* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download No Slide Title
Bra–ket notation wikipedia , lookup
Particle filter wikipedia , lookup
Double-slit experiment wikipedia , lookup
Fictitious force wikipedia , lookup
Velocity-addition formula wikipedia , lookup
Elementary particle wikipedia , lookup
Modified Newtonian dynamics wikipedia , lookup
Jerk (physics) wikipedia , lookup
Relativistic angular momentum wikipedia , lookup
Path integral formulation wikipedia , lookup
Brownian motion wikipedia , lookup
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
Equations of motion wikipedia , lookup
Four-vector wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Classical mechanics wikipedia , lookup
Laplace–Runge–Lenz vector wikipedia , lookup
Rigid body dynamics wikipedia , lookup
Matter wave wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Recall: z A path in R3 can be represented as c(t) = (x(t),y(t),z(t))= x(t)i + y(t)j + z(t)k (the position vector) y x The tangent or velocity vector to c(t) is c (t) = (x (t),y (t),z (t)) = x (t)i + y (t)j + z (t)k If c(t) represents the path of a moving particle, then we often let v(t) = c (t) represent its velocity vector, a(t) = v (t) = c (t) represent its acceleration vector, s = ||v(t)|| = ||c (t)|| represent its speed. Let b(t) = (b1(t) , b2(t) , b3(t)) and c(t) = (c1(t) , c2(t) , c3(t)). Let b(t) = (b1(t) , b2(t) , b3(t)) and c(t) = (c1(t) , c2(t) , c3(t)). d —[b(t) + c(t)] = dt d —(b1(t) + c1(t) , b2(t) + c2(t) , b3(t) + c3(t)) = dt (b1 (t) + c1 (t) , b2 (t) + c2 (t) , b3 (t) + c3 (t)) = (b1 (t) , b2 (t) , b3 (t)) + (c1 (t) , c2 (t) , c3 (t)) = b (t) + c (t) . d d —[p(t)c(t)] = —(p(t)c1(t) , p(t)c2(t) , p(t)c3(t)) = dt dt (p (t)c1(t) + p(t)c1 (t) , p (t)c2(t) + p(t)c2 (t) , p (t)c3(t) + p(t)c3 (t)) = (p (t)c1(t) , p (t)c2(t) , p (t)c3(t)) + (p(t)c1 (t) , p(t)c2 (t) , p(t)c3 (t)) = p (t)(c1(t) , c2(t) , c3(t)) + p(t)(c1 (t) , c2 (t) , c3 (t)) = p (t)c(t) + p(t)c (t) . d d —[b(t) • c(t)] = —[b1(t) c1(t) + b2(t) c2(t) + b3(t) c3(t)] = dt dt [b1 (t)c1(t) + b1(t)c1 (t)] + [b2 (t)c2(t) + b2(t)c2 (t)] + [b3 (t)c3(t) + b3(t)c3 (t)] = [b1 (t)c1(t) + b2 (t)c2(t) + b3 (t)c3(t)] + [b1(t)c1 (t)+ b2(t)c2 (t) + b3(t)c3 (t)] = b (t)•c(t) + b(t)•c (t) . d —[b(t) c(t)] = dt d —(b2(t)c3(t) – b3(t)c2(t) , b3(t)c1(t) – b1(t)c3(t) , b1(t)c2(t) – b2(t)c1(t)) = dt ( [b2 (t)c3(t) + b2(t)c3 (t)] – [b3 (t)c2(t) + b3(t)c2 (t)] , ]–[ [ [ ], ]–[ ]) = ( [b2 (t)c3(t) – b3 (t)c2(t)] + [b2(t)c3 (t) – b3(t)c2 (t)] , [ ]+[ [ ], ])= ]+[ (b2 (t)c3(t) – b3 (t)c2(t) , , )+ (b2(t)c3 (t) – b3(t)c2 (t) , , )= b (t) c(t) + b(t) c (t) . d —[c(q(t))] = dt d —( c1(q(t)) , c2(q(t)) , c3(q(t)) ) = dt ( c1 (q(t)) q (t) , c2 (q(t)) q (t) , c3 (q(t)) q (t) ) = q (t) ( c1 (q(t)) , c2 (q(t)) , c3 (q(t)) ) = q (t) c (q(t)) . All of the preceding derivative formulas, except the one concerning cross product, can be extended to Rn for any n. These are the differentiation rules displayed on page 262 of the text. Suppose the position vector c(t) describes a path. (a) Find d d —||c(t)||2 = — [c(t) • c(t)] . dt dt d —[c(t) • c(t)] = c (t) • c(t) + c(t) • c (t) = 2 c(t) • c (t) . dt (b) If c(t) has constant length, that is, ||c(t)|| is a constant, what kind of path would this describe in R2 or in R3? a path which lies on either a circle or a sphere centered at the origin (c) If c(t) has constant length, then why must c(t) and c (t) be orthogonal for all t? d If ||c(t)|| is a constant, then —||c(t)||2 = 0 dt c(t) • c (t) = 0. Thus, c(t) and c (t) are orthogonal for all t. Suppose c(t) = (x(t) , y(t) , z(t)) describes a path of a particle where a(t) = – k. When t = 0, the position of the particle is (0,0,1), and the velocity is i + j. Find c(t), and find the values of t 0 for which the particle is below the xy plane. a(t) = – k v(t) = p1i + p2j + (– t + p3)k , for some constants p1 , p2 , p3 . v(0) = i + j v(t) = i + j – tk v(t) = i + j – tk c(t) = (t + q1) i + (t + q2) j + (– t2/2 + q3)k , where q1 , q2 , q3 are constants. c(0) = k c(t) = ti + tj + (1 – t2/2)k . The particle is below the xy plane (z < 0) when t > 2. Newton’s Second Law (page 264 of the text) states that if F(t) is the force at time t acting on a particle with mass m, then F(t) = ma(t) . Let us consider a circular path of a planet of mass m orbiting the sun with Note that t = 2r0/s st st mass M, described by r(t) = ( r0cos— , r0sin— ) . is the time after r0 r0 one revolution According to Newton’s Law of Gravitation, the force acting on the planet is described by GmM F(t) = – ——— r(t) , where G is the gravitational constant. ||r(t)||3 Consequently, GmM mr (t) = – ——— r(t) . ||r(t)||3 (Note that the acceleration vector is in the opposite direction of the position vector.) st st The planet’s circular path r(t) = ( r0cos— , r0sin— ) has radius ||r(t)|| = r0 r0 r0 . st st The velocity of the planet is r (t) = (– s sin— , s cos— ) . r0 r0 The speed of the particle is ||r (t)|| = s . 2 2 s st s st The acceleration of the particle is r (t) = (– — cos— , – — sin— ) = r0 r0 r0 r0 s2 st st s2 – — ( r0cos— , r0sin— ) = – — r(t) . r0 2 r0 r0 r0 2 mr (t) = ma(t) is called the centripetal force, and we now see that GmM ms2 – ——— r(t) = mr (t) = – —— r(t) s2 = GM / r0 . r03 r0 2 Since the period for each of sin(kx) and cos(kx) is 2/k, then the period for the orbiting body is T = 2r0/s. We can then derive Kepler’s Law (page 266 of the textbook), which says that the square of the period is proportional to the cube of the radius: (2)2 T2 = r03 —— . GM While we have considered a planet orbiting the sun, all our derivations can be applied to a satellite orbiting the earth. A satellite is said to be in geosynchronous orbit around the earth, if it is always over the same point above the equator. Given G = 6.6710–11 when working with meters, kilograms, and seconds, and given the mass of the earth is 5.981024 kilograms, find the radius necessary for a geosynchronous orbit above the equator. We want the period of the orbit to be 1 day = 606024 seconds = 86,400 seconds. From Kepler’s Law, we find that –11)(5.981024) GM (6.6710 r03 = T2 —— = (86,400)2 —————————— = (2)2 (2)2 7.541022 meters3 r0 = 4.23107 meters 26,200 miles