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Falling outward

  1. Jul 14, 2009 #1
    re: GE = minus GMm/R. It`s said that as R increases M increases to the third power and therefore falling outward is downhill but the increase in M is due to an increase in the vacuum energy which converts to a certain amount of mass. However, the mass of vacuum energy is very small. Could that mass increase less than the increase in R and therefore fall inward?
     
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  3. Jul 14, 2009 #2

    DaveC426913

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    What?
     
  4. Jul 14, 2009 #3
    it was in a book by Prof Mark Whittle. Since the volume , mass (vacuum energy) of a sphere increases as the third power of the radius increases, then the GE minus(GMm/R) would become more negative as R increases and there fore GE would cause a falling outward. I guess it depends however on how much vacuum energy increases when the R increases. If the increase in vacuum energy cubed was less than the increase in R then GE would become less negative and cause a falling inward.
     
  5. Jul 14, 2009 #4

    negitron

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    I don't understand what it is you're trying to ask.
     
  6. Jul 14, 2009 #5

    DaveC426913

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    We have not read the book. Nor have you told us what book it is.

    You are using too much shorthand for us to understand the issue in a contextual vacuum.
     
  7. Jul 14, 2009 #6
    it`s an outline book called "Cosmology: The History and Nature of the Universe" from one of the "Great Courses" DVD`s.
    Basically i`m asking if the increased mass as a result of the vacuum energy (mass) created when the sphere expands can be less than the increase in Radius. Vacuum energy is small and if it is small enough the soution would result in smaller negative GE and therfore a falling inward rather than outward.
     
  8. Jul 14, 2009 #7
    according to this book GE is a negative value and that falling goes in the direction of greater negative value of GE.
     
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