EP 222: Classical Mechanics Tutorial Sheet 3 This tutorial sheet contains exercises related to the central force problem. 1. Suppose a satellite is moving around a planet in a circular orbit of radius r0. Due to a collision with another object, satellite’s orbit gets perturbed. Show that the radial l position of the satellite will execute simple harmonic motion with ! = 2 , where l is mr0 the initial angular momentum of the satellite. Soln: Because it is a small perturbation, we can Taylor expand the potential energy of the satellite around r0 = rmin 0 2 0 0 dV 1 2 d V V (r) = Vmin + (r − rmin) + (r − rmin) + ··· dr 2 dr2 r=rmin r=rmin Noting that 0 dV = 0 dr r=rmin 2 0 2 2 4 4 3 3 3 4 d V 3l 2k 3l m k m k m k = − = − 2k = dr2 mr4 r3 m l8 l6 l6 r=rmin min min But, balance of force condition for the circular orbit yields mv2 k = 2 ; rmin rmin 2 from which, using the fact that v = l=mrmin, we obtain k = l =mrmin. Therefore, 2 0 3 8 2 d V m l l = = dr2 l6 m4r4 mr4 r=rmin min min but rmin = r0 2 0 l 2 V (r) ≈ Vmin + 4 (r − r0) 2mr0 Radial equation of motion of the perturbed orbit dV 0 (r) l2 mr¨ = − = − 4 (r − r0) dr mr0 Define x = r − r0, we obtain from above x¨ + !2x; l where ! = 2 , and equation above denotes simple harmonic motion about r = r0, mr0 with frequency !. 1 C 2. A particle of mass m is moving under the influence of a central force F(r) = − r3 ^r, with C > 0. Find the nonzero values of angular momentum l for which the particle will move in a circular orbit. Soln: For this, the potential energy can be obtained as Z r Z r 0 r 0 0 dr C C V (r) = − F (r )dr = C 03 = − 2 = − 2 : 1 1 r 2r 1 2r The effective potential energy for this case 2 0 l C V (r) = − : 2mr2 2r2 We know that for the circular orbit, the total energy must be equal to the minimum of the effective potential energy, which can be found by dV 0 (r) l2 C = − 3 + 3 = 0 dr pmr r =) l = mC: Thus, if the system has this angular momentum, circular orbit of any radius is possible. 3. In the lectures, we obtained the equation of the orbit (an equation connecting r and θ) for the Kepler’s problem (V (r) = −k=r), by solving a second order differential equation for the variable u = 1=r. Show that one gets the same result if one directly integrates the integral connecting the r and θ coordinates, derived in the lectures. Soln: In the lectures it was shown that the equation of the orbit is given by u Z du θ = θ0 − q ; 2mE 2mV 2 2 − 2 − u u0 l l where u = 1=r. For the Kepler problem V (r) = −k=r =) V (u) = −ku, so that the integral becomes u Z du θ = θ0 − q : (1) 2mE 2mku 2 2 + 2 − u u0 l l Integral of Eq. (1) can be performed using the standard integral Z dx 1 β + 2γx = p cos−1 − p ; (2) pα + βx + γx2 −γ q where, q = β2 − 4αγ, and identifying 2mE 2mk α = ; β = ; γ = −1; l2 l2 2 we obtain in Eq. (1) 2 8 93 2mk −1 < l2 − 2u = θ − θ0 = − cos 4− q 5 ; : 4m2k2 8mE ; l4 + l2 which can be simplified to 1 mk 0 = 1 + e cos(θ − θ ) ; r l2 where r 2El2 e = 1 + : mk2 Clearly, the orbit is a conic section of eccentricity e. 4. Two particles move about each other in circular orbits under the influence of gravita- tional forces, with a period τ. Their motion is suddenly stopped at a given instant of time, and they are thenp released and allowed to fall into each other. Prove that they collide after a time τ=4 2. Solution: Since the two bodies are interacting under a central force (gravitational force), their motion can be seen as that of a single particle of mass µ = mim2 , about mi+m2 the center of mass. So, here we have for circular orbits Figure 1: µv2 k = 2 ; r0 r0 s k v = ; r0µ So, the time period 2πr τ = 0 v 3 r r3µ τ = 2π 0 ; (3) k 1 kτ 2 3 r = : 0 4π2µ If the particle are momentarily stopped at this position and then released, the law of conservation of energy for subsequent motion tell us (motion is purely radial) 1 k k