The timing circuit (diode Dhold and timing capacitor Chold) determines how long the motor is energized.  The voltage on the capacitor Chold and the Vsense network is “held up” even as the voltage on the motor falls.  The capacitor Chold discharges only through the operating current of the voltage monitor, say 150nA.  

You can use the formula F=A*S/V to calculate how long in seconds before the voltage on Chold capacitance (say 0.047uF) falls to the hysteresis voltage of the voltage monitor (Vth - 50mV) at the current of the voltage monitor (150nA.)  The unknown in the formula is S in seconds.  V is 50mV (delta V.)

That determines how long the voltage monitor and mosfet energize the motor from the storage capacitor.  You can use the same formula to crudely determine how far the voltage will drop on the storage capacitor.  The unknown is V (or delta V.)

In practice, rather than do the calculations, I just experiment.  I suppose you might also be able to simulate the circuit.

You can use such an excessively large Chold (say 10uF) that the storage capacitor completely discharges.  See below for more discussion.

The voltage on the motor is the Vth (1.1V) of the voltage monitor plus the voltage drop of the diode Dhold.  I have specified a Schottky diode, with a voltage drop around 0.2V.  Thus the voltage on the motor is about 1.3V.  The usual motor is rated to turn at 1.5V, so this design is pushing the limits.

Note that the Vf (forward voltage drop) of a diode depends on the current.  Usually one assumes a largish current and a voltage drop of 0.6V for a silicon PN diode junction or a voltage drop of 0.2V for a Schottky diode.  In this case, the current is so low (1uA or less) that the voltage drop of the diode is less.  You can consult the data sheet for the diode to find the Vf versus current curve, extrapolate it (usually), and estimate the actual voltage drop Vf in this circuit.

You might need to choose a different combination of voltage monitor threshold Vth and diode voltage drop Vf.  For example, you might try a 1N4841 silicon diode with a Vf that might be around 0.2V at such a low current.

The size (capacitance) of the storage capacitor and the system voltage (say 1.3V) partially determines how much the motor turns when it does turn.  (The timing circuit also factors in.)  You can use a larger storage capacitor to make the motor turn more revolutions, but less frequently.  Too small a capacitor will fail to start the motor.  You can use such a small storage capacitor that the motor will only turn a part revolution per turning event, i.e. barely twitches, but does so more often.

This design uses photodiodes for solar cells since they are all that is commonly available in such a small size.  Their generated voltage is small (0.35V Voc per cell.)  The total generated voltage is about 1.4V.  So you don’t have much wiggle room on designing the timing circuit.  And the system only works in bright light.

Solar cells generate the most power at maximum power point (MPP.)  If the design doesn’t discharge the storage capacitor completely, the solar cells will be operating nearer the MPP for longer.  The MPP is typically 80% of the Voc (open circuit voltage) of a solar cell.  In this case, the MPP is about 1.1V (80% of 1.4V Voc).  So if you trigger at 1.3V and let the system voltage fall only to say 0.8V, you would be operating the solar cell nearer the MPP and the motor would turn more for the same light conditions.

The Voc (voltage open circuit) of a solar cell is the most voltage the solar cell would ever produce (for a given light condition), when it is feeding an “open” or in this case a fully charged storage capacitor.  At that voltage, it is producing little power (because there is little current.)