Sample Solutions for Assignment 3.

Transcription

1 AMath 383, Autumn Sample Solutions for Assignment 3. Reading: Chs Eercise 7 of Chapter 3. If $X is invested toda at 3% interest compounded continuousl, then in ears it will be worth Xe (.3 ) = Xe.6. This will be $ million if X = 6 /e.6 = 548, I guess a fair price for it would be somewhat less than this since the buer would not get the mone for ears, while if he invested at 3% he could withdraw his mone at an time.. Eercise of Chapter 4. (a) Let (t) be the concentration of Pb. While in the ore, it decas according to d/dt = λ, but it is also replenished b r from 6 Ra decaing to Pb. Hence, in the ore, d/dt = λ + r. Outside the ore, it decas at the same rate but is not replenished, so after manufacture d/dt = λ. (b) Inside the ore, there is radioactive equilibrium; that is d/dt =, or, λ = r. If t is the last instant before removal from the ore, then λ(t ) = r = per minute. (c) Since (t ) = r/λ, (t) = e λ(t t ) (t ) = e λ(t t ) r/λ. (d) Differentiating the epression in (c), and equating it to the measured value of d/dt, we find d dt = λe λ(t t ) r/λ = 8.5 e λ(t t ) = 8.5/r. Taking logs on both sides and solving for t t ields t t = ln(8.5/r)/λ. Since the half-life of Pb is ears, the deca rate λ is ln /.35 per ear. The epression for t t is an increasing function of r, so it will be largest for r =. If substituting this value for r and the value of λ indicates that t t < 3, then the painting must be a fake. Making these substitutions we find t t =.4. The painting is at most about ears old and so must be a forger.

2 3. Eercise of Chapter 5. Let r be the distance of the moon from the center of the earth in km and let v be its velocit in km/s. We are told that it travels around a near circular orbit of radius r in 8 das, or, seconds. Since the circumference of the circle is πr, this means that its velocit is v = πr/(.4 6 ) km/s. () Equating the acceleration due to gravit and the centrifugal force, we are told in (b) that GM earth /r = v /r. Substituting the epression in () for v and solving for r, we find GM earth r = 4π r.4 r3 (GM earth ).48. () We are told in part (c) that the radius of the earth is a = 64 km and that the gravitational pull of the earth on an object near its surface is 98 cm/s, or km/s. That is, GM earth a = Substituting a for (GM earth ) in (), we find r = , and taking the cube root, which is about 6 earth radii. r km, 4. This question concerns the motion of a satellite around the sun: the satellite could be a planet, a comet or an unpowered spacecraft, whose mass is assumed to be negligible compared with the mass of the sun. The motion of such a satellite takes place in a plane, so we can describe its coordinates b just two space dimensions and. Assume that the sun is at the origin (, ). The sstem of differential equations describing the satellite s motion is Newton s law of gravitation, discovered b Newton in the late 7th centur, namel = ( + ) 3, = ( + ) 3. (a) This sstem of two second-order differential equations can be simplified to a sstem of four first-order differential equations b introducing two new variables z = (t) (the speed of the satellite in the space direction) and w = (t) (the speed of the satellite in the space direction). Write down this sstem of equations.

