During a short, is it possible for the voltage potential of the ground to change? Say for example, we short a 9V battery, how does the ground stay at 0V and not become 9V?
Ground is a reference point so it is always regarded as 0 volts.
Of course in a lightning strike the localized area around the strike can create massive potentials per metre so if you value your life (and the lives of any future children you wish to bear), keep your feet close together if there's a risk of lightning.
Circuit ground is just some point from which we measure or reference voltages. If you like, it's the point where you connect the black lead of your multimeter. All other voltages can then be measured with respect to that point.
It should be clear from above that if you connect the red lead of your multimeter to the circuit ground that it will always read 0 V no matter what is happening to the battery. (In real life it can be a little more complex if there is a current flowing in the ground circuit and there is some resistance between the two points.)
Figure 1. Both VM1 and VM2 will read zero volts. Battery internal resistance limits the current flow.
The resistance of standard PCB foil is 0.0005 ohms (500 microOhms) per square of foil ---- for any size square of foil.
Each square is 0.0005 ohms. To exit the entry point, there are 8 surrounding squares, thus R of that first ring of square is 0.0005/8 = 0.00006 ohms. Then the next (3x larger) ring adds another 0.00006 ohms. Etc. But in the larger rings, the current is mostly heading toward the DESTINATION output node, and not all squares get equal current density.
Ground is just a reference point, it does not need to be an actual connection to the earth.
When you short a 9 V battery, things which you can normally ignore (like the battery's series resistance) come into play. This prevents the ground potential from changing depending on what you choose as your ground reference point.
In the left circuit, the 1 ohm series resistance of the battery can be ignored as the resistance of the load is 100 times higher. R1 doesn't do much really.
In the right circuit R3 does come into play, it limits the current which will flow. The full 9V will be across R1. The ground will still be 0 V, it cannot even change as ground means that it is 0 V by definition.
Theoretically if you had an ideal battery with zero series resistance, an infinite current would flow if you shorted that ideal battery. But also then ground remains 0 V.
The term "Ground" has historically meant ground literally: an electrical connection to the earth itself, typically via a rod driven into the earth. This was used in power and telecom systems to obtain a voltage reference, and sometimes as a current return path. In electronic circuits, the term has persisted and is typically used to denote the negative side of the power supply, even in cases where the circuit is isolated from the actual earth ground. In circuit analysis, "Ground" is often used to specify a reference point from which other circuit voltages are measured; therefore it is by definition at zero volts.
In the 9V battery example, suppose you specify the negative terminal as Ground. If you short the positive terminal to Ground, a current flows through the short circuit, and the positive terminal will be (nearly) at zero volts, since a physical battery, unlike an ideal voltage source, has an internal resistance.
Ground and Earth are often confused. Personally I only refer to Earth as a connection to to the local electricity supply earth which is literally connected to the planets ground as pointed out by others.
Ground on the other hand means a circuit reference and may, but is not required to, be connected to earth.
Consider the circuit below we have two volt meters what do they measure?
VM1 if sensitive enough should measure a positive voltage because there is current flowing in the connection between its inputs. Theoretically we have zero resistance so no matter how many amps are flowing out of the shorted battery it should be zero but we never, except possibly with super-conductors, really have zero resistance.
VM2 does not see the drop from the battery current so sees 0V as expected.
We talk about 'local grounds' for this reason each bit of circuit cares about the point it is referenced too which may not be the same everywhere if significant currents are involved or at 'high' frequencies due to inductive effects.