Baixe Resolução Marion [Mec.Clássica] - Capítulo 8 e outras Manuais, Projetos, Pesquisas em PDF para Mecânica, somente na Docsity! CHAPTER 8 Central-Force Motion 8-1. x3 m2 m1 x2 r1 r2 x1 In a uniform gravitational field, the gravitational acceleration is everywhere constant. Suppose the gravitational field vector is in the x direction; then the masses and have the gravitational potential energies: 1 1m 2m ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 1 1 1 1 2 2 2 2 1 2 1 g g U F x m x U F x m x α α = − = −   = − = −  (1) where ( ) ( ) ( )( )1 1 11 1 2 3, ,x x x=r and where α is the constant gravitational acceleration. Therefore, introducting the relative coordinate r and the center of mass coordinate R according to ( ) 1 2 1 1 2 2 1 2m m m m = −   + = +  r r r r r R (2) we can express r and r in terms of r and R by 1 2 2 1 1 2 1 2 1 2 m m m m m m = + +   = − + +  r r r r R R (3) 233 234 CHAPTER 8 which differ from Eq. (8.3) in the text by R. The Lagrangian of the two-particle system can now be expressed in terms of r and R: ( ) ( ) ( ) ( ) 2 2 1 2 1 1 2 2 2 2 2 1 1 2 1 2 1 2 2 1 1 2 1 2 1 2 1 1 2 2 1 1 2 2 g gL m m U U U m m m m U r m m m m m m m x X m x m m m m α α = + − − − + − + − + + X = +    + + + −  +  + +    r r r r R r R  (4) where x and X are the components of r and R, respectively. Then, (4) becomes 1x ( ) ( ) ( ) 2 22 22 1 1 2 1 1 2 1 2 1 2 1 2 1 2 1 2 1 1 1 2 m m L m m m m m m m m m m m m U r x m m X m m α α     = + +   + +    − − + + + + r r 2+ R2 2 (5) Hence, we can write the Lagrangian in the form ( ) ( ) ( )22 1 2 1 2 1 1 2 2 r m m m mL U Xµ α= − + + + +r R (6) where µ is the reduced mass: 1 2 1 2 m m m m µ = + (7) Therefore, this case is reducible to an equivalent one-body problem. 8-2. Setting 1u = r , Eq. (8.38) can be rewritten as 2 2 2 2 2 du E k u u θ µ µ = − + − ∫ (1) where we have used the relation ( )21du r dr= − . Using the standard form of the integral [see Eq. (E.8c), Appendix E]: 1 2 2 1 2 sin const. 4 dx ax b aax bx c b ac −  − +=  −+ + −  +∫ (2) we have 2 1 2 2 2 2 2 const. sin 2 8 k r k E µ θ µ µ −    − +  + =     +     (3) CENTRAL-FORCE MOTION 237 2 2 2 1 2 2 T r r µ µ = − + (10) and the time average is 2 2 3 0 0 1 1 2 k T T dt Tr a τ πµ dθ τ π = − =∫ ∫ (11) Part of this integral is trivial, ( ) 2 2 2 3 0 1 2 2 k T rr d a πµ µ π θ π µ   = +    ∫ (12) To evaluate the integral above, substitute the expression for r and make a change of variable ( ) ( ) ( ) 2 22 2 12 2 2 2 0 0 1 sin 1 2 2 11 cos d x rr d x π π θ θε ε θ µ µ εε θ − −    = =       ++ ∫ 2 dx1 1 ∫ ∫ (13) The reader is invited to evaluate this integral in either form. The solution presented here is to integrate by parts twice, which gives a third integral that can be looked up in a table: ( ) ( ) ( ) 1 1 12 2 2 2 1 11 1 1 1 11 1 1 x dx x dxx xx x xε ε εε ε− −− − − = − − ++ − + ∫∫ (14) ( ) 1 11 1 2 11 sin sin1 1 1 x x x d x xε ε ε − − −− x   = − − + +   ∫ (15) ( ) ( ) ( ) ( ) 1 1 1 2 2 1 1 11 2 sin tan 1 11 x x x ε ε εε − − −  − − = − + + + −  (16) 2 2 