The capacitor charges through R1 to +9V - Vbe = 8.3V when the switch is open.
When you close the switch the left end of the capacitor becomes 0V, so the right jumps to -8.3V (exceeding the absolute maximum rating of -6V on Q1, by the way).
Ideally the transistor does not instantly die from this abuse and the capacitor begins to charge towards 0V - Vbe = -0.6V (putting reverse bias on the polarized capacitor, also frowned upon in some circles).
The time constant is \$\tau = R_2 C_1\$. Time Constant has a specific meaning- it is not the same as the time for the transistor to switch because the threshold is not at 63% discharge but more like 50%. The discharge follows an exponential curve. (as pointed out in the comments, the vertical axis is not really right, but the shape is correct).
![enter image description here](https://i.stack.imgur.com/JAHuZ.gif)
To clarify the actual discharge curve measured at the right-hand side of the capacitor, so relative to ground, (and ignoring the transistor base for now) can be shown to be \$v(t) = 9 - 17.2e^{-t/\tau}\$ where t>=0 is the time since the switch was closed. The transistor will switch (and the curve will deviate from the ideal since the base clamps it) when v(t) is about +0.7 so that is at \$t = \tau\cdot \ln(0.483)\$ or about \$0.73 R_2 C_1\$. In this particular case C1 = 470uF and R2 = 22K, so the time should be ~7.5 seconds. It may vary a bit from that because the transistor needs some current to operate the LED and also because the 470uF capacitor probably has a large tolerance.
You can easily simulate this in circuitlab to verify the design- the top curve shows the LED current, and the bottom curve the voltage at the base of Q1.
![schematic](https://i.stack.imgur.com/kcHsI.png)
simulate this circuit – Schematic created using CircuitLab
![enter image description here](https://i.stack.imgur.com/mkME9.png)
Best Answer
simulate this circuit – Schematic created using CircuitLab
Figure 1. The equivalent circuit using a relay.
Note that the circuit is a little inefficient. Not only does it draw power when the LED is off but it draws more power in that condition as the full voltage is across R1.