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radius at that point is always the same at every point on the orbit. Eor example,
the closer the planet is to the sun, the faster it is going, but by how much? By
the following amount: if instead of letting the planet go around the sun, we were
to change the direction (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,
the new speed would be the same as the speed it had in the actual orbit, because
this is just another example of a complicated path. So long as we come baeck to
the same distance, the kinetic energy will be the same. 5o, whether the motion is
the real, 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.
Thus, when we make a numerical analysis of the motion of the planet in is
orbit, as we did earlier, we can check whether or not we are making appreciable
errors by calculating this constant quantity, the energy, at every step, and I1§
should not change. For the orbit of Table 9-2 the energy does change,* it changes
by some 1.5 percent from the beginning to the end. Why? Either because for
the numerical method we use fñnite intervals, or else because we made a slight
mistake somewhere in arithmetic.
Let us consider the energy in another case: the problem of a mass on a spring.
When we displace the mass from its balanced position, the restoring fÍorce is
* 'The energy per unit mass is Hơi: + 92) — 1/zr in the units of Table 9-2.
--- Trang 260 ---
proportional to the displacement. In those circumstances, can we work out a law
for conservation of energy? Yes, because the work done by such a force is
H5 %
w= Paz= | —k# dư = —šk#Ÿ. (13.13)
'Therefore, for a mass on a spring we have that the kinetic energy of the oscillating
mass plus skz? 1s a constant. Let us see how this works. We pull the mass down;
1 is standing still and so is speed is zero. But zø is not zero, + is at is maximum,
so there is some energy, the potential energy, of course. Now we release the mass
and things begin to happen (the details not to be discussed), but at any instant
the kinetie plus potential energy must be a constant. Eor example, after the mass
1s on its way past the original equilibrium point, the position + equals zero, but
that is when it has its biggest ø2, and as it gets more #2 it gets less 02, and so
on. So the balance of z7 and œ2 is maintained as the mass goes up and down.
'Thus we have another rule now, that the potential energy for a spring is skz, 1Í
the force is —k#.
13-3 Summation of energy
Now we go on to the more general consideration of what happens when there
are large numbers of objects. Suppose we have the complicated problem of many
objects, which we label ¿ = 1, 2, 3,..., all exerting gravitational pulls on each
other. What happens then? We shall prove that if we add the kinetic energies
of all the particles, and add to this the sum, over all pøirs of particles, of their
mutual gravitational potential energy, —GMm/r;¡;, the total is a constant:
1 2 Gm¿m;
» 51n;U; + » TT g = cCOnSf. (13.14)
? (pairs 27) Ñ
How do we prove it? We diferentiate each side with respect to time and get
zero. When we diferentiate 1m02, we fnd derivatives of the velocity that are
the forces, Just as in Eq. (13.5). We replace these forces by the law of force that
we know from Newton”s law of gravity and then we notice that what is left is
minus the time derivative of
» Gmm;
palrs Tj
--- Trang 261 ---
The time derivative of the kinetic energy is
d 1 2 dù;
n2 5n =2 min X“.
=) Fiui (13.15)
Gm1n;T';
D j 1ÿ
The time derivative of the potential energy is
d Gm¿m; _— Gm¿m; đĩ;;
3 ` xa, '
pairs palrs +2
Tịj = Vị — #7)” + (Mì — 9)” + (3í — 2):
so that
đĩ;; 1 d+; d+;
—=“=_—_— |2(z;—z;)| —-_— “”
dE — 2ny | ứ “0Í dt — đi )
đụi — đụ;
2(— ;)| — —--Sˆ
+36, =1) (SE — Sự)
đzi¿ — dz;
2(z;¿T—z;)| - °
+3 Hi dt — dị )|
¿ — Ðÿ
—= ¿7 * ————————
— +22 T¡j Xà) Ti
since T¡j — TTj¡, while T¡j — Ti. 'Thus
d Gm¿m; Gm1m;1T'; Gm;1n¿T;¡
dt » có Tụ 2> | Tử _v® THỊ 3.16)
pairs pairs +2 4!
Now we must note carefully what 3 ){Š)} and 3` mean. In Eq. (13.15), 3 {5`}
? 3 pairs Ũ 3
means that ? takes on all values ? = 1, 2, 3,... in turn, and for each value oŸ ¿,
--- Trang 262 ---
the index 7 takes on all values except ?. Thus if ¿ = 3, 7 takes on the values 1, 2,
In Eq. (13.16), on the other hand, Ề` means that given values of ? and 7
occur only once. 'Phus the particle pair 1 and 3 contributes only one term to the
sum. 'To keep track of this, we might agree to let ¿ range over all values 1, 2,
3,..., and for each ¿ let 7 range only over values greø‡er than ¿. Thus iŸ ¿ = 3, 7
could only have values 4, 5, 6,... But we notice that for each z, j7 value there are
©wo contributions to the sum, one involving ¿, and the other ø;, and that these
terms have the same appearance as those of Eq. (13.15), where ai values of ?
and 7 (except 2 = 7) are included in the sum. 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 time derivative of the kinetic plus potential energy is
indeed zero. 'Thus we see that, for many objects, ¿he kứnetic energụ is the sum
0ƒ the contributions [rom cach ?ndiuidual ob7ect, and that the potential energy