1 1 1 π ε ε   = − +  −  (17) Substituting this into (13) and then into (12), we obtain the desired answer, 2 k T a = (18) This explicitly verifies the virial theorem, which states that for an inverse-square law force, 1 2 T = − U (19) 238 CHAPTER 8 8-5. a m1 m2 Suppose two particles with masses and move around one another in a circular orbit with radius a. We can consider this motion as the motion of one particle with the reduced mass µ moving under the influence of a central force 1m 2m 2 1 2G m m a . Therefore, the equation of motion before the particles are stopped is 2 1 22 m m a G a µ ω = (1) where 1 2 1 1 1 m mµ = + , 2π ω τ = (2) The radius of circular motion is 1 32 1 2 24 G m m a τ π µ   =     (3) After the circular motion is stopped, the particle with reduced mass µ starts to move toward the force center. We can find the equation of motion from the conservation of energy: 21 2 1 2 1 2 m m m m G x G a x µ− = − (4) or, 1 2 1 22 1 1G m mx x aµ   = −     (5) Therefore, the time elapsed before the collision is 0 1 22 1 1a dx t dt G m m x aµ = = −  −   ∫ ∫ (6) where the negative sign is due to the fact that the time increases as the distance decreases. Rearranging the integrand, we can write 0 1 22 a a x t G m m a x µ = − −∫ dx (7) Setting ( )2 2x y dx y dy≡ = , the integral in (7) becomes 0 0 2 2 2 a a yx I dx a x a y = = − − ∫ ∫ dy CENTRAL-FORCE MOTION 239 Using Eq. (E.7), Appendix E, we find 0 2 12 sin 2 2 a y a y ya a a 2 I π− −   + = −     (8) = − Therefore, 1 22 2 a a t G m m µ π = or, 4 2 t = τ (9) 8-6. m1 m2 x1 x2O r r x x= −2 1 When two particles are initially at rest separated by a distance , the system has the total energy 0r 1 20 0 m m E G r = − (1) The coordinates of the particles, and , are measured from the position of the center of mass. At any time the total energy is 1x 2x 2 2 1 21 1 2 2 1 1 2 2 m m E m x m x G r = + − (2) and the linear momentum, at any time, is 1 1 2 2 0p m x m x= + = (3) From the conservation of energy we have E 0E= , or 2 21 2 1 21 1 2 2 0 1 1 2 2 m m m m G m x m x G r r − = + − (4) Using (3) in (4), we find 1 1 2 0 1 2 1 0 2 1 1 2 1 1 G x v m M r r G x v m M r r   = = −      = = − −    (5) 242 CHAPTER 8 ( ) 2 2 2 1 2 2 1 dx r k E x x x θ µ µ = + − ∫ (2) Using Eq. (E.10b), Appendix E, 1 2 1 2 sin 4 dx bx c cx ax bx c x b ac − 2  + =   −+ + −   ∫ (3) and expressing again in terms of r, we find ( ) 2 2 1 02 2 2 4 2 1 1 sin 2 E r r E k r µ θ θ µ µ −   −   =   +     + (4) or, ( ) 2 0 22 2 2 1 n 2 1 1 E rk k E E µ θ θ µ µ 1 − = − + + 2 si (5) In order to interpret this result, we set 2 2 2 1 k E E ε µ α µ  + ≡ ′     ≡ ′   (6) and specifying 0 4θ π= , (5) becomes 2 1 cos 2r α ε θ′ = + ′ (7) or, ( )2 2 2 2cos sinr rα ε θ= + −′ ′ θ (8) Rewriting (8) in x-y coordinates, we find ( )2 2 2 2x y x yα ε= + + −′ ′ (9) or, 22 1 1 1 yx α α ε ε = +′ ′ + −′ ′ (10) Since a′ > 0, ε′ > 1 from the definition, (10) is equivalent to CENTRAL-FORCE MOTION 243 22 1 1 1 yx α α ε ε = +′ ′ + −′ ′ (11) which is the equation of a hyperbola. 