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  • $\begingroup$ That's the effective potential, and Carroll is giving you the potential per unit mass. See the discussion in Could we send a man safely to the Moon in a rocket without knowledge of general relativity?. $\endgroup$
    – John Rennie
    Commented Jun 5, 2016 at 16:34
  • $\begingroup$ @JohnRennie, But that still doesn't make sense because if Wikipedia derivation is right then the last term should be $-\frac{G(M+m)L^{2}}{m\,r^{3}}$ $\endgroup$
    – user3741635
    Commented Jun 5, 2016 at 16:38
  • $\begingroup$ @JohnRennie, I am just not sure why there is $-\frac{G(M+m)L^{2}}{r^{3}}$ term on wikipedia while in your derivation you have $-\frac{GM L^{2}}{m r^{3}}$ Why are the masses arranged differently. Could you explain this please. Thanks. $\endgroup$
    – user3741635
    Commented Jun 5, 2016 at 16:48
  • $\begingroup$ From a quick glance, Carroll and I are assuming an infinitely small test mass so the central mass can be taken as stationary. The Wikipedia calculation assumes an orbiting mass high enough to be significant. Note that Wikipedia uses the reduced mass $\mu$ in the equation. $\endgroup$
    – John Rennie
    Commented Jun 5, 2016 at 16:54
  • $\begingroup$ @JohnRennie, Thanks for help. By "infinitely small test mass" do you mean the mass of the moon compared to the mass of the earth? $\endgroup$
    – user3741635
    Commented Jun 5, 2016 at 17:03