2. A mass M is in the shape of a thin uniform disk of radius R. Let the =-axis represent the symmetry axis of the disk as indicated in Figure 4. The Milky Way Galaxy is modeled as such a mass disk to a first approximation de (a)Find the gravitational field of the disk at a coordinate along the symmetry axis of the disk? Fig 4 (b) What is the expression for g for >>R in part(a)? (c)Let o be the surface mass density of the disk(the number of kilograms per square meter of the disk, so that as↓ zr. what is the gravitational field in part (a)as :>0 along the positive c-axis? olution (a)The mass Mis a uniform disk, so the surface mass density of the disk ISoM Choose a circular differential mass dM shown in figure. dM= 2mrdr.o The gravitational field established by the ring at a coordinate z is GzdM Gzo·2mdr Thus the gravitational field of the disk at a coordinate along the symmetry axis of the disk is -k =>0 g==-=+7)yk M -1-,k:<0 (2+R2) (b) For =>>R, the disk can be treated as a point mass, so 8= (c)In part(a) 2GM R When 20, the gravitational field is 8=-2GMK R2. A mass M is in the shape of a thin uniform disk of radius R. Let the z-axis represent the symmetry axis of the disk as indicated in Figure 4. The Milky Way Galaxy is modeled as such a mass disk to a first approximation. (a) Find the gravitational field of the disk at a coordinate z along the symmetry axis of the disk? (b) What is the expression for g r for z >> R in part (a)? (c) Let σ be the surface mass density of the disk (the number of kilograms per square meter of the disk), so that 2 R M π σ = . What is the gravitational field in part (a) as z → 0 along the positive z-axis？ Solution: (a) The mass M is a uniform disk, so the surface mass density of the disk is 2 R M π σ = . Choose a circular differential mass dM shown in figure, dM = 2πrdr ⋅σ The gravitational field established by the ring at a coordinate z is k z r Gz r r k z r Gz M g ˆ ( ) 2 d ˆ ( ) d d 2 2 3 / 2 2 2 3 / 2 + ⋅ = − + = − v σ π Thus the gravitational field of the disk at a coordinate z along the symmetry axis of the disk is ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ < + − − − > + − − = + ⋅ = = − ∫ ∫ 0 ˆ] ( ) 2 [ 1 0 ˆ] ( ) 2 [1 ˆ ( ) 2 d g d 2 2 2 1/ 2 2 2 2 1/ 2 0 2 2 3 / 2 k z z R z R M G k z z R z R M G k z r Gz r r g v v R σ π (b) For z >> R , the disk can be treated as a point mass, so k z GM g ˆ 2 = − r (c) In part (a) ( ) k R z z R GM g ˆ 1 2 2 1 2 2 2 ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ + = − − r , When z → 0 , the gravitational field is k R GM g ˆ 2 2 = − r 0 z R M k ˆ Fig.4 dθ dr r