Comments
Description
Transcript
#13 , 2014 14
#13 , 2014 14 F~ = 3x2 yx̂ + (x3 + 1)ŷ + 9z 2 ẑ .x = 2 .x2 + y 2 > 1 , x2 + y 2 + z 2 = 9 x2 + y 2 + z 2 = 4 S .z ,F~ = (y, x, z) .S .1 . . .2 S .3 .n̂ = (x, y, z)/2 F~ · n̂ = = = .1 (x, y, z) z2 (y, x, z) · = 2 2 ¨ ˆ 2⇡ ˆ 5⇡/6 4 cos2 ✓ F~ · n̂dS = 4 sin ✓d✓d = 2 S 0 ⇡/6 ˆ 5⇡/6 5⇡/6 p cos3 ✓ 16⇡ cos2 ✓ sin ✓d✓ = 16⇡ = 4 3⇡ 3 ⇡/6 ⇡/6 (1) (2) (3) F~ · n = yx . ~ · F~ r = = .2 xy = 0 n̂ = (x, y, 0) ˜ ˝ ~ · F~ dV , , S F~ · n̂dS = V r ˚ ˚ ~ · F~ dV = 1) r dV = V = V V ¨ ¨ p F~ · n̂dS + F~ · n̂dS = 4 3⇡ S .3 (4) (5) cylinder a, b ,F~ = (ax2 y + y 3 + 1)x̂ + (2x3 + bxy 2 + 2)ŷ , . .F~ = rf a, b , f (x, y) .x = et cos t, y = et sin t, 0 t ⇡ C 1 ´ C F~ ·d~r .1 .2 .3 ,@x Fy = @y Fx @ x Fy @ y Fx = = .1 6x2 + by 2 2 ax + 3y (6) 2 (7) (8) ) a = 6, b = 3 .2 f (x, y) = ˆ fy0 (x, y) = 2x3 + 3xy 2 + c0 (y) = Fy = 2x3 + 3xy 2 + 2 ) c(y) = 2y + const Fx dx + c(y) = x f (x, y) = 3 ˆ (6x2 y + y 3 + 1)dx + c(y) = 2x3 y + xy 3 + x + c(y) (9) 3 2x y + xy + x + 2y + const ,( e⇡ , 0) , ˆ C . : F~ · d~r = f ( e⇡ , 0) = 2x v = J = x+y 1 ux uy vx vy dudv = V = = = y)2 + (x + y .z = (2x u f (1, 0) = 1)2 e ⇡ (1, 0) C .3 z=4 (13) (14) = 2 1 1 1 (15) =3 1 3dxdy ) dxdy = dudv 3 ¨ (4 (2x y)2 (x + y (2x y)2 +(x+y 1)2 <4 ¨ 1 (4 u2 v 2 ) dudv = 3 u2 +v 2 <4 ˆ 2⇡ ˆ 2 ˆ 2⇡ 2 2 1 4 2 1 (4 r ) rdrd✓ = r r 3 3 12 0 0 0 (16) 1)2 )dxdy = 2 d✓ = 0 ˜ R R C (17) (18) ˆ 0 .F~ = .x2 + y 2 + z 2 = 1 (11) (12) 1 y .x2 + y 2 + z 2 = 1 (10) 2⇡ 4 8⇡ d✓ = 3 3 (19) 2xz x̂ + y 2 ẑ ~ ⇥ F~ .r .1 ~ ⇥ F~ · n̂dS = 0 r ¸ F~ · d~r = 0 .2 C .3 .1 x̂ @x 2xz ~ ⇥ F~ = r ŷ @y 0 ẑ @z y2 = 2yx̂ (20) 2xĵ .2 n̂ = ~ ⇥ F~ ) · n̂ (r = (21) (x, y, z) (2y, 2x, 0) · (x, y, z) = 0 ) ¨ R ~ ⇥ F~ · n̂dS = 0 r ˜ ¸ ~ ⇥ F~ · n̂dS = R r F~ · d~r , R C ˜ ¸ ~ ⇥ F~ · n̂dS = . Rr F~ · d~r = 0 C C .f (x, y) = x + 4y + (22) .3 , 2 xy .1 2 =0 x2 y 2 =0 xy 2 fx = 1 fy = 4 2 ) xy 2 = )x = 2 1 2 4y ) 1 y= , x=2 2 x y = (23) (24) (25) (26) (27) (28) (29) .(x, y) = (2, 1/2) .2 fxx = A = 4 4 2 , fyy = 3 , fxy = fyx = 2 2 x3 y xy x y fxx (2, 1/2) = 1, C = fyy (2, 1/2) = 16, B = fxy (2, 1/2) = 2 ) AC 2 B = 12 > 0, A > 0 ) (2, 1/2)IS MINIMUM (1.02)2 · (1 p (30) (31) (32) 3.98) f (x, y) = x2 (1 p y) (33) .x = 1.02, y = 3.98 f (x, y) ⇡ f (1, 4) = fx0 (1, 4) = f (x0 , y0 ) + fx0 (x0 , y0 )(x x0 ) + fy0 (x0 , y0 )(y p 12 (1 4) = 1 2 = 1 p 2x(1 y)|(1,4) = 2 2 fy0 (1, 4) = f (1.02, 3.98) ⇡ (x0 , y0 ) = (1, 4) x p |(1,4) = 2 y 1 2(1.02 1 4 1) y0 ) (34) (35) (36) (37) 1 (3.98 4 4) = 1.035 (38) Gauss law for gravity. Let ~g be the gravitational field. The following equation is satisfied by the field ~ · ~g (~r) = r 4⇡G⇢(~r) where G is the universal gravitational constant, and ⇢(~r) is the mass density at each point of space, carrying units of mass/volume. Using the terminology we learned in class, the mass (multiplied by 4⇡G) is the source for the gravitational field, and the divergence of the field is the source density. 