If a purely inductive circuit is started in presence of gravity, will the power source ever run out? I have read that purely inductive circuits do no work. So if an Inductor with an AC source, coming from an inverter is set up, will the source of power ever run out due to work against gravity?
This operates under the assumption that the coil and connecting wires have 0 resistance. You might say that the work done in transferring electrons up the wire, will be regained as potential when it descends. But what happens in a non uniform gravitational field? 
 A: For Newtonian gravith it's a conservative force, so there is no EMF around the loop due to gravity.
Similarly for an electrostatic force.
Suppose that current is delivered by mobile electrons. Just as a surface charge distribution on the surface of the conducting wires could arrange itself to oppose a constant electric field, so a proportional surface charge distribution can arrange itself to produce an electric field sufficient to suspend a mobile electron in a gravitational field. That's why if you lift up a piece of wire the mobile electrons don't all fall to the bottom of the wire (basically, some of them do until their field keeps the rest in check).
Now when you hook the wires up there will still be that electric field supporting the mobile electrons. The other parts of the wire are supported by a stress caused by a strain of the lattice making up the solid.
However as the current starts to flow, additional surface charge (and a Hall voltage) will have to develop to counter any magnetic forces (to enforce that the flowing current remain in the wire).
Note that when the gravitational field is non uniform, it can still be countered by electrostatic forces if it is conservative. And Newtonian gravity is a conservative force.
Since the electrostatic charge distribution counters the force of gravity on the mobile electrons you can ignore both together.
Issues would happen if the circuit moves because then the surface charge would move and because the gravitational field would change so you'd need changing surface charges to counter it.
