Assume it's an circular orbit. Object A orbits around object B. Take object B as frame of reference.
.$E=KE_a + GPE$
.$E=\frac 12m_av_a^2 +(-\frac {GM_bm_A}r)$
.$E=\frac 12m_a(GM_br)+(-\frac {GM_bm_a}r)$
.$E=-\frac {GMm}{2r} < 0$
What does negative total energy at any instant of time mean?
Negative energies are totally fine, because you had to pick a zero-point for energy. In your calculation you picked it to be at infinity. You could have chosen the zero-point for potential energy in such a way that your system had zero energy, or whatever. Only changes in energy are meaningful, in general.
Consider this: what happens if you add energy to this system? It gets closer to zero, and zero for us is the point where the particle is at rest, but is infinitely far away from the other particle. So negative energy represents the fact that to "free" the particle from the central potential requires you to add energy. This comes up a lot in quantum mechanics--the ground state energy of the hydrogen atom is -13.6 eV.
As another answer points out, a constant can be added to the potential energy without affecting the equations of motion. Often, we impose the boundary condition that the potential energy is zero 'at infinity'.
For the case of a central gravitational (attractive) force, imposing the "zero at infinity" boundary condition means that the gravitational potential energy is negative for non-zero $r$.
Since the kinetic energy is always positive, it is possible that the total energy of the particle can be negative, zero, or positive.
Considering purely radial motion:
Considering 2D motion:
Since a circle is a degenerate ellipse, it follows that the total energy must be negative for a circular orbit.
Basically negative energy doesn't mean it is less than zero. It just implies that the orbiting object need that amount of energy to be added so that it comes to stable equilibrium Or say zero energy
