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1.2 0.362 —0.362
—0.949
1.3 0.267 —0.267
—0.976
1.4 0.169 —0.169
—0.993
1.5 0.070 —0.070
—1.000
1x ....
--- Trang 187 ---
like this: the position later is equal to the position before plus e times the velocity
d‡ the từme in the rmniddle oƒ the interudl. Simllarly, the velocity at this halfway
point is the velocity at a tỉme e before (which is in the middle of the previous
interval) plus e tỉimes the acceleration at the time ứ. That is, we use the equations
z(£ + e) = z() + cu( + c/2),
0(£ + /2) = u(t — c/2) + ca(t), (9.16)
a(£) = —z(Ð).
There remains only one slipght problem: what is 0(c/2)? At the start, we are
given 0(0), not ø(—e/2). To get our calculation started, we shall use a special
equation, namely, 0(e/2) = (0) + (e/2)a(0).
Now we are ready to carry through our calculation. EOor convenience, we
may arrange the work in the form of a table, with columns for the time, the
position, the velocity, and the acceleration, and the in-between lines for the
velocity, as shown in Table 9-1. Such a table is, of course, just a convenient way
Of representing the numerical values obtained from the set of equations (9.16),
and in fact the equations themselves need never be written. We just fill in the
various spaces in the table one by one. 'Phis table now gïves us a very good idea
of the motion: it starts from rest, fñrst picks up a little upward (negative) velocity
and it loses some of its distance. The acceleration is then a little bit less but
1t is still gaining speed. But as it goes on it gains speed more and more slowly,
until as it passes ø = 0 at about ý = 1.50 sec we can confidently predict that it
will keep goïing, but now it will be on the other side; the position # will become
negative, the acceleration therefore positive. Thưus the speed decreases. Ít 1s
interesting to compare these numbers with the function = cos¿, which is done
in Eig. 9-4. The agreement is within the three significant fgure accuracy of our
calculationl We shall see later that ø = cosứ is the exact mathematical solution
of our equation of motion, but i1 is an impressive illustration of the power of
numerical analysis that such an easy calculation should gïve such precise results.
9-7 Planetary motions
'The above analysis is very nice for the motion of an oscillating spring, but can
we analyze the motion of a planet around the sun? Let us see whether we can
arrive at an approximation to an ellipse for the orbit. We shall suppose that the
sun is infñnitely heavy, in the sense that we shall not inelude its motion. Suppose
--- Trang 188 ---
1.0
0.5
ọ 0.5 1.0 1.5é £ (sec)
Fig. 9-4. Graph of the motion of a mass on a spring.
a planet starts at a certain place and is moving with a certain velocity; it goes
around the sun in some curve, and we shall try to analyze, by Newton's laws of
motion and his law of gravitation, what the curve is. How? At a given moment it
1s at some position in space. lf the radial distance from the sun to this position
is called r, then we know that there is a force directed inward which, according
to the law of gravity, is equal to a constant times the product of the sun”s mass
and the planet's mass divided by the square of the distance. 'To analyze this
further we must fnd out what acceleration will be produced by this force. We
shall need the componenfs of the acceleration along two directions, which we call
z and . Thus iŸ we specify the position of the planet at a given moment by
giving z and (we shall suppose that z is always zero because there is no force
in the z-direction and, if there is no initial velocity 0;, there will be nothing to
make z other than zero), the force is direcbed along the line joining the planet to
the sun, as shown in Fig. 9-5.
y F„ PLANET (x,y)
Fig. 9-5. The force of gravity on a planet.
--- Trang 189 ---
trom this fgure we see that the horizontal component of the force is related
to the complete force in the same manner as the horizontal distance z is to the
complete hypotenuse z, because the bwo triangles are similar. Also, IÝ # is positive,
F, is negative. That is, F„/|F| = —z/r, or F„ = —|F|z/z == —GMmz/r3. Ñow
we use the dynamical law to fnd that this force component is equal to the mass
of the planet times the rate of change of its velocity in the z-direction. 'Phus we
ñnd the following laws:
m(du„/dt) = —GMma/rẺ,
m{(duy/đt) = —GMmy/rẺ, (9.17)
r= V+2 +92.
This, then, is the set of equations we must solve. Again, in order to simplify
the numerical work, we shall suppose that the unit of time, or the mass of the
sun, has been so adjusted (or luck is with us) that GẢM = 1. Eor our specifc
example we shall suppose that the initial position of the planet is at z = 0.500
and = 0.000, and that the velocity is all in the, g-direction at the start, and
1s Of magnitude 1.630. Now how do we make the calculation? We again make
a table with columns for the time, the #-position, the z-velocity „, and the
-acceleration œ„; then, separated by a double line, three columns for position,
velocity, and acceleration in the -direction. In order to get the accelerations we
are going to need Ed. (9.17); it tells us that the acceleration in the z-direction
is —#/rỞ, and the acceleration in the z-direction is —#/rỞ, and that z is the
square root of z2 +”. Thus, given ø and ¿, we must do a little calculating on the
side, taking the square root of the sum of the squares to fnd r and then, to get
ready to calculate the two accelerations, it is useful also to evaluate 1/r3. This
work can be done rather easily by using a table of squares, cubes, and reciprocals:
then we need only multiply # by 1/r, which we do on a slide rule.
Our calculation thus proceeds by the following steps, using time intervals e =
0.100: Initial values at # = Ú:
+(0) = 0.500 (0)=_ 0.000
0„(0) = 0.000 0„(0) = +1.630
trom these we flnd:
r(0)= 0.500 1/r(0) = 8.000
ø„ = —4.000 a„ = 0.000
--- Trang 190 ---
Thus we may calculate the velocities 0„(0.05) and 0„(0.05):