Physics 181 Assignment 4 Solutions 1. A sphee has within it a gavitational field given by g = g, whee g is constant and is the position vecto of the field point elative to the cente of the sphee. This field has constant magnitude and is diected towad the cente. a. Find the potential () at points in the sphee, taking (0) = 0. Check that you answe obeys " = #g. b. Find the mass density () fom Gauss s law " g = #4G. c. Let the sphee have adius. Find its total mass. d. Find the field g() at a point outside the sphee ( > ). e. Find the potential at a point outside the sphee. f. Detemine 0 so that () " 0 as ". # d a. We have () " 0 = " g # d == g 0. But d = d, so we find 0 () = 0 + g. b. " g = # i g i = g# i ( i /). But i ( i /) = ( i i ) "(1/) + i " i (1/), whee i i = and i (1/) = "(1/ 2 ) #( i ) = "(1/ 2 ) #( i /). Using i i = 2 and putting the pieces togethe we find "g = #2g/. Using Gauss s law we have = g 1 2"G. c. The mass is given by M = " dv whee the integal coves the volume of the sphee. Since depends only on the distance fom the cente, we can calculate easily the mass in a thin spheical shell of adius and thickness d: dm = " dv = g 1 2#G " 4#2 d = 2g " d. We integate this ove to get G the total mass: M = 2g G d = g2 0 G. d. As with all spheically symmetic cases, at points outside all the mass the fields ae just those of a point mass with the total mass, placed at the cente of symmety. In this case, fo >, g = GM = g2. 1
e. Since we have fixed the potential at = 0 to be 0, we must stat thee in calculating the potential elsewhee. Fist we find the value at the suface of the sphee: () = 0 " g # d = 0 0 + g d = 0 0 + g. (We could use the answe to (a) fo this pat.) Then fo points outside the sphee we find () = () " g # d = 0 + g + g 2 d = 0 + 2g " g 2 /. f. If " 0 as ", then 0 = "2g. 2 2. A sphee with unifom mass density and adius has a spheical cavity in it, of adius 0, centeed at distance a fom the cente of the massive sphee. We ae inteested in the field within the cavity. To get at it, we will calculate the potential in the cavity fist. a. Conside fist the sphee without the cavity. Use the integal fom of Gauss s law, " g nds = #4G" dv to find g() at points inside the sphee. [Choose fo the suface a sphee of adius about the cente of the actual sphee. Be caeful about diections and signs.] b. Find the potential () at a point inside the sphee, taking to vanish at infinite distance. [The integal is in two pats, one fo fom to the suface ( = ), the othe fom to the location of the field point.] The fields in the cavity can be thought of as a supeposition of two fields: (1) that of the solid sphee of density ; (2) a solid sphee of negative density ". The densities cancel, giving the cavity. (It s a tick, of couse.) c. The dawing shows a close-up view of the cavity and an abitay field point in it, which is at y distance fom the cente of the solid sphee and at x a distance fom the cente of the cavity. By the law of cosines we have a 2
2 = 2 + a 2 " 2a cos#. Find the total potential at the field point, as a function of and, o of x and y. (You may lump all the additive constants in the potential into one.) d. Use g = "# to find the field at the field point. You should find that it is unifom, and entiely in the negative x-diection. e. Obtain the same esult diectly by using the answe to (a) and adding the g fields of the positive and negative density sphees. a. By the spheical symmety, g has the same magnitude at all points on the sphee suggested, and is pointing diectly inwad, so g nds = "g()ds. We have " g nds = #g()" ds = #g() 4 2. The othe side of the equation is 4"G times the mass enclosed by the sphee of adius, which is 4 ". We find theefoe g() = 4 G" #. Since the field is towad the cente, the vecto fomula is g() = 4 "G#. b. Since we chose the potential at infinity to be zeo, we must stat thee to find its value elsewhee. So we calculate () = " g # d " g # d. Fo the fist integal we use the field outside the sphee, which is g( > ) = GM = 4 "G #. Fo the second integal we use the answe to (a). The esult is 2 "G#(2 2 ). c. At the point shown the contibution of the sphee of positive density is 1 () = 2 "G#2 + const. That of the sphee of negative density is 2 ( ") = # 2 G " 2 + const. The total is = 2 "G#(2 2 ) + const. Using the law of cosines we find 2 " 2 = 2acos# a 2, and using the fact that cos = x we find = 4 "G#ax + const. d. Fom g = "# we find easily g = 4 "G#a i. This is in the negative x- diection (towad the cente of the solid sphee) and unifom in magnitude. e. We use the same dawing, with vectos indicating the diections of the contibutions. We have g = g 1 + g 2 = 4 G"( + #) = 4 G"a. Since a = ai, this is the same as (d). a
. A satellite has speed v p at peigee and v a at apogee. a. Find the eccenticity of the obit in tems of these speeds. b. It is desied to make a small incease in the enegy of the obit. To do this the engine can be fied in a shot bust poviding a foce F in any diection. Explain why the lagest incease in the enegy is accomplished by fiing the engine when the satellite is at peigee, and diecting the foce paallel to the satellite s velocity at that point. c. Let the enegy added be de, a small faction of the magnitude E of the oiginal obit enegy. Show that to fist ode in small quantities, the factional change da/a of the semi-majo axis is equal to de/ E. [Since the total enegy is negative, inceasing it means making it close to zeo.] d. Show that the obit becomes less cicula, i.e., that inceases. [Hint: thee is essentially no change in the peigee distance p duing the bief time while the engine is fied.] e. Suppose that when this satellite eaches apogee it encountes a second satellite moving in a cicula obit of adius equal to a. Which satellite will be moving faste? Explain. [You may assume