Question

For this problem assume that the stars in the Milky Way are in circular orbits in the plane of the galaxy with speedy(r), where r measures the distance from the galactic center. 1. The inner portions of the Milky Way rotate as a solid body, meaning that the angular velocity ω is constant. ωΞω0 What does this imply about how the orbital (rotational) velocity as a function of distance from the galactic cen Using Newton's laws of motion (balance centripetal and gravitational a. ter? b. acceleration-a=GM (r)) show that this 1s consistent with a spherically 3a) symmetric mass distribution of constant density, p 2. In the outer portions of the Milky Way, by contrast, the rotation curve is flat meaning that the orbital velocity is constant, vVo a. What does this imply about the angular velocity ω as a function of distance from the galactic center? spherically symmetric mass distribution with density, p- HINT: If you assume Mp even though p is not constant, you will be off in your answer by a factor of 3. Instead find* and use the fact that b. Again using Newton's laws of motion, show that this is consistent with a 4TGr2 dM dr dM

Answer 1

(a) For the inner portion, w = v_theta/r = constant v_theta proportional r Hence, orbital velocity increases (proportional to r) as we go away from the Galactic centre. v_theta proportional r (b) v^2_theta/r = G M (r)/r^2 w^2 r^2/r = G M (r)/r^2 w^2 r^3 = GM (r)... (1) if the inner portion of the Galaxy has a spherically symmetric mass distribution, with constant density rho_0, then M (r) = 4/3 pi r^3 rho from (1) w^2 r^3 = G 4/3 pi r^3 rho_0 rho_0 3 w^2_0/4 pi G 2) In the outer portion, v_theta = v_0 = constant Hence from w = v_theta/r, w proportional 1/r, w will fall off on (1/r) on we go away from Galactic center. From Newton's law, v^2_0/r = G M (r)/r^2 M (r) = r v^2_0/h d M (r)/dr = v^2_0/G For a spherical distribution, d M (r) = 4 pi r^2 rho (r) dr