where s is the horizontal range, in this case 50 yards. Note that as long as the initial speed of the bullet is great enough to let it hit a target at

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1 1 PHYS 31 Homework Assignment Due Friday, 13 September A monkey is hanging from a tree limbat height h above the ground. A hunter 50 yards away from the base of the tree sees him, raises his gun and fires. Just as he pulls the trigger, the monkey (who, when he saw the hunter, said Oh, oh. Trouble!") lets go of the branch and drops, hoping to make the hunter miss. But the hunter has anticipated the monkey's evasive maneuver, and has aimed accordingly. Where did the hunter aim, in order to hit the falling monkey? The hunter fires the bullet at an angle to the horizontal. Let us suppose the gun is a height y 0 from the ground. In time t the bullet travels horizontally x = v cos t; its vertical position is y = y 0 + v sin t gt ; finally, the monkey's vertical position is y m = h gt For the bullet to hit the monkey, wemust have or y = y m ; y 0 + v sin t gt = h gt We notice that the term involving g is identical on both sides, so it drops out. In other words, the answer is the same as if there were no gravity that is, as if neither the monkey nor the bullet falls. But this means the hunter must aim directly at the monkey, giving tan = h y 0 s ;

2 where s is the horizontal range, in this case 50 yards. Note that as long as the initial speed of the bullet is great enough to let it hit a target at a height h y 0 at a distance s, it is irrelevant to the qualitative conclusion that the hunter must aim directly at the monkey.. A damped, driven harmonic oscillator satisfies the differential equation ẍ +fl _x +Ω x = f(t) If the driving force is periodic, f (t) =f (t +ß/!) ; and has the sawtooth form f (t) = ( 1+!t; ß 3!t ß 0 <t<ß/! ; ; ß/! <t<ß/! find the solution x(t) for times sufficiently far in the future that the system has forgotten" its initial conditions. Use the operator method and the method of superposition suggested in B&O, Problem 1-6. To use the superposition method, expand f(t) infourier series 1 f (t) = a 0 + 1X n=1 [a n cos (n!t)+b n sin (n!t)] where (we set! = 1 to simplify the algebra) and a n = 1 ß Z ß 0 f (t) cos (nt) dt b n = 1 ß Z ß 0 f (t) sin (nt) dt We see that the coefficients are a n = ( 0; n =k 8 ; ß (k+1) n =k +1 1 This is possible for a wide class of periodic functions. See Ch. 10 of the math-physics notes at the link http// phys.pdf

3 3 b n =0 Next we solve the equation for a sinusoidal driving term z +fl _z +Ω z = fe int If we want a cosine driving term we take the real part of z(t), and for the sine we take the imaginary part. Here the sine is irrelevant since b n =0,sowe take the result from B&O where x (t) = 1X k=0 fl (k +1) f k cos (k +1)t arctan Ω (k +1) f k = 8 q ß (k +1) (Ω (k +1) ) +4fl (k +1) Now doing it by the operator method, we express the solution as x = 1» Z t! Im e (i! fl)t dfie (fl i!)fi f (fi) ; where! = q Ω fl 0 I guarantee that if one works out the sickening details for large t one gets the same result as before. 3. An open coal car is coasting along the track and passes under a coal chute, from which a steady stream of coal falls vertically into the car. If the car has length L, mass M, and initial speed v, and if the rate at which the coal is falling is _m, what is the speed of the car when it has gone past the hopper? We assume conservation of linear momentum. The vertical restraint force provided by the track precisely matches the increased gravitational force, not to mention impulse, from the (vertically) falling coal

4 4 (at least, if the track does not fail!). Thus we find the horizontal momentum of the car to be constant, or Thus _p x = _Mv + M _v =0 _M M + _v v =0 so that M(t)v(t) =M(0)v(0). Since M(t) =M(0) + _mt, wehave v (t) = v 0 1+ _mt/m 0 We can look at this problem from a somewhat different point of view in a frame co-moving with the coal car, the falling coal appears to have x-component of velocity v(t). Thus, in that frame, the car experiences a horizontal force F x = dp x dt = v (t) dm dt = _mv (t) In a time dt the increment of velocity (in the co-moving frame) is therefore dv x = v x _mdt M but since the speeds are small compared with the speed of light, the Galilean formula for addition of velocities applies, and therefore dv x j Lab = dv x j co moving = _M M v xj Lab dt which gives the relation we got from momentum conservation in the laboratory (track) frame, _M M + _v v =0