1. For an arbitrary closed surface S show that the following holds for the gravitational field: ¨ ~g (~r) · n̂(~r) dS = 4⇡GMS S where MS is the total mass enclosed within the surface S. ~ g (~r) = 0, and it goes to zero at infinity. Using previous 2. Gravity is a conservative force, i.e. r⇥~ results (question 1), prove Newtonian gravity: show that for a point mass M located at the origin the following holds r̂ ~g (~r) = GM 2 r Hint: assume from symmetry considerations that the gravitational field is radial, i.e. that ~g (~r) = g(r)r̂, and choose the surface S wisely. 3. Repeat the previous question for a spherical mass of constant mass density ⇢: what is the gravitational field as function of the distance from the center of the mass? (use the result from question 1). 4. The Earth is hardly point-like; however, let us assume it is approximately a sphere. Its density is not constant either - however we can assume it is constant within each layer; we shall further assume that it changes linearly with the distance from the center of the Earth, where the density is maximal. The density at the crust is approximately 3 g/cm3 . What is the density at the core? (the mass of the earth is 5.97 · 1024 Kg and its radius is 6378 Km). 5. Using the previous results (questions 1 and 4), calculate the gravitational field along a straight line starting at the center of the earth and ending in an airplane some 10km above China. Draw a graph showing the magnitude of the gravitational field |~g (rr̂)| as function of r, the distance from the center of the earth. Indicate your units on your x-axis and y-axis. 6. USS Enterprise encounters an infinite cylindrical shape of constant radius R and a constant mass density ⇢ (within the region of space 1 < z < 1, r R), pulling the starship towards it! Captain Kirk’s orders: “Mr. Spock, quickly calculate the gravitational field as function of the distance from the center of the shape! Both for r < R and r > R” (assume you are Mr. Spock). 