James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 13.4 Motion in Space: Velocity and Acceleration 877
Then
a 3 h − 2GM
r 2 u 3 sr 2 u 3 u9d − 2GM u 3 su 3 u9d
− 2GM fsu ? u9du 2 su ? udu9g (by Theorem 12.4.11, Property 6)
But u ? u − | u | 2 − 1 and, since | ustd |
− 1, it follows from Example 13.2.4 that
u ? u9 − 0
Therefore
and so
a 3 h − GM u9
sv 3 hd9 − v9 3 h − a 3 h − GM u9
Integrating both sides of this equation, we get
11 v 3 h − GM u 1 c
z
h
c
¨
r
v
y
where c is a constant vector.
At this point it is convenient to choose the coordinate axes so that the standard
basis vector k points in the direction of the vector h. Then the planet moves in the
xy-plane. Since both v 3 h and u are perpendicular to h, Equation 11 shows that c lies
in the xy-plane. This means that we can choose the x- and y-axes so that the vector i lies
in the direction of c, as shown in Figure 8.
If is the angle between c and r, then sr, d are polar coordinates of the planet. From
Equation 11 we have
x
FIGURE 8
u
r ? sv 3 hd − r ? sGM u 1 cd − GM r ? u 1 r ? c
− GMr u ? u 1 | r | | c | cos − GMr 1 rc cos
where c − | c | . Then r −
r ? sv 3 hd
GM 1 c cos − 1
GM
r ? sv 3 hd
1 1 e cos
where e − cysGMd. But
r ? sv 3 hd − sr 3 vd ? h − h ? h − | h | 2 − h 2
where h − | h | . So r − h 2 ysGMd
1 1 e cos − eh 2 yc
1 1 e cos
Writing d − h 2 yc, we obtain the equation
12 r −
ed
1 1 e cos
Comparing with Theorem 10.6.6, we see that Equation 12 is the polar equation of a conic
section with focus at the origin and eccentricity e. We know that the orbit of a planet is a
closed curve and so the conic must be an ellipse.
This completes the derivation of Kepler’s First Law. We will guide you through
the der ivation of the Second and Third Laws in the Applied Project on page 880. The
proofs of these three laws show that the methods of this chapter provide a powerful tool
for describing some of the laws of nature.
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