What is the logic behind "negative binding" energy??

$$ Tom [He between his error-bars] Roberts ] wrote:
In order for a system consisting of multiple constituents to be
bound as a single system, it must not fly apart on its own. That
is, in order to disassemble the system you must somehow reach
into it and _pull_ one or more constituents out of the system
(of course this requires you to somehow hold on to some of the
other constituents). In pulling it apart you of course do work,
which is putting energy into the constituents of the system. The
total amount of work you must do to completely disassemble the
system is minus the binding energy of the system. That _is_ the
way it was originally defined, and since you must do positive
work to disassemble it, binding energy is inherently negative.
Equivalently, when assembling the system an amount of energy
must be released, which is also minus the binding energy.

For example, in Newtonian gravitation, the gravitational
potential \phi for a point mass M is -kM/r (k is Newton's
gravitational constant). The minus sign is essential, because
the force _must_ be -grad phi. The minus sign in this last
equation is essential, because the force _must_ be directed
away from regions of higher potential energy. Remember that
gravitation is attractive and energy is conserved -- these two
properties completely determine those minus signs. So a system
that is gravitationally bound has negative potential energy,
and it is quite appropriate to equate this to the binding
energy of the system (disassembling it means separating the
masses to infinity, where the gravitational potential energy
is zero, as is the binding energy).

In relativity, applying E=mc^2 in the rest frame of the system,
and always disassembling it into components at rest far away in
that frame, the mass of the system is simply the sum of: a) the
masses of its constituents, b) their kinetic energies (while
part of the system), and c) the binding energy. Note if binding
energy were positive there would be a funny minus sign in there.

Tom [He between his error-bars] Roberts ].

$$ You lack a distinction between "Total ENERGY" & "Total ENTHALPY".
$$ [LaGrangian L; intrinsic REST energy eM; Volt*Amp*sec energy eV].
$$
$$ GUESS iSS STANDARD TOTAL ENTHALPY E = m*c^2 + pL*c + pA*fA
$$ = eM + L + eV
$$ = eM + L + nA*hbar*fA
$$ = eM + Kinetic energy eK
$$ ..Note, *total-OTHERwise-ENERGY*... = eM + L - (m*v^2 / 2).
$$
$$ Where kinetic energy eK is the photoelectric "iONization energy".
$$ [This energy is NOT ONLY particular to the photoelectric effect].
$$
$$ STANDARD Total ENTHALPY ..NO "funny minus sign in there", Dimwit.