Solução_Calculo_Stewart_6e

F.
320 ¤ CHAPTER 18 SECOND-ORDER DIFFERENTIAL EQUATIONS ETX.10
CHAPTER 17
(b) The particle is acted upon by a varying gravitational force during its motion. By Newton’s Second Law of Motion,
m d2 y
dt 2
= Fy = −GMm
R 3
g = GM
R 2 . Therefore k2 = g R .
y,soy 00 (t) =−k 2 y (t) where k 2 = GM
R 3
. At the surface, −mg = FR = −GMm ,so
(c) The differential equation y 00 + k 2 y =0has auxiliary equation r 2 + k 2 =0. (This is the r of Section 18.1 [ ET 17.1],
not the r measuring distance from the earth’s center.) The roots of the auxiliary equation are ±ik,soby(11)in
Section 18.1 [ ET 17.1], the general solution of our differential equation for t is y(t) =c 1 cos kt + c 2 sin kt. Itfollowsthat
y 0 (t) =−c 1 k sin kt + c 2 k cos kt. Nowy (0) = R and y 0 (0) = 0,soc 1 = R and c 2 k =0.Thusy(t) =R cos kt and
y 0 (t) =−kR sin kt. This is simple harmonic motion (see Section 18.3 [ ET 17.3]) with amplitude R, frequency k, and
phase angle 0. The period is T =2π/k. R ≈ 3960 mi = 3960 · 5280 ft and g =32ft/s 2 ,so
k = g/R ≈ 1.24 × 10 −3 s −1 and T =2π/k ≈ 5079 s ≈ 85 min.
(d) y(t) =0 ⇔ cos kt =0 ⇔ kt = π 2 + πn for some integer n ⇒ y0 (t) =−kR sin π
2 + πn = ±kR. Thusthe
particle passes through the center of the earth with speed kR ≈ 4.899 mi/s ≈ 17,600 mi/h.
R 2