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We shall assume that the initial velocity is in a radial direction away from Earth s center. According to Newton s laws of motion, the acceleration of a particle is inversely proportional to the square of the distance of the particle from the center of Earth. If r represents that distance, R the radius of the earth (approximately 3,960 miles), v the velocity of the particle, and a its acceleration, then a = dv = k2 . At Earth s surface dt r gR 2 (r = R), a = g, where g = 32.16 ft/sec2 = .00609 mi/sec2. It follows that k = gR 2, so a = 2 . Since r dr a = dv and v = dt , by the chain rule we have a = dv = dv dr = v dv . If v0 represents the escape velocity, dt dt dr dt dr dv = gR 2 with initial condition v = v when r = R. we are led to the differential equation v 0 dr r2 2 2 DSolve[{v[r]v'[r] g R /r , v[R] v0}, v[r], r]
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2 g r R +2 g R2 +r v02 2 g r R +2 g R2 +r v02 , v[r] v[r] r r
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Since the velocity is positive at the surface of Earth (r = R), and must remain positive for the duration of 2 the flight, we reject the first solution. Furthermore, v(r) will remain positive if and only if 2 gR + v0 0, so v0 2gR . 2 g R /.{g .00609, R 3960} 6.94498 The escape velocity is 6.94498 mi/sec.
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11.5 According to Newton s law of cooling, the temperature of an object changes at a rate proportional to the difference in temperature between the object and the outside medium. If an object whose temperature is 70 F is placed in a medium whose temperature is 20 , and is found to be 40 after 3 minutes, what will its temperature be after 6 minutes
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Ordinary Differential Equations
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If u(t) represents the temperature of the object at time t, du = k (u 20) . The initial condition is dt u(0) = 70. solution1 = DSolve[{u'[t] k(u[t] 20), u 70}, u[t], t]
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{{ u[t] 10 (2 + 5 ) }}
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u[t_]= solution1[[1, 1, 2]] 10(2 + 5 k t) We determine k using the information about the temperature 3 minutes later. Since we are using Solve for the transcendental function e x , Mathematica supplies a warning that may safely be ignored. solution2 = Solve[u 40, k] Solve ifun : Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information.
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k 1 Log 5 3 2
u /. k solution2[[1, 1, 2]] 28 The temperature 6 min later is 28 F.
11.6 If air resistance is neglected, a freely falling body falls with an acceleration g, which is approximately 32.16 ft/sec2. If air resistance is considered, its motion is changed dramatically. If an object whose mass is 5 slugs is dropped from a height of 1000 ft, determine how long it will take to hit the ground (a) neglecting air resistance and (b) assuming that the force of air resistance is equal to the velocity of the object. Draw a graph of the height functions with and without air resistance.
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Let h(t) represent the height of the object at time t, v(t) its velocity, and a(t) its acceleration. Recall that v (t ) = h '(t ) and a (t ) = v '(t ) = h ''(t ) and, by Newton s law, the sum of the external forces acting upon the object is equal to its mass times its acceleration: m a(t ) = F . (a) If air resistance is neglected, the only force acting on the object is gravity, so m a(t ) = m g . We can divide by m and solve the differential equation h ''(t ) = g with initial conditions h '(0) = 0, h(0) = 1000. (Note: We take up to be the positive direction.) g = 32.16; solution = DSolve[{h''[t] g, h' 0, h 1000}, h[t], t]; height1[t_]= solution[[1, 1, 2]] 1000 16.08t2 When the object reaches the ground its height will be 0. Solve[height1[t] 0, t]
{{t 7.886}, {t 7.886}}
It takes 7.886 sec to reach the ground. (b) If air resistance is taken into account, there is an external force acting upon the object in addition to gravity, equal to v(t). The differential equation becomes
m a(t ) = mg v(t ) m h ''(t ) = mg h '(t )