Capacitor size for battery operated device?
I suggest using double spring loaded contacts, so that the batteries can float a bit without losing contact.

Otherwise, measure the disconnect time for many drops, and then calculate the capacitor based on this time.

Let's say you find out that the disconnection does not last for more than 50ms.  Let's assume 100ms to be on the safe side.  Let's also assume that your system can tolerate a voltage drop to 3.0V.

$$
Q_{max} = I_{max} * t_{max} = 500mA * 100ms = 50mC \\
C_{min} = \frac{Q_{max}}{V_{drop,max}} = \frac{50mC}{3.6V - 3.0V} = 83333\mu F
$$

Simulation

The voltage source is the batteries, the switch simulates the drop event at \$t=0\$.  Observe in the time domain simulation how the capacitor voltage drops linearly, reaching 3.0V after 100ms.







simulate this circuit &ndash; Schematic created using CircuitLab

Conclusion

Some of the assumptions in this calculation lead to unfeasible capacitor values.  Maybe the time the contacts break typically is much shorter.  If not, reconsider whether your device will use 500mA during a drop condition.

If you still get to such large capacitor values, consider soldering in the batteries.  Or use double spring floating contacts, as suggested above.