they have the same mass, because the mass doesn t matte in detemining the speed at a cetain point.] a. By consevation of angula momentum, m p v p = m a v a, whee m is the satellite s mass, p is the peigee distance and a is the apogee distance. Thus v p /v a = a / p. Now by definition p = a(1 ") and a = a(1 + ), whee a is the semi-majo axis. So we find v p /v a = (1 + )/(1 " ). Solving, we find = v p " v a v p + v a. b. The powe supplied by a foce F acting on a paticle that is moving with velocity v is P = F v. Fo a given foce this is clealy lagest when v is lagest (which is tue at peigee) and when F is paallel to v. c. Use the fomula E = GMm 2a to find de = + GMm 2a 2 da da = E a. d. Easiest: p = a(1 "), so = 1 " p /a. Since p is fixed while a inceases, gets close to 1. e. The value of a fo the cicula obit is geate than fo the elliptical one, so its total enegy is geate (close to zeo). Since the two ae at the same point thei potential enegies ae equal (if the masses ae), so the kinetic enegy (and thus the speed) of the one in the cicula obit is lage. 4
4. The example woked out in the notes to Unit can be solved using odinay Newtonian mechanics if one uses effective gavity. y ecall the situation. A small block of x mass m slides without fiction down a y wedge of mass M, which is on a fictionless floo. The wedge also slides x acoss the floo. The figue shows the θ aangement. a. Let the acceleation of the wedge be a w. We will analyze the situation in a efeence fame moving with the wedge. In that fame, find the components of g eff in the x-y set of axes shown. b. The motion of the block is moe easily descibed in the x -y set of axes shown. Find the components of g eff in that set. [This involves otating the axes; see the notes on linea algeba.] c. Daw the fee body diagam showing the foces acting on the block (in the efeence fame we ae using). d. Find the acceleation of the block a b (in tems of a w and known quantities). e. Find the nomal foce exeted on the block by the wedge. f. If the fame we ae using, the wedge is at est. elate the nomal foce in (e) to the acceleation a w. [etun to the x-y axes fo this.] g. Solve fo a w and a b in tems of given quantities. [You should get the same answes as in the notes on Unit.] a. In vectos g eff = g a w, so we have g eff = a w i gj. In column fom: " g eff = # a w g & '. # b. The tansfomation matix is M = so we have # cos "sin sin & # cos ' ( "a w "g cos "sin sin & cos ' (, & # ' ( = "a w cos " gsin a w sin " gcos & ' (. In vecto notation g eff = (a w cos" gsin") i # + (a w sin" gcos") j #. 5
c. Dawing as shown. The component of g eff along the slope is the block s acceleation. The component of mg eff pependicula to the slope cancels the nomal foce N. d. The magnitude of the block s acceleation down the slope is a b = gsin + a w cos. e. We have fom the fee body diagam N = m(gcos " a w sin). f. The hoizontal component of the nomal foce exeted by the block on the wedge (the eaction to the one in the fee body diagam) is cancelled by the hoizontal component of Mg eff, so we have N sin = Ma w, and eaanging this we find (M + msin 2 )a w = mgsin cos. mgsin cos g. We have fom (f) a w = M + msin 2. Then fom (d) we find afte some M + m algeba a b = M + msin 2 gsin. These ae the same answes we got using the Lagangian appoach. N mg eff 5. Two questions about the Coiolis foce. a. In some locations on the eath s equato thee ae demonstations fo touists. Standing noth of the line, wate is obseved to otate counteclockwise as it goes down a dain; a few steps away, south of the line, the wate otates clockwise going down the dain. Is it the Coiolis foce that makes this happen? Explain. b. Thee wee two majo battles between the Geman and Bitish navies in Wold Wa I. The fist took place nea Jutland, about 55 N latitude, the last majo battle between lage battleships. The othe took place nea the Falkland Islands, about 50 S latitude. Conside a shell fied due noth in the Battle of Jutland, with initial speed v 0 = 800 m/s and elevation angle 45. Use a coodinate system with oigin whee the shell is fied fom the suface of the eath; let the x-axis un noth along the suface, and the y-axis be upwad pependicula to the suface. (The z-axis will un east along the suface.) 1. Wite (in components) the eath s angula velocity, the shell s velocity v s (t) (neglecting the Coiolis effect), and the acceleation a c (t) poduced by the Coiolis effect. 2. Find the contibution v c (t) of the Coiolis acceleation to the shell s velocity. 6
. Find the deflection c (t) caused by the Coiolis effect. 4. How long (in s) is the shell in the ai? (Neglect the Coiolis effect.) 5. Find the total deflection (in m) due to the Coiolis effect. a. No. The dains ae designed to poduce this diffeence, but many touists believe the scam. The Coiolis foce is exactly zeo at the equato, so it is negligbly small nea it. In fact it is too small to poduce any significant effect on wate going down dains anywhee. b. Shown is the situation. We see that without the Coiolis effect the motion would be in the x-y plane. 1. The shell s velocity is v s = v 0 cos " i + (v 0 sin # gt) " j, whee = 45. The eath s angula velocity is = cos" # i + sin" # j. The acceleation poduced by the Coiolis foce is a c = 2v s " = 2"[v 0 sin(# ) + cos# & gt)]&k. 2. We integate a c with espect to time to find v c = 2[v 0 sin(" # ) t + 1 2 cos" gt2 ]k.. Integate again to find the displacement. (The time of flight is T.) T c = v c dt = "[v 0 sin(# ) &T 2 + 1 gcos# &T ]&k. 0 4. The time of flight is twice the time to each maximum height, o T = 2v 0 sin g " 115 s. 5. Using = 2" /(24 # 600) 7.27 10 &5 s &1 and the given data we find c 40.7 m. This is a small angula deflection (the ange is about 65 km) but it is enough to make the shell miss its taget. x v 0 = 55 y 7