Phys 7221, Fall 2006: Homework # 5

Transcription

1 Phys 722, Fal006: Homewor # 5 Gabriela González October, 2006 Prob 3-: Collapse of an orbital system Consier two particles falling into each other ue to gravitational forces, starting from rest at a istance a. The system has zero angular momentum, with the energy given by E T + V 2 mṙ2 r a where m is the reuce mass of the system, an r is the istance between the masses. Notice that the value of the energy, /a, calculate from the initial conition ṙ 0, r a, is not that of a Kepler s orbit, /2a, because l 0. We can erive an equation for r as usual: r 2 E V m 2 m r a ma rr 2 a r 2ma a u 2 u where we use the substitution u 2 a r, an use the fact that r/ < 0 to a a negative sign when taing the square root of ṙ 2. We can integrate the equation from the initial time when u 0, to the collapse time when u a, obtaining the time of the fall: 2ma a 2ma t 0 a u 2 πa ma u 4 π If the masses were in a circular orbit of raius a, the perio is τ 2π ma 3 /, so the time of the fall can be expresse as t 0 τ/4 2.

2 Prob 3-2: A moifie Kepler s potential Consier a central potential of the form V r /r + h/r 2. The orbit equation 3.34 for uθ /rθ is 2 u θ 2 + u m u V m 2 u θ 2 + u u + hu2 m + 2mh The solution to this equation is of the form u m u m + A cosβθ θ 0 2mh u with β 2 + 2mh/. This is the equation of a Kepler orbit parabola, ellipse or hyperbola in a coorinate system where the angular coorinate is θ βθ. A revolution aroun the origin sweeps a θ angle equal to 2π. If β,there are many raial oscillations in one revolution; if β, the orbit shows a slow precession. If the energy is negative an 2mh/, the orbit is a precessing ellipse. In a cycle of the perioic motion with perio τ, the raial coorinate returns to the original value when βθ θ 0 2π, or θ θ 0 2π β 2π 2π Ωτ + 2mh/ The precession spee is then Ω 2π τ 2πmh + 2mh/ τ This means orbit precession can be use as a test of Newton s theory for the gravitational force being erive from a potential /r. Using ma e 2, we obtain an expression for Ω in terms of the perturbation parameter of Kepler s potential η h/a, an orbital parameters: Ω 2πmh τ 2πmh ma e 2 τ 2πη e 2 τ The effect is more pronounce for eccentric an long orbits. The perihelion of Mercury is observe to precess after correcting for nown planetary perturbations by 43 arc-secons per century: 43 2π/360 /3600 ra Ω ra/yr 00yr 2

3 an thus η e2 τ Ω π This iscrepancy is also an better explaine by General Relativity. Prob 3-23: Mass ratio of Sun an Earth The perio an the semi-major axis of elliptical orbits in Kepler s potential are relate by τ/2π 2 a 3 µ/ where µ m m 2 /m + m 2 is the reuce mass of the system, an Gm m 2. When m m 2, we have τ/2π a 3 /Gm. For the Earth-Sun system, For the Earth-Moon system, Taing ratios, we obtain τes 2π 2 a 3 es GM s. τem 2 a 3 em. 2π GM e τes 2 a3 em M e M 3 s aes τem M e a em τ es τ em a 3 em M s The measure value, from is ±0.02; our estimate is within 3% of this value. Prob 3-24: Kepler s equation The energy in Kepler s motion is E 2 mṙ2 + l2 2mr 2 r For negative energy an elliptical orbits, the energy is E 2a, an the angular momentum is ma e 2, thus ṙ 2 2 E l2 m 2mr 2 + r 2 m 2a a e2 2r 2 + r r 2 mar 2 a 2 e 2 + 2ar a 2 mar 2 e 2 r a 2 3

4 Since the perio of the motion is τ 2π/ω, with ω /ma 3, we obtain ṙ 2 r mar 2 a 2 e 2 r a 2 ω 2 a2 r 2 a 2 e 2 r a 2 ω a r a 2 e 2 r a 2 aω which we can use to integrate tr. r ea sin ψψ we obtain r r a 2 e 2 r a 2 Using the orbit equation r a e cos ψ, an ω r r a a 2 e 2 r a 2 a e cos ψea sin ψψ a a 2 e 2 ae cos ψ 2 e cos ψ ψ which can be now trivially integrate into Kepler s equation: ωt ψ e sin ψ Prob 3-33: A particle in a paraboloi of revolution A particle with coorinates r x, y, z is constraine to move in a paraboloi f revolution. i.e., z r 2 /a x 2 + y 2 2 /a. We will use generalize polar coorinates r, θ to escribe the motion of the particle, with z r 2 /a. The inetic energy is The potential energy is The Lagrangian is T 2 mṙ2 + r 2 θ2 + ż 2 2 mṙ2 + r 2 θ2 + 4 r2 a 2 ṙ2 V mg r +mgz mg a r2 L T V 2 mṙ2 + 2 mr2 θ2 + 2m r2 a 2 ṙ2 mg a r2 We see that the coorinate θ is cyclic, so the z-component of the angular momentum is conserve associate with the symmetry of rotation about the z-axis: mr 2 θ l constant 4

5 Lagrange s equation for the coorinate r is mṙ + 2m r2 a 2 ṙ mr θ 2 4m rṙ2 a 2 + 2mg a r r2 a 2 r l2 m 2 r 3 + 2g a r 0 There are solutions for circular orbits, with r 4 r0 4 l2 a/2gm 2. If the orbit is circular with raius r 0, the angular momentum is relate to the raius an the spee, l mr 0 v. We can then fin the conition between the spee an the raius for circular orbits: v 2 2gr0 2 /a. If the orbit is only approximately circular, we fin an approximate equation for the perturbation δr r r 0 : + 2 r 0 + δr 2 δr a r2 0 a 2 m 2 r 3 0 δr 2g a δr + δr/r 0 3 2g a r 0 + δr This equation has a perioic solution, with perio τ 2π/ω, with ω 2 2g/a/+2r 2 0 /a2. 5