1. Several planets(the gas giants Jupiter, Saturn, Uranus and Neptune) possess nearly circular surrounding ring erhaps composed of material that failed to form a satellite. dM In addition, many galaxies contain ring-like structures Consider a homogeneous ring of mass M and radius R(a) Find an expression for the gra the ring on a particle of mass m located a distance x from e center of the ring along its axis. See Fig 3.(b) Suppose that the particle falls from rest as a result of the attraction Fig 3 of the ring of matter. Find an expression for the speed with which it passes through the center of the ring Solution (a)Set up the coordinate system shown in figure. Choose a pointlike differential masses dM shown as in figure 3, the field produced by dm is GdM GdM d g =-dg cose cos0 dg R2+x2 dg, -dgsin =GdM sine According to the symmetry of the ring-like structures 0 So the gravitational force is g= dg, i According to the graph, we get cos0=r GxdM R+x (R2+y2)3/2 Thus the gravitational force is g=dg i ('+r)/adMi=-GrM (b) Apply Newtons second law of motion, we have MMx GMx dy dv dx d Due to g (R2+x2)2 dt dx dt dx We can get the speed GMx 2GM( R√R1. Several planets (the gas giants Jupiter, Saturn, Uranus, and Neptune) possess nearly circular surrounding rings, perhaps composed of material that failed to form a satellite. In addition, many galaxies contain ring-like structures. Consider a homogeneous ring of mass M and radius R. (a) Find an expression for the gravitational force exerted by the ring on a particle of mass m located a distance x from the center of the ring along its axis. See Fig.3. (b) Suppose that the particle falls from rest as a result of the attraction of the ring of matter. Find an expression for the speed with which it passes through the center of the ring. Solution: (a) Set up the coordinate system shown in figure. Choose a pointlike differential masses dM shown as in figure 3, the field produced by dM is 2 2 d d R x G M g + = ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ + = = + = − = − ⇒ θ θ θ θ sin d d d sin cos d d d cos 2 2 2 2 R x G M g g R x G M g g y x According to the symmetry of the ring-like structures, d = 0 ∫ g y . So the gravitational force is g g i x d ˆ ∫ = v . According to the graph, we get 2 2 cos R x x + θ = , so 2 2 3 / 2 ( ) d d R x Gx M gx + = − Thus the gravitational force is i R x GxM Mi R x Gx g g ix ˆ ( ) d ˆ ( ) d ˆ 2 2 3 / 2 2 2 3 / 2 + = − + = = − ∫ ∫ v (b) Apply Newton’s second law of motion, we have i R x GMmx F mg ˆ ( ) 2 2 3 / 2 + = = − v v Due to x v v t x x v t v R x GMx g d d d d d d d d ( ) 2 2 3 / 2 = = = + = , We can get the speed ) 1 1 d 2 ( ( ) d 2 2 0 2 2 3 / 2 0 R R x x v GM R x GMx v v x v + ⇒ = − + = ∫ ∫ m R M Fig.3 x θ x dM y