7. In Terry Pratchett’s Discworld the world is an infinite, perfectly flat disc, of constant mass density (carrying units of mass/area). Calculate the gravitational field as function of the distance r away from the disc. What should be so that g = 9.8 m/s2 ? (Hint: use a surface that is a cylinder of radius R extending both above and below the disc, r z r, where the disc is at z = 0) . You can neglect the elephants and the turtle. Solutions to Ex 8 ˜ ˝ ~ ·~g dV = 1. S ~g (~r) · n̂(~r) dS = V r within S. 4⇡G ˝ 4⇡GMS where MS is the mass trapped ⇢(~r) = V 2. Take S to be a sphere of radius r, so that n̂ = r̂. Since ~g (~r) = g(r)r̂ we get ¨ g(r)r̂ · r̂ = 4⇡r2 g(r) = 4⇡GM (39) S and therefore g(r) = 3. For r > R same as (2), with M = 4⇡ 3 3 R ⇢. 1 r2 GM (40) For r < R we get 4⇡r2 g(r) = 4⇡G 4⇡ 3 r ⇢ |3{z } (41) M (r) and therefore 4⇡G⇢ r 3 g(r) = (42) 4. We take (⇢c - density at core, ⇢s -density at surface) ⇢(r) = ⇢c (1 (43) ↵r) We are given that ⇢s = ⇢(R) = ⇢c (1 ↵R) (44) The mass of the earth is ME = ˆ R 2 ⇢(r)4⇡r dr = 0 ˆ R ↵r)4⇡r2 dr ✓ ◆ 4⇡ 3 3 ⇢c R 1 ↵R 3 4 ⇢c (1 (45) 0 = and therefore 4⇡ ME = ⇢s R3 3 Plugging in 1 1 3 4 ↵R ↵R = 1 1 ME 4⇡ 3 3 ⇢s R ! (47) = 1.831 (48) 3R 4 ↵ R↵ and hence ⇢c = 1 (46) ↵R = 0.769 (49) ⇢s ' 13g/cm3 0.769 (50) 5. First r < R: The mass within a ball of radius r is 4⇡ 3 r 3 M (r) = ⇢c r 1 (↵R) 3 4 R (51) and using Gauss’ law for a spherical Gauss surface g(r)4⇡r2 we get g(r < R) = = 4⇡GM (r) 4⇡ G ⇢c r 1 3 3 r (↵R) 4 R (52) (53) For r > R it is the same as (3). 6. We take a cylindrical Gauss surface of height h and radius r. Notice that the two flat surfaces at the top and bottom do not contribute as the normal is orthogonal to the field. For r < R: (2⇡r)hg(r) = 4⇡G ⇢(⇡r2 )h | {z } (54) M (r) and hence g(r < R) = For r > R: (2⇡r)hg(r) = 2⇡G⇢r 4⇡G ⇢(⇡R2 )h | {z } (55) (56) M (R) and therefore g(r > R) = 2⇡G⇢ R2 r (57) #14 , 2015 19 x = f (t), y = g(t), ti t tf .(f (tf ), g(tf )) (f (ti ), g(ti )) x = t2 , y = t + 1, 1<t<1 , . , x = t2 = (y 1)2 . . . a . 1 P P . t P x = at + a cos ✓, y = a + a sin ✓ t + ✓ = 3⇡/2 x = y = t ✓ 3⇡ t) = at a sin t 2 3⇡ a + a sin( t) = a + a sin t 2 at + a cos( .#7 S : I C F~ (~r) · d~r = ZZ S ~ ⇥ F~ ) · n̂dS (r ,S C . , ,S . n̂ C , I Fx dx + Fy dy = C ZZ ✓ D @Fy @x , @Fx @y ◆ dxdy 3 ~v (~r) = (ex sin y 3y + x2 , ex cos y x2 + y 2 = 2x . 1, 0) R C C ,2014 ~v (~r) · d~r , f (~r0 + ~r) = f (~r0 ) + ⇣ ⌘2 ⇣ ⌘n ~ (~r0 ) + 1 ~ f (~r0 ) + ... 1 ~ ~r · rf ~r · r ~r · r f (~r0 ) + ... 2! n! 2014 p 2 .f (x, y) = sin(x) + sin(y)2 .(x = 320 , y = 320 ) , y 00 (t) + p(t)y 0 (t) + q(t)y(t) = g(t) y(t) = c1 y1 (t) + c2 y2 (t) Y (t) = u1 (t) = u2 (t) = W (y1 , y2 ) = u1 (t)y1 (t) + u2 (t)y2 (t) Z y2 (t)g(t) dt W (y1 , y2 ) Z y1 (t)g(t) dt W (y1 , y2 ) y1 y20 y10 y2 2014 ẍ(t) + 4ẋ(t) + 4x(t) = et x(0) = 0 ẋ(0) = 1