8-9. (a) By the virial theorem, 2U= −T for a circular orbit. The firing of the rocket doesn’t change U, so f iU U= But ( )2 21 2 2f i T m v v= + = T So 2 0f i i i iE T U U U= + = − + = 0f i E E = The firing of the rocket doesn’t change the angular momentum since it fires in a radial direction. 1f i = (b) E = 0 means the orbit is parabolic. The satellite will be lost. ( ) ( )0 e sGM mE r U r r = = − ( ) e sGM mT r E U r = − = ( ) ( ) 2 2 2 22 2 e sGM mV r U r r r rµ µ = + = − + Behavior of V(r) is determined by 2 22 for small for large e s r r GM m r r µ   − 244 CHAPTER 8 Energy V(r) U(r) T(r) E(r) = 0 r0 Minimum in V(r) is found by setting 0 dV dr = at 0r r= 2 2 3 0 0 0 e s GM m r rµ = − + 2 0 e s r GM mµ = − 8-10. For circular motion 2 2 1 2 e e T m ω= r s e e GM m U r = − We can get 2ω by equating the gravitational force to the centripetal force 22 s e e e e GM m m r r ω= or 2 3 s e GM r ω = So 2 3 1 1 2 2 s es e e e e GM mGM T m r r r = ⋅ = = − 2 U 1 2 E T U U= + = If the sun’s mass suddenly goes to 1 2 its original value, T remains unchanged but U is halved. CENTRAL-FORCE MOTION 247 3 2 2 E E E k r τ µ π ′ = (7) Substituting (7) into (5), we find ( )2 2 2 1 1 3 2 E E k T k τµ β β µ π ′ = + − (8) where sk GM µ= and s Ek GM µ=′ . Therefore, ( ) ( )1 2 1 1 2 3 E T β β τ π = − + (9) where 1 yearEτ = . Now, Mercury 0.387Er rβ = = . Therefore, ( ) ( )1 2 1 0.387 1 2 0.387 365 days 3π + × ×T = − so that 76 daysT = (10) 8-13. Setting 1≡u we can write the force as r 22 3 k 3F ku u r r λ λ= − − = − − (1) Then, the equation of orbit becomes [cf. Eq. (8.20)] ( ) 2 2 2 2 2 1d u u ku d u µ λ θ + = − − − 3u (2) from which 2 2 21 d u k u d 2 µλ µ θ  + − =   (3) or, 2 2 2 2 2 1 1 1 d u k u d µλ µ µλθ     0+ − −     −   = (4) If we make the change of variable, 2 2 1 1 k v u µ µλ= − − (5) we have 2 2 21 d v v d µλ θ   0+ −   = (6) 248 CHAPTER 8 or, 2 2 2 0 d v v d β θ + = (7) where 2 21β µλ= − . This equation gives different solutions according to the value of λ. Let us consider the following three cases: i) 2λ µ< : For this case and the solution of (7) is 2 0β > ( )cosv A βθ δ= − By proper choice of the position θ = 0, the integration constant δ can be made to equal zero. Therefore, we can write 2 1 cos k A r µβθ µλ = + − (9) When β = 1 (λ = 0), this equation describes a conic section. Since we do not know the value of the constant A, we need to use what we have learned from Kepler’s problem to describe the motion. We know that for λ = 0, ( )2 1 1 cos k r µ ε θ= + and that we have an ellipse or circle (0 ≤ ε < 1) when E < 1, a parabola (ε = 1) when E = 0, and a hyperbola otherwise. It is clear that for this problem, if E ≥ 0, we will have some sort of parabolic or hyperbolic orbit. An ellipse should result when E < 0, this being the only bound orbit. When β ≠ 1, the orbit, whatever it is, precesses. This is most easily seen in the case of the ellipse, where the two turning points do not have an angular separation of π. One may obtain most constants of integration (in particular A) by using Equation (8.17) as a starting point, a more formal approach that confirms the statements made here. ii) 2λ µ= For this case and (3) becomes 2 0β = 2 2 d u k d 2 µ θ = (10) so that 22 1 2 k u r µ θ θA B= = + + (11) from which we see that r continuously decreases as θ increases; that is, the particle spirals in toward the force center. CENTRAL-FORCE MOTION 249 iii) 2λ µ> For this case and the solution (7) is 2 0β < ( )2coshv A β θ δ= − − (12) δ may be set equal to zero by the proper choice of the position at which θ = 0. Then, ( )2 21 cosh kAr µβ θ µλ = − + − (13) Again, the particle spirals in toward the force center. 8-14. The orbit equation for the central-force field is [see Eq. (8.17)] 2 4 2 2 2 2 dr r E U d r µ θ µ 2    = − −      (1) But we are given the orbit equation: 2r kθ= (2) from which 2 2 24 dr k d θ θ   =   (3) Substituting (2) into (3), we have 2 24 4 dr r k d kθ   = =   kr (4) From (1) and (4), we find the equation for the potential U: 4 2 2 2 4 2 r kr E U r µ µ 2   = − −    (5) from which 2 2 3 2 1 2 k U E r rµ µ = − − 2 1 (6) and ( )F r U r= −∂ ∂ . Therefore, ( ) 2 4 3 6 1k F r r rµ  = − +   (7) 252 CHAPTER 8 3 1 2 22 2 2 E b EE E E E b δ δ δ δ µ µ µ   δ δ   = = + − = −    − −    −  (7) Easily enough, we can show that v vδ δ= and therefore 2a b E a b E v vδ δ δ δ = = = − (8) 8-17. The equation of the orbit is 1 cos r α ε θ= + (1) from which 1 cos r α ε θ = + (2) where 2 kα µ= and 2 2 2 1 E mk ε = + . Therefore, the radial distance r can vary from the maximum value ( )1α ε− to the minimum value ( )1α ε+ . Now, the angular velocity of the particle is given by 2r ω µ = (3) Thus, the maximum and minimum values of ω become max 22 min min 22 max 1 1 r r ω µ αµ ε ω µ αµ ε = =    +     = =       −   (4) Thus, 2 max min 1 1 n ω ε ω ε + = = −  (5) from which we find 1 1 n n ε − = + (6) 8-18. Kepler’s second law states that the areal velocity is constant, and this implies that the angular momentum L is conserved. If a body is acted upon by a force and if the angular momentum of the body is not altered, then the force has imparted no torque to the body; thus, CENTRAL-FORCE MOTION 253 the force must have acted only along the line connecting the force center and the body. That is, the force is central. Kepler’s first law states that planets move in elliptical orbits with the sun at one focus. This means the orbit can be described by Eq. (8.41): 1 cos with 0 r 1 α ε θ ε= + < < (1) On the other hand, for central forces, Eq. (8.21) holds: ( ) 2 2 2 2 1 1d r F r d r r µ θ   + = −   (2) Substituting 1 from (1) into the left-hand side of (2), we find r ( )22 1 r F r a µ = − (3) which implies, that ( ) 2 2F r rαµ = − (4) 8-19. The semimajor axis of an orbit is defined as one-half the sum of the two apsidal distances, and [see Eq. (8.44)], so maxr minr [ ]max min 2 1 1 2 2 1 1 r r 1 α α α ε ε ε  + = + = + − −  (1) This is the same as the semimajor axis defined by Eq. (8.42). Therefore, by using Kepler’s Third Law, we can find the