we go on the other side of thesheet it is reversed in sign.
因此，由于 r2 = ρ2 + a2,ρdρ= r dr. 所以：
(13.17)
这样，力就是独立于距离a的！为什么? 我们是否犯了错？ 有人可能会想，我们走的越远，力就会越弱。但是，非也！如果我们离得近，大多数物质以一个不利的角度吸引平板；如果我们离得远，大多数物质吸引时，位置更合适。在任何距离，最有效的物质，都处于某个圆锥体中。当我们比较远时，由于反比于距离平方，力比较小，但在同一个圆锥中，在同一个角度，则有更多的物质，比距离的平方要大。在任何给予的圆锥中，积分得到贡献，事实上是独立于距离的，因为，来自于一个被给予的物质的力的强度，与一个圆锥中所包含的物质的量、和变化的距离有关，而后两者的关系，则是此消彼长；注意到这一点，可以让这个分析，变得严格。当然，力并不真正就是常数，因为，当我们来到板的另一侧时，符号就相反了。