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Point mass model

If we assume there is no dark halo, then it is easy to find a relation between the circular velocity $ V_c$ and the escape speed $ v_e$. If we assume all mass $ M$ of the MW is interior to the solar radius $ R_\odot$ 7.3, then $ V_c^2=GM/R_\odot$, and the potential $ \Phi=-GM/R_\odot=-V_c^2$. The velocity of a star just able to escape from the gravitational well is the escape speed, $ v_e$. So since that star has zero total energy,

$\displaystyle 0=E={1\over 2}v_e^2+\Phi={1\over 2} v_e^2-V_c^2 .$ (7.1)

So in this case, $ v_e=2^{1/2} V_c\approx 311{\hbox{\rm km s$^{-1}$}}$ (Using $ V_c=220{\hbox{\rm km s$^{-1}$}}$.) So for a point mass model of the MW, these high velocity stars cannot be bound to the MW - suggestingthe presence of more mass in the outskirts of the MW than expected, given the rapid decrease in the distribution of light.



Tom Theuns 2003-04-28