semimajor axis of Ceres in astronomical units: 2 2 2 2 4 4 C C C C EE E E k a ka τ π µ τ π µ      =      (2) where c sk M mcγ= , and 1 1 1 c sM mµ = + c Here, sM and are the masses of the sun and Ceres, respectively. Therefore, (2) becomes cm 1 32 C s c c E s e E a M m a M m τ τ   + =   +    (3) from which 254 CHAPTER 8 ( ) 1 3 2 1 333, 480 8,000 4.6035 333, 480 1 C E a a  +  =   +     (4) so that 2.767C E a a ≅ (5) The period of Jupiter can also be calculated using Kepler’s Third Law: 1 22 3 1 23 2 3 4 4 J J J J Js E EE s E E a k aM m M m a a k π µ τ π µτ     J E   + = =     +        (6) from which ( ) 1 2 3333, 480 1 5.2028 333, 480 318.35 J E τ τ +  =  +  (7) Therefore, 11.862J E τ τ ≅ (8) The mass of Saturn can also be calculated from Kepler’s Third law, with the result 95.3s e m m ≅ (9) 8-20. Using Eqs. (8.42) and (8.41) for a and r, we have 44 2 0 1 cos1 cos cos 1 a dt r τ ε θ θ θ τ ε +   =     − ∫ (1) From Kepler’s Second Law, we can find the relation between t and θ: ( ) 2 2 1 2 1 cos dt dA d ab ab τ τ α θ π π ε θ = = + (2) since ( ) 21 2dA r dθ= . Therefore, (1) becomes ( ) ( 4 22 2 42 0 1 1 cos cos 1 cos 21 a a d r ab πτ )θ θ ε θ τ πε = − ∫ θ+    (3) It is easily shown that the value of the integral is 2πε. Therefore, CENTRAL-FORCE MOTION 257 In order to investigate the stability of a circular orbit in a 3r1 force field, we return to Eq. (8.83) and use ( ) 3g r k rµ= . Then, we have ( ) ( ) 2 32 3 31 1 k x xµ ρ ρ µρ ρ − = − 3x   + +     (11) or, ( ) 2 33 1 0 1 x k xµ µρ ρ   + − ⋅ =     +  (12) Since 0r pr = = , Eq. (8.87) shows that 2k µ= . Therefore, (12) reduces to 0x = (13) so that the perturbation x increases uniformly with the time. The circular orbit is therefore not stable. We can also reach the same conclusion by examining the basic criterion for stability, namely, that 2 20 and 0 r r V V r rρ ρ= = ∂ ∂ = > ∂ ∂ The first of these relations requires 2k µ= while the second requires 2 kµ > . Since these requirements cannot be met simultaneously, no stable circular orbits are allowed. 8-23. Start with the equation of the orbit: 1 cos r α ε θ= + (1) and take its time derivative 2 sin sin r r r2 ε ε θ θ α αµ = = θ (2) Now from Equation (8.45) and (8.43) we have 2 2 2 1 a ab µ πµ α τ π ε = ⋅ = − (3) so that from (2) max 2 2 1 a r ε π ε µ α τ ε = ⋅ = − (4) as desired. 258 CHAPTER 8 8-24. r (b) (a) θ b ra rp a) With the center of the earth as the origin, the equation for the orbit is 1 cox r α ε θ= + (1) Also we know ( )min 1r a ε= − (2) ( )max 1r a ε= + 6min 300 km 6.67 10 mp er r r= = + = × r r 6max 3500 km 9.87 10 ma er= = + = × ( ) 61 8.27 10 m2 a pa r r= + = × Substituting (2) gives ε = 0.193. When θ = 0, min 1 a r ε= + which gives . So the equation of the orbit is 67.96 10 mα = × 67.96 10 m 1 0.193 cos r θ × = + When θ = 90°, 67.96 10 mr α= = × T he satellite is 1590 km above the earth. b) b a – rmin β θ CENTRAL-FORCE MOTION 259 1 min 1 min tan Using tan 101 b a r b a a a r θ π β π α α θ π − − = − = − − = = − ° − Substituting into (1) gives ; which is 68.27 10 mr = × 1900 km above the earth 8-25. Let us obtain the major axis a by exploiting its relationship to the total energy. In the following, let M be the mass of the Earth and m be the mass of the satellite. 