Physics 35 Chapter 8 Chapter 8 Centra-Force Motion In this Chapter we wi use the theory we have discussed in Chapter 6 and 7 and appy it to very important probems in physics, in which we study the motion of two-body systems on which centra force are acting. We wi encounter important exampes from astronomy and from nucear physics. Two-Body Systems with a Centra Force Consider the motion of two objects that are effected by a force acting aong the ine connecting the centers of the objects. To specify the state of the system, we must specify six coordinates (for exampe, the (x, y, z) coordinates of their centers). The Lagrangian for this system is given by L = 1 m 1 1 r 1 + m r U ( r1 r ) Note: here we have assumed that the potentia depends on the position vector between the two objects. This is not the ony way to describe the system; we can for exampe aso specify the position of the center-of-mass, R, and the three components of the reative position vector r. In this case, we choose a coordinate system such that the center-of-mass is at rest, and ocated at the origin. This requires that R = The reative position vector is defined as m 1 m r 1 + r = 0 m 1 + m m 1 + m r = r 1 r The position vectors of the two masses can be expressed in terms of the reative position vector: r 1 = m m 1 + m r The Lagrangian can now be rewritten as r = m 1 m 1 + m r - 1 -
Physics 35 Chapter 8 L = 1 m m 1 m 1 + m r + 1 m m 1 m 1 + m r U ( r ) = 1 µ r U ( r ) where µ is the reduced mass of the system: µ = m 1m m 1 + m Two-Body Systems with a Centra Force: Conserved Quantities Since we have assumed that the potentia U depends ony on the reative position between the two objects, the system poses spherica symmetry. As we have seen in Chapter 7, this type of symmetry impies that the anguar momentum of the system is conserved. As a resut, the momentum and position vector wi ay in a pane, perpendicuar to the anguar momentum vector, which is fixed in space. The three-dimensiona probem is thus reduced to a twodimensiona probem. We can express the Lagrangian in terms of the radia distance r and the poar ange θ: L = 1 µ ( r + r θ ) U ( r) The generaized momenta for this Lagrangian are p r = L r = µ r p θ = L θ = µr θ The Lagrange equations can be used to determine the derivative of these momenta with respect to time: p r = d dt L r = L r = µr θ U r p θ = d dt L θ = L θ = 0 The ast equation tes us that the generaized momentum p θ is constant: - -
Physics 35 Chapter 8 = µr θ = constant The constant is reated to the area veocity. Consider the situation in Figure 1. During the time interva dt, the radius vector sweeps an area da where da = 1 r dθ The area veocity, da/dt, is thus equa to Figure 1. Cacuation of the area veocity. da dt = 1 dθ r dt = 1 r µr = µ = constant This resut is aso known as Keper's Second Law. The Lagrangian for the two-body system does not depend expicity on time. In Chapter 7 we showed that in that case, the energy of the system is conserved. The tota energy E of the system is equa to E = T +U = 1 µ ( r + r θ ) +U r = 1 µ r + 1 µr +U r ( ) µr ( ) = 1 µ r + r +U r ( ) = Two-Body Systems with a Centra Force: Equations of Motion If the potentia energy is specified, we can use the expression for the tota energy E to determine dr/dt: dr dt = ± µ E U ( ) µ r - 3 -
Physics 35 Chapter 8 This equation can be used to find the time t as function of r: t = dt = ± 1 µ E U r ( ( )) µ r dr However, in many cases, the shape of the trajectory, θ(r), is more important than the time dependence. We can express the change in the poar ange in terms of the change in the radia distance: dθ = dθ dt dt dr dr = θ r dr Integrating both sides we obtain the foowing orbita equation θ ( r) = θ r dr = ± µr ( µ E U ) dr = ± µ r r µ E U µr dr The extremes of the orbit can be found in genera by requiring that dr/dt = 0, or ( µ E U ) µ r = µ E U r ( ) µr = 0 In genera, this equation has two soutions, and the orbit is confined between a minimum and maximum vaue of r. Under certain conditions, there is ony a singe soution, and in that case the orbit is circuar. Using the orbita equation we can determine the change in the poar ange when the radius changes from r min to r max. During one period, the poar ange wi change by Δθ = r max r min r µ E U µr dr If the change in the poar ange is a rationa fraction of π then after a number of compete orbits, the system wi have returned to its origina position. In this case, the orbit is cosed. In a other cases, the orbit is open. - 4 -
