From: keithstein111@gmail.com   
      
   On 03/05/2018 08:03, Phillip Helbig (undress to reply) wrote:   
   > In article , Keith Stein   
   >  writes:   
   >   
   >> I suspect there could be a simple relationship of the form:   
   >   
   > On what grounds, other than just trying to fit the data?   
   >   
   >>       ( constant orbital speed )^2 = K * mass of galaxy   
   >>   
   >> I arrive at this by assuming that for spiral galaxies there is, in   
   >> addition to the usual Newtonian inverse square gravitation, an   
   >> additional inverse linear relationship.   
   >   
   > This means that there is a logarithmic potential, which doesn't go to   
   > zero at infinity.  Difficult.   
   >   
   >> At normal distances the linear term is clearly very small and can be   
   >> ignored, but obviously at very large distances the linear term will   
   >> dominate, and at sufficiently large distances the inverse square term   
   >> can be ignored.   
   >>   
   >> At large distance from the galactic center (R) we may therefore write:   
   >>   
   >>          Centripetal Force = m * v^2 / R = K * M * m / R   
   >>   
   >> Which gives:                   v^2     = K * M   
   >>   
   >> Note v is independent of R, so this would nicely explain the   
   >> flat orbital velocity curves of spiral galaxies.   
   >   
   > The empirical relationship, though, is v^4 = K * M, where K=GMa_0 and   
   > a_0 is the empircally determined MOND acceleration.  It is approximately   
   > equal to the speed of light times the Hubble constant. No-one knows what   
   > this means.   
      
   Thank you Mr.Helman.  When i first read this I thought it implied the   
   empirical relationship did indeed match my suggestion, but I have noted   
   your correction to K = G*a_0.   
      
   I suppose I could now suggest a linear gravitational force of the form:   
      
                           F = K' * M^.5 * m / R   
      
   and since i was "just trying to fit the data", this is perhaps no worse   
   than my original suggestion, and  would yield the required:   
      
   			V^4 = K * M   
      
   with the additional information (thanks to Mr.Helman) that:   
      
                            K  = K'^2  ~=  G * H * c   
      
   Thanks again,   
      
   keith stein   
      
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