Q: re: simple Newtonian System-Mass estimator ?

[Widdekind]

From simplistic assumptions of circular orbits, dominant central masses, etc., one can derive, from Newton's laws, Kepler's equations, including:

 

[math]\frac{\left( \frac{r}{1 \, AU} \right)^3}{\frac{m}{M_{\odot}}} = \left( \frac{t}{1 \, yr} \right)^2[/math]

Now, knowing that [math]1 \, ly \approx 64,000 \, AU[/math], cubing that value on the LHS, moving the result (~2.6e14) to the RHS, and calculating the square root, wht:

 

[math]\frac{\left( \frac{r}{1 \, ly} \right)^3}{\frac{m}{M_{\odot}}} = \left( \frac{t}{16 \, Myr} \right)^2[/math]

Thus, if the mass of the Local Group is roughly [math]5 \times 10^{12} M_{\odot}[/math] (PF), and if its 'effective radius' is roughly [math]3 Mly[/math], then a characteristic Local Group orbital period ('super-year') would be [math]\sqrt{(3e6)^3/5e12} \approx 2300 \times 16 Myr \approx 37 Gyr[/math]. Is that physically reasonable -- the Milky Way galaxy has yet to complete even a single orbit about the Local Group ?

 

Likewise, if the Local Super-Cluster is roughly [math]1 \times 10^{15} M_{\odot}[/math], and if its 'effective radius' is roughly [math]100 Mly[/math] (Wiki), then a characteristic Local Super-Cluster orbital period ('hyper-year') would be [math]\sqrt{(100e6)^3/1e15} \approx 3200 \times 16 Myr \approx 500 Gyr[/math]. Is that physically reasonable -- the Local Group has yet to complete even a few percent of an orbit about the Local Super Cluster ??