2 1 2 2 p p GMm GMm E mv a r = − = = (1) where pr and pv are the radius and velocity of the satellite’s orbit at perigee. We can solve for a and use it to determine the radius at apogee by 1 2 2 2a p p p p GM r a r r r v 1 −   = − = −     (2) Inserting the values 11 2 26.67 10 N m kgG − −= × ⋅ ⋅ (3) (4) 245.976 10 kgM = × (5) 66.59 10 mpr = × 37.797 10 m spv 1−= × ⋅ (6) we obtain , or 288 km above the earth’s surface. We may get the speed at apogee from the conservation of angular momentum, 61.010 6.658 10 ma pr r = × a a p pmr v mr v= (7) giving . The period can be found from Kepler’s third law 127,780 km hrav −= ⋅ 2 3 2 4 at GM π = (8) Substitution of the value of a found from (1) gives τ = 1.49 hours. 262 CHAPTER 8 ( ) ( ) ( ) ( ) max max min min min max min max max min 0 1 1 1 1 2 2 r ve r e v v e v e v v v v e v v e + = = − + = − + = − = 0 v e v = 8-30. To just escape from Earth, a velocity kick must be applied such that the total energy E is zero. Thus 22 1 0 2 eGM mmv r − = (1) where 2 24 11 2 2 6 6 velocity after kick 5.98 10 kg 6.67 10 Nm /kg 200 km 200 km 6.37 10 m 6.57 10 m e e v M G r r − = = × = × = + = + × = × Substituting into (1) gives v . 2 11.02 km/sec= For a circular orbit, the initial velocity v is given by Eq. (8.51) 1 1 7.79 km/sec eGMv r = = Thus, to escape from the earth, a velocity ck of 3.23 km/sec must be applied.ki CENTRAL-FORCE MOTION 263 Since E = 0, the trajectory is a parabola. parabolic escape orbit circular orbit Earth 8-31. From the given force, we find ( ) ( ) 3 2 4dF r k k F r dr r r5 ′ = = +′ (1) Therefore, the condition of stability becomes [see Eq. (8.93)] ( ) ( ) ( ) ( ) 2 5 2 4 2 2 3 01 k k F F k k ρ ρ ρ ρ ρ ρρ ρ + ′ ′ 3 + = − + ′ + > (2) or, ( ) 2 2 0 k k k k ρ ρ ρ − ′ > + ′ (3) Therefore, if , the orbit is stable. 2k kρ > ′ 8-32. For this force, we have ( ) ( ) 3 2 3 2 2 r a r a r a dF r k k F r e e dr r ar k r e r a − − − = = +′  = +   (1) Therefore, the condition of stability [see Eq. (8.93)] becomes ( ) ( ) 2 3 3 0 r F r a F r r r  − + + ′  + = > (2) This condition is satisfied if r < a. 264 CHAPTER 8 8-33. The Lagrangian of the particle subject to a gravitational force is written in terms of the cylindrical coordinates as ( )2 2 2 21 2 L T U m r r z mgzθ= − = + + − (1) From the constraint , we have 2 4r a= z 2 rr z a = (2) Therefore, (1) becomes 2 2 2 2 2 1 1 2 4 4 mgr L m r r r a a θ    = + + −      2 (3) Lagrange’s equation for θ is ( )2 0L d L d mr dt dt θ θ θ ∂ ∂ − = − = ∂ ∂ (4) This equation shows that the angular momentum of the system is constant (as expected): (5) 2 const.mr θ = = Lagrange’s equation for r is 2 2 2 2 14 2 4 mgL d L m d r rr mr r m r r dt r a a dt a θ   ∂ ∂ − = + − − + =  ∂ ∂    2 0 (6) from which 2 2 2 2 2 214 2 4 2 mgm r rr mr r m r rr a a a a θ   + − − + − =    2 0 m (7) After rearranging, this equation becomes 2 2 2 2 2 3 1 1 4 4 2 mgr m rr r a a a m r   + + + −    0=m r (8) For a circular orbit, we must have 0r r= = or, r = ρ = constant. Then, 2 32 mg a m ρ ρ = (9) or, 2 2 2 m g a 4ρ= (10) Equating this with 2 2 4m 2ρ θ= , we have 2 2 4 2 4 2 m g m a ρ θ = ρ (11) CENTRAL-FORCE MOTION 267 8-35. If we write the radial distance r as , cor x nst.