µr_2 − = − ; 2 r r0 s dr 2k 1 1 = − − ; dt µ r r0 negative sign for radial speed implies that radial distance is decreasing with time. Thus the time taken for the particle to collide with each other is 0 r µ Z dr T = − r ; 2k 1 1 r0 − r r0 0 p r µ Z rdr T = − r0 p ; 2k r0 − r r0 2 substituting r = r0 sin θ and dr = 2r0 sin θ cos θdθ. So that π 0 p 2 1 Z Z 2 rdr r0 sin θ2r0 sin θ cos θdθ − p = 1 ; r0 − r r 2 cos θ r0 0 0 π Z2 2 = 2r0 sin θdθ; 0 π Z2 = r0 (1 − cos 2θ) θdθ; 0 π 1 2 = θ − sin 2θ r0 ; 2 0 π = r : 2 0 So, r µ π T = r r ; 2k 0 2 0 4 π rµ T = p r3; 2 2 k 0 using Eq. 1, we get τ T = p : 4 2 5. Show that the motion of a particle in the potential field k h V (r) = − + r r2 is the same as that of the motion under the Kepler potential alone when expressed in terms of a coordinate system rotating or precessing around the center of force. For negative total energy, show that if the additional potential term is very small compared to the Kepler potential, then the angular speed of precession of the elliptical orbit is 2πmh Ω_ = : l2τ The perihelion of Mercury is observed to precess (after correction for known planetary perturbations) at the rate of 4000 of arc per century. Show that this precession could be accounted for classically if the dimensionless quantity h η = ka (which is a measure of the perturbing inverse-square potential relative to the gravi- tational potential) were as small as 7 × 10−8. (The eccentricity of Mercury’s orbit is 0.206, and its period is 0.24 year.) Solution: The equation of the orbit is given by u Z du θ = θ0 − q ; 2mE 2mV 2 2 − 2 − u u0 l l since k h V (r) = − + ; r r2 V (u) = −ku + hu2; So from above we get u Z du θ = θ0 − q ; 2mE 2mku 2mh 2 2 + 2 − 2 + 1 u u0 l l l but we know that Z dx 1 β + 2γx = p cos−1 − p ; pα + βx + γx2 −γ q 5 where, q = β2 − 4αγ: Here, we have 2mE 2mk α = ; β = ; l2 l2 2mh + l2 γ = − ; l2 so, 4m2k2 8mE (2mh + l2) q = β2 − 4αγ = + ; l4 l4 4m2k2 + 16m2Eh + 8mEl2 q = : l4 So, we get 2 l −1 2mk − 2 (2mh + l ) u θ = θ0 − p cos −p ; 2mh + l2 4m2k2 + 16m2Eh + 8mEl2 2 2 3 2mh+l u − 1 l −1 mk θ = θ0 − p cos 4−q 5 ; 2mh + l2 2E 2 mk2 (2mh + l ) + 1 2 r 2 2mh+l 3 1 + 2mh mk u − 1 (θ − θ ) = − cos−1 ; 2 0 4q 5 l 2E 2 mk2 (2mh + l ) + 1 or by further simplifying we get, 1 = c [1 + cos αθ] ; (4) r where 2mk c = ; 2mh + l2 s 2E (2mh + l2) = 1 + ; mk2 r 2mh α = 1 + : l2 Eq. 2 is the equation of an ellipse, where the particle makes an angle θ from a fixed direction in space, but it makes an angle αθ from the semi-major axis. 6 Figure 2: Clearly, this corresponds to a semi-major axis that is rotating in space, or the ellipse is precessing. This is can be written as αθ(t) = θ(t) − Ω_ t; where, Ω_ is angular velocity of precession of the semi-major axis. αθ_ = θ_ − Ω_ ; r ! 2mh Ω_ = (1 − α) θ_ = θ_ 1 − 1 + ; l2 for small h, θmh_ Ω_ = : l2 For small h, Ω_ will be small and hence we can get a good estimate of it by putting average value of θ_, i.e. 2π θ_ = ; τ 2πmh Ω_ = : l2τ Because h is small, we can approximate by e, the eccentricity of the perfect ellipse r 2El2 ≈ e = 1 + ; mk2 and using the fact that for perfect elliptical orbit E = −k=2a, we obtain from above m 1 = ; l2 (1 − 2) ka leading to 2π h 1 Ω_ = ; τ ka 1 − 2 7 2π 1 Ω_ = η ; τ 1 − 2 Ω_ (1 − 2) τ η = ; 2π where = 0:206, τ = 0:24 year and Ω_ = 4000 per century. By substituting all these above, we obtain η = 7:0 × 10−8: 6. A geostationary orbit is one in which a satellite moves in a circular orbit at the given height in the equatorial plane, so that its angular velocity of rotation around earth is same as earth’s angular velocity, thereby, making it look stationary when seen from a point on equator. Assuming that the earth’s rotational velocity, and radius, respec- 2π tively, are Ωe = 86400 rad/s, andRe = 6400 km, calculate the altitude of the satellite, and its orbital velocity.