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to 2, which we want to calculate, can be evaluated as the work done in goiỉng
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from 1 to plus the work done in going from ? to 2, because the forces are
conservative and the work does not depend upon the curve. Now, the work done
in goïng from position ? to a particular position in space is a function of that
position in space. Of course it really depends on ? also, but we hold the arbitrary
point P fñxed permanently for the analysis. IÝ that is done, then the work done
in goiïng from point ? to poiïnt 2 is some function of the ñnal position of 2. It
depends upon where 2 is; if we go to some other point we get a different answer.
We shall call this function oŸ position —(z, g, z), and when we wish to refer
%o some particular point 2 whose coordinates are (Za, 2, Z2), we shall write (2),
as an abbreviation for (zas,a,za2). The work done in going from point 1 to
point ? can be written also by going the ø/her t0øy along the integral, reversing
all the ds”s. That is, the work done in going from 1 to ? is mnus the work done
in going from the point P to 1:
P 1 1
J Esds= [ E+(—dg) =— F- ds.
1 P P
Thus the work done in going from to 1 is —U(1), and from P to 2 the work
is —U(2). Therefore the integral from 1 to 2 is equal to —U(2) plus [—U(1)
backwards]l, or +U(1) — U(2):
0q) == [ T- ds, U@) =~ [ T- ds,
II t-ds = U(1) — UD(). (14.1)
The quantity U(1) — (2) is called the change in the potential energy, and
we call Ư the potential energy. We shall say that when the object is located
at position 2, it has potential energy (2) and at position 1 it has potential
energy (1). If it is located at position ?, it has zero potential energy. IÝ we had
used any other point, say Q, instead o£ P, it would turn out (and we shall leave it
to you to demonstrate) that the pofenfial energụ ¡s changed onhụ bụ the addition
öoƒ a cons‡ơønt. Since the conservation of energy depends only upon chønges, 1Ề
does not matter if we add a constant to the potential energy. Thus the poïint ?
1s arbitrary.
Now, we have the following two propositions: (1) that the work done by a force
is equal to the change in kinetic energy of the particle, but (2) mathematically,
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for a conservative force, the work done is minus the change in a function which
we call the potential energy. Âs a consequence of these two, we arrive at the
proposition that #ƒ on conseruatiue jorces aœcl, the kinetlic energu T' pÌus the
potential energụ Ù remains constant:
7'+U = constant. (14.2)
Let us now discuss the formulas for the potential energy for a number oÝ cases.
TỶ we have a gravitational field that is uniform, iŸ we are not going to heights
comparable with the radius of the earth, then the force is a constant vertical
force and the work done is simply the force times the vertical distance. 'Phus
D{(z) = mụgz, (14.3)
and the point P? which corresponds to zero potential energy happens to be any
point in the plane z = 0. We could also have said that the potential energy
1s rmwg(z — 6) iŸ we had wanted to—all the results would, of course, be the same in
our analysis except that the value oŸ the potential energy at z = 0 would be —rng6.
lt makes no diference, because only đjfƒerences In potential energy count.
The energy needed to compress a linear spring a distance z from an equilibrium
point 1s
U(œ) = škzŸ, (14.4)
and the zero of potential energy is at the point z = 0, the equilibrium position of
the spring. Again we could add any constant we wish.
'The potential energy of gravitation for point masses ⁄ and rn, a distance z
apart, 1s
U(r) =—GMm/r. (14.5)
The constant has been chosen here so that the potential is zero at inñnity. Of
course the same formula applies to electrical charges, because it ¡is the same law:
U(r) = qiqa/4mcqr. (14.6)
Now let us actually use one of these formulas, to see whether we understand
what it means. Question: How fast do we have to shoot a rocket away from the
earth in order for it to leave? Solutzon: The kinetie plus potential energy must
be a constant; when it “leaves,” it will be millions of miles away, and ïŸ it is just
barely able to leave, we may suppose that it is moving with zero speed out there,
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Just barely going. Let œ be the radius of the earth, and Mƒ its mass. The kinetic
plus potential energy is then initially given by D1 — ŒGmM/a. At the end of
the motion the two energies must be equal. The kinetic energy is taken to be zero
at the end of the motion, because it is supposed to be just barely drifting away
at essentially zero speed, and the potential energy is GmMĩ divided by infnity,
which is zero. 5o everything is zero on one side and that tells us that the square
of the veloeity must be 2ŒGÄ/a. But ŒAf/a2 is what we call the acceleration of
gravity, g. Thus
UŠ = 2ga.
At what speed must a satellite travel in order to keep going around the earth?
We worked this out long ago and found that øŸ = GŒA//a. Therefore to go øa
from the earth, we need v⁄2 times the velocity we need to just go arownd the
carth near its surface. We need, in other words, tướce œs rnuch energu (because
energy goes as the square of the velocity) to leave the earth as we do to go around
it. Thherefore the first thíng that was done historically with satellites was to get
one to øo around the earth, which requires a speed of five miles per second. The
next thing was to send a satellite away from the earth permanently; this required
twice the energy, or about seven miles per second.
Now, continuing our discussion of the characteristics of potential energy, let
us consider the interaction of bwo molecules, or two atoms, ÿwo oxygen atoms
for instance. When they are very far apart, the Íorce is one of attraction, which
varies as the inverse seventh power of the distance, and when they are very close
the force is a very large repulsion. If we integrate the inverse seventh power to
fnd the work done, we fnd that the potential energy Ứ, which is a function of
the radial distance between the two oxygen atoms, varies as the inverse sixth
power of the distance for large distances.
TỶ we sketch the curve of the pobential energy (7) as in Fig. 14-3, we thus
start out at large r with an inverse sixth power, but IŸ we come in sufficiently
near we reach a point đ where there is a minimum of potential energy. “The
minimum of potential energy at r = đ means this: if we start at đ and move
a small distance, a very small distance, the work done, which is the change In
potential energy when we move this distance, is nearly zero, because there is
very little change in potential energy at the bottom of the curve. Thus there is