Classical Mechanics LECTURE 16: ORBITS : CENTRAL FORCES Prof. N. Harnew University of Oxford HT 2017 1 OUTLINE : 16. ORBITS : CENTRAL FORCES 16.1 Central force: the equation of motion 16.2 Motion under a central force 16.2.1 Motion in a plane 16.2.2 Sweeping out equal area in equal time 16.3 Central force : the total energy 16.3.1 The potential term (inverse square interaction) 16.3.2 Example 2 16.1 Central force : the equation of motion I Recall the acceleration in polar coordinates a = ¨r = (¨r − r θ_2) ^r + (2r_ θ_ + r θ¨) ^θ I If F = f (r) ^r only, then Fθ = 0 ! Fθ = m(2r_ θ_ + r θ¨) = 0 2 ! Fr = m(¨r − r θ_ ) = f (r) d 2 _ _ _ 2 ¨ I Consider dt (r θ) = 2rr θ + r θ 1 d 2 _ Hence r dt (r θ) = 0 ! (r 2θ_) = constant of motion I The angular momentum in the plane : J = m r × v = m r × (r_ ^r + r θ_ ^θ) = (mr 2 θ_)n ^ where r × ^r = 0 and n^ = ^r × ^θ dJ I Torque about origin : τ = dt = r × F = 0 (F acts along r) Angular momentum vector is a constant of the motion 3 16.2 Motion under a central force 16.2.1 Motion in a plane I J = m r × v I Angular momentum is always perpendicular to r and v I J is a constant vector ; J · r = 0 ; J · v = 0 Motion under a central force lies in a plane 4 16.2.2 Sweeping out equal area in equal time GMm I jF j = Central force example : planetary motion : r r 2 I Angular momentum is conserved ! jJj = mr 2 θ_ = constant 1 2 I dA ≈ 2 r dθ dA 1 2 _ I dt = 2 r θ dA J nd dt = 2m = constant (Kepler 2 Law) Orbit sweeps out equal area in equal time 5 16.3 Central force : the total energy I Total energy = kinetic + potential : 1 2 E = T + U(r) = 2 mv + U(r) = constant 2 I v = r_^r + rθ_^θ ! jvj = (^r + rθ_^θ) · (^r + rθ_^θ) ! jvj2 = r_ 2 + r 2θ_2 (since ^r · ^θ = 0) 1 1 _ 2 2 _2 I E = 2 mr + 2 mr θ + U(r) I No external torque: angular momentum is conserved ! jJj = mr 2 θ_ = constant 2 1 _2 J E = 2mr + 2mr 2 + U(r) I Potential energy for a central force U(r) = − R r F · dr = − R r f (r) dr rref rref 6 16.3.1 The potential term (inverse square interaction) A A I F = − ^r ! ( ) = − r 2 f r r 2 [Attractive force for A > 0 ! signs are important !] R r I U(r) = − F · dr rref = − R r f (r) dr rref A A I U(r) = − + r rref Usual to define U(r) = 0 at rref = 1 A ! U(r) = − r F = − GMm ^r ! ( ) = − GMm Newton law of gravitation : r 2 U r r 7 16.3.2 Example A projectile is fired from the earth’s surface with speed v at an angle α to the radius vector at the point of launch. Calculate the projectile’s subsequent maximum distance from the earth’s surface. Assume that the earth is stationary and its radius is a. 8 16.3.2 Example : solution GMm I U(r) = − r I jJj = mjr × vj = mav sin α 1 2 J2 I = _ + + ( ) Energy equation : E 2 mr 2mr 2 U r 2 2 2 ! = 1 _2 + ma v sin α − GMm E 2 mr 2r 2 r 1 2 GMm _ I At r = a : E = 2 mv − a . At maximum height : r = 0 1 2 GMm ma2v 2 sin2 α GMm ! mv − = 2 − (1) 2 a 2rmax rmax 2 2GM 2 2 2 2 ! v − a rmax + 2GM rmax − a v sin α = 0 I Solve and take the positive root I Note from Equ.(1) : When r_ ! 0 as rmax ! 1 , the rocket just escapes the earth’s gravitational field q 1 2 GMm 2GM i.e. 2 mv − a ! 0 ; vesc = a (independent of α) 9.