5 5 4. How much energy must be given to a space probe of mass m, initially in a low parking orbit about the Moon, to put it into an orbit that will escape from the Moon's gravitational well? The lunar mass is kg and the lunar radius is m, so its surface gravity is 0169g. Now, when a spacecraft is in a low parking orbit, the centripetal acceleration must balance the Moon's gravitational acceleration v R M = GM M R M The total energy is therefore E 0 = 1 mv GmM M R M = GmM M R M On the other hand, the same probe, at infinite distance, travelling with zero speed, has zero energy. The total energy change is thus E = E f E 0 =0 GmM M R M = GmM M R M = 1 mg MR M 5. Communications satellites are commonly placed in geosynchronous orbits (that is, they orbit above the Equator, with an orbital period that just matches the Earth's rotational period so they always remain above one point on the Earth's surface). A space shuttle is launched to service a geosynchronous satellite. It first attains a circular, low-earth parking orbit, then must accelerate to match the satellite's orbit. What is the minimum number of times the pilot must turn on the rocket engines, to perform this matching maneuver? How much change of velocity, and in what direction, is needed at each maneuver? (Assume each application of thrust is essentially instantaneous.) As discussed in class, the equations describing a small mass in orbit about a much larger one are _r + a r b r = c

6 6 and mr _ = `; where a = ` m ; b = MG; c = E m Here E and ` are constants. We want tomove theshuttle, from its initial low-earth circular orbit, with r = r < = constant, to a circular orbit of much larger radius r = r >. (The orbits have the same rotational direction, i.e., _ has the same sign in both.) We do this in two stages (a) First fire the rockets appropriately, to place the shuttle in an elliptical orbit whose inner radius is r < and whose larger radius is r >. Clearly the impulse must be tangential to the inner orbit, since if it were not the inner radius would change. (b) Then fire the rockets again to convert the ellipse into a circular orbit at the larger radius. Again the impulse must be tangential or else the final radius could not be r >. We see that in the intermediate elliptical orbit, 1 1 = p b r < r > a ac and = b r < r > a We get these relations by solving the equation _r =0. We also note that in the low circular orbit, r < _ < v < = b r < and in the geosynchronous orbit, v > = b r >

7 7 (The latter relations simply state that centripetal acceleration equals gravitational acceleration, in a circular orbit.) Combining these formulas we see that 1 MG + 1 = r < r > 1 ` m = _ r4 < 1 = r> 4 _ or, noting that MG = gr E,wefind v < = gr E r < v > = gr E r > v 1 = gr E r < v = gr E r > = v< r < r > = v> r < r > 1 1+r < /r > 1+r > /r < > v < < v > In other words, at both perigee and apogee we need a velocity change tangent to the circular orbit, and in the direction of the orbital rotation. The initial velocity change is v init = v 1 v < = v < " and the final one is v fin = v v > = v > "1 1+r < /r > 1/ 1 # # 1+r > /r < 1/ Manifestly, each velocity increase raises the both the energy of the orbit and its angular momentum. Putting in numbers, assuming the radius of the low-earth orbit is R E km = km

8 8 and that of the geosynchronous orbit is we obtain and r > = 3 qgr E (T /ß) = km ; v init =47 km=sec v fin =149 km=sec (Note that escape speed" is 11km=sec so the change-of-velocity needed to escape the Earth's gravitation, starting from low-earth orbit, is much less than that needed starting from Earth's surface.)

Satellite Orbital Maneuvers and Transfers. Dr Ugur GUVEN

Satellite Orbital Maneuvers and Transfers Dr Ugur GUVEN Orbit Maneuvers At some point during the lifetime of most space vehicles or satellites, we must change one or more of the orbital elements. For example,