Physics 35 Chapter 8 The orbita motion is specified above in terms of the potentia U. Another approach to study the equations of motion is to start from the Lagrange equations. In this case we obtain an equation of motion that incudes the force F instead of the potentia U: d 1 dθ r + 1 r = µr F( r) This version of the equations of motion is usefu when we can measure the orbit and want to find the force that produces this orbit. Exampe: Probem 8.8 Investigate the motion of a partice repeed by a force center according to the aw F(r) = kr. Show that the orbit can ony be hyperboic. The genera expression for θ(r) is [see Eq. (8.17) in the text book] θ ( r) = ( r )dr (8.8.1) µ E U µr where U = kr dr = kr in the present case. Substituting x = r and dx = rdr into (8.8.1), we have θ ( r) = 1 dx x µk x + µe (8.8.) x 1 Using Eq. (E.10b), Appendix E, dx = 1 bx + c x ax + bx + c c sin 1 (8.8.3) x b 4ac - 5 -
Physics 35 Chapter 8 and expressing again in terms of r, we find θ ( r) = 1 sin 1 µe r 1 µ E r + µk 4 + θ 0 (8.8.4) or, sin ( θ θ 0 ) = 1 1 + k µe 1 µe r 1 + k µe (8.8.5) In order to interpret this resut, we set 1 + k µe ε µe α (8.8.6) and specifying θ 0 = π/4, (8.8.5) becomes α r = 1 + ε cos θ (8.8.7) or, α = r + ε r ( cos θ sin θ) (8.8.8) Rewriting (8.8.8) in x-y coordinates, we find α = x + y + ε x y ( ) (8.8.9) or, - 6 -
Physics 35 Chapter 8 1 = x + y α α 1 + ε 1 ε (8.8.10) Since α ' > 0, ε ' > 1 from the definition, (8.8.10) is equivaent to 1 = x + α 1 + ε y α 1 ε (8.8.11) which is the equation of a hyperboa. Soving the Orbita Equation The orbita equation can ony be soved anayticay for certain force aws. Consider for exampe the gravitationa force. The corresponding potentia is -k/r and the poar ange θ is thus equa to ( ) = ± θ r / r µ E + k r µr dr Consider the change of variabes from r to u = /r: ( ) = ± θ r ( / r) / d u / µ E + k r 1 µ u = ± r µ E + k u 1 u du = µ u = ± 1 µ E + k u 1 µ u du The integra can be soved using one of the integras found in Appendix E (see E8.c): - 7 -
Physics 35 Chapter 8 ( ) = ± θ r k u + µ 1 u + µ k du = ±sin 1 u + µe µ k µ k = ±sin 1 u µ k + µe µk = ±sin 1 r µk ( ) + µ E + 8µE + C = µ k + C = ±sin 1 r + C = µ k + µe + C This equation can be rewritten as sin( θ + constant) = µk r ( µk) + µ E We can aways choose our reference position such that the constant is equa to π/ and we thus find the foowing soution: cos( θ) = µk r ( µk) + µ E We can rewrite this expression such that we can determine the distance r as function of the poar ange: r = µk ( µk) + µ E cos( θ) = µk 1 1+ µ k E cos( θ) Since cosθ varies between -1 and +1, we see that the minimum (the pericenter) and the maximum (the apocenter) positions are - 8 -
Physics 35 Chapter 8 r min = r max = µk 1+ 1+ µ k E µk 1 1+ µ k E The equation for the orbit is in genera expressed in terms of the eccentricity ε and the atus rectum α: ε = 1+ µ k E α = The possibe orbits are usuay parameterized in terms of the eccentricity, and exampes are shown Figure. µk Figure. Possibe orbits in the gravitationa fied. The period of the orbita motion can be found by integrating the expression for dt over one compete period: - 9 -
Physics 35 Chapter 8 τ = dt = µ da = µ πab ( ) = µ π k E µ E = π k µ E 3/ When we take the square of this equation we get Keper's third aw: µ τ = π k E 3 µ = π k a k 3 = 4π µ a 3 k The Centripeta Force and Potentia In the previous discussion it appears as if the potentia U is modified by the term /(µr ). This term depends ony on the position r since is constant, and it is interpreted as a potentia energy. The force associated with this potentia energy is ( ) F c = U c r = µr = µr θ = µ r θ 3 r This force is often caed the centripeta force (athough it is not a rea force), and the potentia is caed the centripeta potentia. This potentia is a