I have found that when I'm probing a circuit using more than one oscilloscope probe, clipping the ground of one of the probes to the board seems sufficient to give me stable readings from all of the probes. And this also reduces the chance of accidentally connecting two different voltage potentials through the ground clips of different probes.
I realize that the longer "ground loop" for the probes not directly connected to the circuit can affect some of the readings, but how? And at what acquisition speed does it become a concern? (I vaguely remember the original, analog scope that I was trained on in the military some 30 years ago only having one ground lead, separate from the probes. But I have no idea how "fast" that scope was.)
Best Answer
One simple example
Let's say you have two scope probes. Probe A is correctly grounded to the circuit you are measuring but probe B doesn't have a ground connection. Let's also say that probe A is measuring a 1Vp-p squarewave on the circuit and probe B is looking at a DC level.
Despite scopes having a 1M\$\Omega\$ input impedance, there will be currents that flow through the probe due to the edges of the square wave - the scope input impedance is also likely to comprise 20pF capacitance (smaller for x10 probes usually).
This capacitance "takes" a pulse of current down the probe lead every time the square wave alternates causing a small volt-drop down the lead. This won't matter for probe A but, this will appear as a small "spike" on channel B's display because the glitch "becomes" in series with the probe B signal due to it being ungrounded.
If probe B were correctly grounded at the circuit this "glitch" won't be present on the scope B waveform.
What point/frequency does this become significant? I would answer that by saying just don't take short cuts. If you are probing a circuit for whatever reason, use best practise and ground both probe ends at the circuit you are measuring or expect anomalies.
If probe A is looking at 1Vp-p squarewave and probe B is looking at a 5mV analogue signal you'll have troubles on channel B.