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1s also simple, it being also just a sum of contributions, the energies between
all the pairs. We can understand œh¿ it should be the energy of every pair this
way: Suppose that we want to fnd the total amount of work that must be done
to bring the objects to certain distances from each other. We may do this in
several steps, bringing them in from infinity where there is no force, one by one.
First we bring in number one, which requires no work, since no other objects
are yet present to exert$ force on i§. Next we bring in number two, which does
take some work, namely W/1a = —Œmmyma/r+s. NÑow, and thìs is an important
point, suppose we bring in the next object to position three. Ất any moment the
force on number 3 can be written as the sum of two forces—the force exerted by
number 1 and that exerted by number 2. 'Therefore fhe tuork done is the sum oƒ
the tuorks done bụ cach, because 1Ÿ F'z can be resolved into the sum of two forces,
tạ = Fla + F›a,
then the work is
[Fi-dẽ= Í Fúycdst | Ea cds= Ha ti
That is, the work done is the sum of the work done against the fñrst force and the
second force, as if each acted independently. Proceeding in this way, we see that
the total work required to assemble the given confguration of objects is precisely
the value given in Eq. (13.14) as the potential energy. It is because gravity obeys
--- Trang 263 ---
the prineiple of superposition of forces that we can write the potential energy as
a sum over each pair of particles.
13-4 Gravitational ñeld of large objects
Now we shall calculate the fñelds which are met in a few physical circumnstances
involving đistributions oƒ mass. We have not so far considered distributions of
mass, only particles, so it is interesting to calculate the forces when they are
produced by more than just one particle. Pirst we shall fnd the gravitational
force on a mass that is produced by a plane sheet of material, infñnite in extent.
'The force on a unit mass at a given point , produced by this sheet of material
(Fig. 13-5), will of course be directed toward the sheet. Leb the disbance of the
point from the sheet be ø, and let the amount of mass per unit area of this huge
sheet be u. We shall suppose / to be constant; it is a uniform sheet of material.
Now, what small fñeld đŒ is produced by the mass đm lying between ø and ø+ đo
from the point Ó of the sheet nearest point: P? Answer: đŒ = —GŒ(dmr/r3). But
this field ¡is directed along ?, and we know that only the z-component of it will
remain when we add all the little vector đŒ”s to produce Œ. “The z-component
Of dC is
dŒ, =—G = =-G CHỊ
Now all masses đi. which are at the same distance r from will yield the
same đŒ„, so we may at once write for đmn the total mass in the ring between /ø
and ø + đo, namely đừn = u2mp dp (27p dp 1s the area oŸ a rỉng oŸ radius ø and
width đø, if đo < ø). Thus
đŒy = —GMu27p ng,
~IdPF— ø —IO
dm ` a
Fig. 13-5. The gravitational field C at a mass point produced by an
Iinfinite plane sheet of matter.
--- Trang 264 ---
Then, since rŸ = øŸ + a”, odo =rdr. Therefore,
Œy = —>nGua | là = ->nGua(^ — —) = —27GU. (13.17)
“ T a ®°
Thus the force is independent of distance al! Why? Have we made a mistake?
One might think that the farther away we go, the weaker the force would be. But
nol TỶ we are close, most of the matter is pulling at an unfavorable angle; if we
are far away, more of the matter is situated more favorably to exert a pull toward
the plane. At any distance, the matter which is most efective lies in a certain
cone. When we are farther away the force is smaller by the inverse square, but
in the same cone, in the same angle, there 1s much rnore matter, larger by just
the square of the distancel “This analysis can be made rigorous by just noticing
that the diferential contribution in any given cone is in fact independent of the
distance, because of the reciprocal variation of the strength of the force from a
given mass, and the amount oŸ mass included in the cone, with changing distance.
The force is not really constant of course, because when we go on the other side
of the sheet it is reversed in sign.
We have also, in effect, solved an electrical problem: if we have an electrically
chargcd plate, with an amount ø of charge per unit area, then the electric feld
at a poinÈ outside the sheet is equal to ø/2eo, and is in the outward direction If
the sheet is positively charged, and inward ïf the sheet is negatively charged. To
prove this, we merely note that —G, for gravity, plays the same role as 1/47o
for electricity.
Now suppose that we have two plates, with a positive charge +ơ on one and
a negative charge —ơ on another at a distance Ù from the frst. What is the
fñeld? Outside the two plates it is zero. Why? Because one attracts and the other
repels, the force being ?ndependent oƒ đistance, so that the two balanece outl Also,
the fñeld befteen the two plates is clearly twice as great as that from one plate,
namely # = ø/co, and is directed from the positive plate to the negative one.
Now we come to a most interesting and important problem, whose solution
we have been assuming all the time, namely, that the force produced by the earth
at a point on the surface or outside it is the same as if all the mass of the earth
were located at its center. The validity of this assumption is not obvious, because
when we are close, some of the mass is very close to us, and some is farther away,
and so on. When we add the efects all together, it seems a miracle that the net
force is exactly the same as we would get iŸ we put all the mass in the middlel
--- Trang 265 ---
Fig. 13-6. A thịn spherical shell of mass or charge.
We now demonstrate the correctness of this miracle. In order to do so,
however, we shall consider a thin uniform hollow shell instead of the whole earth.
Let the total mass of the shell be rn, and let us calculate the potental energu of
a particle oŸ mass mm“ a distance ?‡ away from the center of the sphere (Eig. 13-6)
and show that the potential energy is the same as it would be if the mass ?n were
a point at the center. (The potential energy is easier to work with than is the
fñeld because we do not have to worry about angles, we merely add the potential
energies of all the pieces of mass.) IÝ we call z the distance of a certain plane
section from the center, then all the mass that is in a slice dz is at the same
distance ? from , and the potential energy due to this rỉng is —Œm đm/r. How
much mass is in the small slice dz? An amount
2 đ 2 d
đĩn —= 2ml ds — “uhet ¬..... 2na_u đa,
sin 8 Ụ
where / = rm/4a? is the surface density of mass on the spherical shell. (It is a