- #36
ubavontuba
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SpaceTiger said:That equation is the energy of a mass starting with zero velocity at [itex]r=r_0[/itex]. It doesn't matter what the form of the potential is, as long as it's spherically symmetric (or, alternatively, one-dimensional), that equation will be valid. If it were true that [itex]V_r(r_0) > V_g(r_0)[/itex], then the orbit would be unbound and the particle would be able to escape.
Think about it this way. A mass that is "free" will have only kinetic energy, which must be positive. If its total energy is negative at any point in the orbit, then conservation of energy demands that it never be "free".
Hmm... Is it a one-dimensional consideration? Did you consider that in this model at [itex]r=r_0[/itex] the DE potential is zero but at any other given distance it is no longer zero?