fictitious potentia and it represents the effect of the anguar momentum about the origin. Figure 3 shows an exampe of the rea potentia, due to the gravitationa force in this case, and the centripeta potentia. The effective potentia is the sum of these two potentias and has a characteristic dip where the potentia energy has a minimum. The resut of this dip is that there are certain energies for which the orbit is bound (has a minimum and maximum distance). These turning points are caed the apsida distances of the orbit. Figure 3. The effective potentia for the gravitationa force when the system has an anguar momentum. - 10 -
Physics 35 Chapter 8 We aso note that at sma distances the force becomes repusive. Exampe: Probem 8. Discuss the motion of a partice moving in an attractive centra-force fied described by F(r) = k/r 3. Sketch some of the orbits for different vaues of the tota energy. For the given force the potentia is F ( r) = k r 3 U ( r) = k (8..1) r and the effective potentia is V ( r) = 1 µ k 1 (8..) r The equation of the orbit is [cf. Eq. (8.0) in the text book] d u dθ + u = µ ( ku 3 ) (8..3) u or, d u dθ + 1 µk u = 0 (8..4) Let us consider the motion for various vaues of. i) = µk : In this case the effective potentia V(r) vanishes and the orbit equation is - 11 -
Physics 35 Chapter 8 d u dθ = 0 (8..5) with the soution u = 1 r = Aθ + B (8.6) and the partice spiras towards the force center. ii) > µk : In this case the effective potentia is positive and decreases monotonicay with increasing r. For any vaue of the tota energy E, the partice wi approach the force center and wi undergo a reversa of its motion at r = r 0 ; the partice wi then proceed again to an infinite distance. Setting 1 µk β > 0 equation (8..4) becomes d u dθ + β u = 0 (8..7) with the soution u = 1 r = A cos ( βθ δ ) (8..8) Since the minimum vaue of u is zero, this soution corresponds to unbounded motion, as expected from the form of the effective potentia V(r). iii) < µk : For this case we set µk 1 G > 0-1 -
Physics 35 Chapter 8 and the orbit equation becomes d u dθ G u = 0 (8..9) with the soution u = 1 r = A cosh ( βθ δ ) (8..10) so that the partice spiras in towards the force center. Orbita Motion The understanding of orbita dynamics is very important for space trave. The orbit in which a spaceship traves is determined by the energy of the spaceship. When we change the energy of the ship, we wi change the orbit from for exampe a spherica orbit to an eiptica orbit. By changing the veocity at the appropriate point, we can contro the orientation of the new orbit. The Hofman transfer represents the path of minimum energy expenditure to move from one soar-based orbit to another. Consider trave from earth to mars (see Figure 4). The goa is to get our spaceship in an orbit that has apsida distances that correspond to the distance between the earth and the sun and between mars and the sun. This requires that and The eccentricity of such an orbit is thus equa to r 1 = a( 1 ε) r = a( 1+ ε) ε = r r 1 a The tota energy of an orbit with a major axis of a = (r 1 + r )/ is equa to E = k a = k ( ) r 1 + r - 13 -
Physics 35 Chapter 8 Since the space ship starts from a circuar orbit with a major axis a = r 1, its initia energy is equa to E = k r 1 Figure 4. The Hofman transfer to trave from earth to mars. The increase in the tota energy is thus equa to k ΔE = ( r 1 + r ) k r 1 = k 1 r 1 ( ) r 1 + r = k r r 1 r 1 r 1 + r ( ) This energy must be provided by the thrust of the engines that increase the veocity of the space ship (note: the potentia energy does not change at the moment of burn, assuming the thrusters are ony fired for a short period of time). The probem with the Hofman transfer mechanism is that the conditions have to be just right, and ony of the panets are in the proper position wi the transfer work. There are many other ways to trave between earth and mars. Many of these require ess time than the time required for the Hofman transfer, but they require more fue (see Figure 5). - 14 -
Physics 35 Chapter 8 Figure 5. Different ways to get from earth to mars. SECTIONS 8.9 AND 8.10 WILL BE SKIPPED! - 15 -