From time to time I see discussions about the expansion of the universe that describe a future state where the view of the sky from Earth (assuming Earth were to still be in existence) will be completely dark. There are two possible explanations of this prediction. 1. The universe will eventually exhaust its supply of hydrogen, the fuel that stars use to make light (and to make other things as well). 2. The expansion of the universe will cause shining stars and galaxies to eventually move outside of the Earth's observable universe. It is explanation (2) I would like some help with. Consider a system of currently shining stars and galaxies of total mass M (including dark stuff) which fits into a sphere of radius R. If the distribution of the mass is approximately radially symmetric, we can calculate the approximate escape velocity Vesc of a test particle at the edge of the sphere. Vesc = Sqrt (2 G M / R) Vesc can then be compared with the velocity V with which a test particle at the periphery of a system will be moving away from the system center due the universe expansion. V = R hinf, where hinf is the calculated limit of the Hubble constant as time approaches infinity. If V < Vesc, then the system would be gravitationally sufficiently bound to avoid disruption by universe expansion (assuming spherical symmetry). Otherwise, not. My questions are: 1. In general, what is the approximate effect of a non-radially symmetric, non-spherical shape on this calculation of Vesc? 2. In particular, if we assume the mass distribution is two-dimensional and circularly symmetric, how would this effect the calculation of Vesc? Below are some examples. Here are the values for G and hinf used for these examples. G = 6.69 x 1011 m3 s-2 kg-1 hinf = 1/17.3 Gy-1 = 1/5.45 x 1017 s-1 https://www.physicsforums.com/threads/the-hypersine-cosmic-model.819954/page-3#post-5197970 Example 1. Laniakea super cluster Laniakea is very much NOT a spherically nor circularly symmetric example. https://en.wikipedia.org/wiki/Laniakea_Supercluster M = 1017 solar masses = 2 x 1047 kg R = 260 million light years = 2.46 x 1025 m Vesc = 1.14 x 106 m/s V = 4.51 x 107 m/s Since V >> Vesc, I would guess that the shape of the Laniakea supercluster would likely have no effect on the system's not being gravitationally bound with respect to the expanding universe. Example 2. Local Group This example is much closer to being symmetrical than Laniakea. http://mnras.oxfordjournals.org/content/384/4/1459.full.pdf+html M = 5.27x 1012 solar masses = 1.05 x 1043 kg https://en.wikipedia.org/wiki/Local_Group R = 5 million light years = 4.73 x 1023 m Vesc = 5.46 x 104 m/s V = 8.68 x 105 m/s Since V > Vesc, the Local Group is also not gravitationally bound with respect to the expanding universe. Example 3. Milky Way This is approximately a circularly symmetric example. https://en.wikipedia.org/wiki/Milky_Way M = 7 x 1011 solar masses = 1.4 x 1042 kg R = 160 thousand light years = 1.51 x 1022 m Vesc = 1.11 x 105 m/s V = 2.78 x 104 m/s Since V < Vesc, the Milky Way IS gravitationally bound with respect to the expanding universe, assuming spherically symmetry. Would it still be calculated as bound if only two dimensional and circularly symmetric? Also, I note it as unexpected that the Milky Way has a greater value for Vesc than does the much large Local Group.