ρ ρ= + = (1) then x obeys the oscillatory equation [see Eqs. (8.88) and (8.89)] 20 0x xω+ = (2) where ( ) ( )0 3g g ρ ω ρ ρ = + ′ (3) The time required for the radius vector to go from any maximum value to the succeeding minimum value is 0 2 t τ ∆ = (4) where 0 0 2π τ ω = , the period of x. Thus, 0 t π ω ∆ = (5) The angle through which the particle moves during this time interval is 0 t πω φ ω ω = ∆ = (6) where ω is the angular velocity of the orbital motion which we approximate by a circular motion. Now, under the force ( ) ( )F r g rµ= − , ω satisfies the equation ( ) ( )2 F r gµρω µ ρ= − = (7) Substituting (3) and (7) into (6), we find for the apsidal angle ( ) ( ) ( ) ( )( ) 0 3 3 g g g g g ρ π ρπω π ω φ ρ ρ ρ ρ ρ ρ = = = ′ + +′ (8) Using ( ) 1n k g r rµ = , we have ( ) ( ) g n g ρ ρ ρ ′ = − (9) Therefore, (8) becomes 3 nφ π= − (10) 268 CHAPTER 8 In order to have the closed orbits, the apsidal angle must be a rational fraction of 2π. Thus, n must be 2, 1, 6,n = − − … n = 2 corresponds to the inverse-square-force and n = –1 corresponds to the harmonic oscillator force. 8-36. The radius of a circular orbit in a force field described by ( ) 2 r akF r e r −= − (1) is determined by equating F(r) to the centrifugal force: 2 2 ak e r m ρ 3ρ − = (2) Hence, the radius ρ of the circular orbit must satisfy the relation 2 ae mk ρρ − = (3) Since the orbit in which we are interested is almost circular, we write ( ) ( )[ ]1r θ ρ δ θ= + (4) where ( ) 1δ θ for all values of θ. (With this description, the apsides correspond to the maximum and minimum values of δ.) We can express the following quantities in terms of δ by using (4): (1 1 1u r )δ ρ = = − (5a) 2 2 2 1 1d d d r d 2 δ θ ρ θ   = −   (5b) ( ) ( ) ( 12 22 1 1 au a F u ku e k e a ρ ρδ ρ δ − − = − ≅ − − + ) (5c) Then, substitution into Eq. (8.20) yields ( ) ( 2 2 2 1 1 1 1 ad mke )p a d ρδ δ ρ θ ρ − − = − δ− + (6) Multiplying by ρ, using (3) and simplifying, (6) reduces to ( ) 2 2 1 d a d δ ρ δ θ 0+ − = (7) CENTRAL-FORCE MOTION 269 This equation obviously has two types of solution depending on whether aρ is larger than or smaller than 1; we consider only ρ < a. (In fact, there is no stable circular orbit for ρ > a.) For the initial condition, we choose 0δ δ= to be a maximum (i.e., an apside) at θ = 0. Then, we have ( )1 20 cos 1 , for aδ δ ρ θ ρ= − a< (8) This solution describes an orbit with well-defined apsides. The advance of the apsides can be found from (8) by computing for what value of θ is δ again a maximum. Thus, 2 1 a π θ ρ = − (9) The advance of the apside is given by ( ) 1 22 2 1 1 aθ π π ρ − ∆ = − = − −  (10) In the particular case in which ρ a we obtain, by extending (10), 2 2 1 2a ρ π π  ∆ ≅ − +   (11) so that a πρ ∆ ≅ (12) 8-37. From the equations in Section 8.8 regarding Hohmann transfers: ∆ = 1 2v v v∆ + ∆ ∆ = 1 21 2t t v v v v v− + − 2 1 1 2 1 2 2 1 2 2 rk k k k v mr r r mr mr mr r r     − + −  + +    12 r ∆ = (1) Substituting ( ) ( )11 2 2 24 1 2 6 6.67 10 Nm /kg 5.98 10 kg initial height above center of Earth 2 final height above center of Earth 3 radius of the Earth 6.37 10 m e e e e k GM m r r r r r −= = × × = = = = = = × gives 1020 m/sv∆ 272 CHAPTER 8 1 2 1 1 1 2 1 2 ;t rk k v v mr r r mr   = = +  2 1 1 2 1 2 2 2 ;t rk k v v mr r r mr   = = +  Substituting ( ) ( )11 2 2 24 6 5 1 8 2 6.67 10 Nm /kg 5.98 10 kg 200 km 6.37 10 m 2 10 m mean Earth-moon distance 3.84 10 m e e k GM m r −= = × × = + = × + × = = r r × gives 3966 m/sv∆ = From Eq. (8.58), the time of transfer is given by 3 2 3 2 1 2 2t r rm m T a k k τ π + = =    Substituting gives 429,000 sec. 5 daysτ = 8-42. r = × 11 2 2 24 5 6 1 2 8 3 6.67 10 Nm /kg 5.98 10 kg 2 10 m 6.37 10 m ? mean Earth-moon distance 3.84 10 m e G M r r −= × = × + × = = = × We can get from Kepler’s Third Law (with τ = 1 day) 2r 1 32 7 2 2 4.225 10 m4 eGMr τ π   = = ×    We know 2E GMm= − r CENTRAL-FORCE MOTION 273 So ( ) ( ) ( ) 11 1 1 10 2 9 3 3.04 10 J 2 4.72 10 J 5.19 10 J eGM mE r r E r E r = − = − × = − × = − × To place the satellite in a synchronous orbit would require a minimum energy of ( ) ( )2 1E r E r− = 112.57 10 J× 8-43. In a circular orbit, the velocity v of satellite is given by 0 2 0 02 mv GMm GM v R R R = ⇒ = where M is the Earth’s mass. Conservation of energy implies 2 2 1 2 2 2 mv mvGMm GMm R R − = − 2 Conservation of angular momentum gives 1 22mRv m Rv= From these equations, we find 1 4 3 GM v R = so the velocity need to be increased by a factor 4 3 to change the orbit. 8-44. The bound motion means that 2 0 2 mv E V= + < where r a k r −= −V e . The orbit of particle moving in this central force potential is given by 274 CHAPTER 8 ( ) ( ) min min 2 2 2 2 2 2 / 2 2 1 2 2 r r r ra r r dr r E V r dr r ke E r r θ µ µ µ µ − =   −   = + − ∫ ∫ In first order of , this is ( / )r a ( ) 2 2 2 2 2 2 2 2 2 2 r dr dr k k k k r E r E r r a a r r µ µ θ µ µ  + − − − + −   ∫ ∫≈ = Now effectively, this is the orbit of particle of total energy k E a − moving in potential  k r − . It is well known that this orbit is given by (see Chapter 8) 1 cos r α ε θ= + where 2 k α µ = and 2 2 2 1 k E k a ε µ  = + −   If 0 1ε< < , the orbit is ellipsoid; if 0ε = , the orbit is circular. 8-45. a) In equilibrium, for a circular orbit of radius r0, 2 00 0 0 F F m r mrφ φ ω ω= ⇒ = b) The angular momentum (which is conserved) of a particle in circular orbit is 20L mr φω= 3 0mFr= The force acting on a particle, which is placed a distance r (r is very close to equilibrium position ) from the center of force is or ( ) ( ) ( ) 3 2 0 03 3 2 2 0 0 0 03 2 4 0 0 0 3 3 L F m r F F mr L L L r r F r r k r r mr mr mr φω= − = − ≈ − − − = − − = − −