3 = z = w z = ( + ) 3 w = ( + ). 3 (b) Use MATLAB routine ode45 to solve the sstem ou wrote down in (a). Tpe help ode45 to find out how to use this routine. You will need to create a.m file that returns the derivative values (t), (t), z (t) and w (t) when given values of t,,, z, and w. It might be in a file called dvdt.m and look like the following function vprime = dvdt(t,v) % Given t and vector v = [(t),(t),z(t),w(t)], this function % returns vector vprime = [ (t), (t),z (t),w (t)]. vprime(,) = vprime(,) = vprime(3,) = vprime(4,) = % Enter epression for (t) % Enter epression for (t) % Enter epression for z (t) % Enter epression for w (t) Start at t = with () = 4, () =, z() =, and w() =.5 and run out to tma = 5. Plot the position of the satellite at each time step returned b ode45. You should see an almost circular orbit. [Note: If ou plot things in MATLAB, be sure to sa ais( equal ), or the aes will be different lengths and our circle will look like an ellipse!] Keeping (), (), and z() fied, tr varing w(). You will have to adjust tma so that ou see a complete orbit. Some recommended values are: w() tma For initial velocities slightl larger than.5, ou should see an elliptical orbit. The fact that planets and comets have elliptical orbits was observed b Kepler in the earl 7th centur, but it was not eplained until Newton developed his law of gravitation 6 ears later. The eccentricit of the ellipse depends on the size of the initial velocit w(). The orbits of planets, such as the Earth, are tpicall not ver eccentric. The look almost like circles. But the orbits of comets, such as Halle s comet and Hale-Bopp, can be ver eccentric. That is wh Halle s 3

4 comet can be seen from Earth onl once ever 76 ears and Hale-Bopp about once ever ears: the rest of the time the are ver far from the sun traveling around their highl eccentric orbits. If ou make the initial velocit large enough, the orbit becomes a hperbola, with the satellite approaching the direction of a straight line heading right out of the solar sstem; in this case the satellite escapes the gravitational attraction of the sun. Man comets pass through the solar sstem on hperbolic orbits: the enter the solar sstem, pass b the sun once and then leave the solar sstem again. Some spacecraft launched from Earth, such as Voager II which was launched in 977 and flew past Pluto in 989, will soon leave the solar sstem on a hperbolic path, never to return. The borderline case between elliptical and hperbolic orbits is when w() is eactl equal to the escape velocit: the value just large enough for the satellite to escape from the solar sstem. In this case the orbit is actuall a parabola, which is the borderline case between an ellipse and a hperbola. For initial velocities less than.5 but greater than, the orbit is again an ellipse. Turn in plots showing the tpes of orbits described, labeled with the value of w() which gives rise to that path. The code for computing the derivative of each variable should read: function vprime = dvdt(t,v) % Given t and vector v = [(t),(t),z(t),w(t)], this function % returns vector vprime = [ (t), (t),z (t),w (t)]. vprime(,) = v(3); % epression for (t) vprime(,) = v(4); % epression for (t) vprime(3,) = -v()/(v()^ + v()^)^(3/); % epression for z (t) vprime(4,) = -v()/(v()^ + v()^)^(3/); % epression for w (t) The following code runs ode45 to solve the sstem of ode s with different initial data and it plots the orbits ( vs coordinates): % Plot orbits corresponding to different initial velocities. v = [4; ; ;.5]; % First set of initial data. [tout,vout] = ode45(@dvdt, [ 5], v); % Use ode45 to solve ivp. subplot(,3,) plot(vout(:,),vout(:,)), ais equal % Plot orbit. label( ), label( ), title( w() =.5 ) v(4) =.6; % Second initial velocit. [tout,vout] = ode45(@dvdt, [ 5], v); % Use ode45 to solve ivp. subplot(,3,) plot(vout(:,),vout(:,)), ais equal % Plot orbit. label( ), label( ), title( w() =.6 ) 4

5 v(4) =.8; % Third initial velocit. [tout,vout] = ode45(@dvdt, [ ], v); % Use ode45 to solve ivp. subplot(,3,3) plot(vout(:,),vout(:,)), ais equal % Plot orbit. label( ), label( ), title( w() =.8 ) v(4) =.4; % Fourth initial velocit. [tout,vout] = ode45(@dvdt, [ 35], v); % Use ode45 to solve ivp. subplot(,3,4) plot(vout(:,),vout(:,)), ais equal % Plot orbit. label( ), label( ), title( w() =.4 ) v(4) =.; % Fifth initial velocit. [tout,vout] = ode45(@dvdt, [ ], v); % Use ode45 to solve ivp. subplot(,3,5) plot(vout(:,),vout(:,)), ais equal % Plot orbit. label( ), label( ), title( w() =. ) w() =.5 w() =.6 w() = w() =.4 w() =