Electronic – Coil Gun – High V, few MFR but very high power (in theory); need opinions

The coil, it seems, is not well-defined and my first thought is that an op-amp won't cut the mustard on driving it successfully. The LM301 is only really suitable for driving kohm loads so I believe you ought to invest in a power transistor to interface the op-amp to the solenoid/coil.

But hold-on, lose the op-amp and use a logic level mosfet driven from the PIC'c PWM output. Connect source to 0V, gate to the PIC output and put the coil between drain and positive supply. This will give you a better drive to the coil and don't forget to put the diode across the coil reverse biased to catch the back emfs when the transistor is turned off: -

The 100k resistor is to ensure the mosfet remains off when the input is not connected to anything (important). The 1k is to limit stress on the PIC output because the MOSFET will likely have input capacitance in the order of 1nF. You need to find a MOSFET that will switch fully on with the positive drive level from the PIC (5V presumably).

In practical terms, what you are doing seems dangerous and I will not endorse it. In theoretical terms, if you were designing a safe discharge bar:

The first parameter I would be concerned with is the maximum permissible current in the target circuit, which may be limited by the traces on the PCB if there is one, or by the capacitor itself. If you're lucky, you'll be able to either see the capacitor's current rating on the can, or identify it and look up its specsheet.

The maximum current will be seen at the first instant that the discharge bar is connected, and will exponentially decay thereafter, via

$$ I(t) = \frac V R e^{-t / \tau} $$

So at time 0, the capacitor will be "transparent" and the current will be limited only by your discharge bar (assuming no lead inductance, which is a whole other thing). Knowing the maximum safe current is a little complicated, so give yourself a very wide safety margin (particularly since you describe yourself as a beginner). Once you have chosen a maximum safe current (say 500mA), 440V/500mA = 880 ohms minimum.

Note that

the formula T = R x C where T is the time it takes to discharge

is NOT true. In one sense, an RC circuit "never discharges completely" - it exponentially approaches zero. At some point the voltage becomes less than the noise voltage and can't be measured anymore. The rate of decay is governed by \$\tau\$ the RC time constant. To say that a capacitor has discharged you need to pick a threshold like 99%.

Even then, the exponential decay curve is the theoretical ideal, and physical construction of the capacitor will produce behaviour that can deviate from this; see comments from @SpehroPefhany.