When the particles that make up cosmic rays get boosted up to the speeds at which we see them, they're ionized. The electrons and nuclei are separated from the start.
Both types of particle are going to be releasing energy every time they accelerate (synchrotron radiation) and when they interact with photons (inverse Compton scattering). When it comes to inverse Compton scattering, the electrons scatter much more than the alpha particles - there's a $\frac{1}{m}$ factor in both the energy exchange and the angle involved.
What this results in is basically the electrons going in all directions, gradually losing energy. Meanwhile, the alpha particles continue on with less deflection and with most of their energy intact. The electron is the ping-pong ball to the alpha particle's bowling ball. (4 bowling balls, really, but you get the idea.)
We still get electrons from space, but more of them are coming in slow, which puts them out of the "cosmic ray" category.
My question is this: does the relative movement of the point of origin for a cosmic ray have a significant impact on that ray's energy as we experience it here on earth, or is this effect dwarfed by the power imparted on particles by the effects of the black holes and super novae that generate them?
The second idea is the correct one. The Earth moves around the Sun at a velocity of about 30 km/s, which revolves around the center of the Milky Way at about 200 km/s. Andromeda is getting closer to us at a speed of about 300 km/s. We know from the microwave background that the total speed of the Earth with respect to the CMB frame is about 370 km/s.
How fast are cosmic rays coming from a supernova? They move at (or slightly below) the speed of light, which is 300000 km/s. You see that the small velocity of the Earth with respect to other celestial objects is 1000 times smaller. Therefore the velocity of the Earth is definitely negligible. The energy of the cosmic rays will be almost the same, whether they smash with the Earth head-on, or they hit it from behind.
If you want to calculate exactly how tiny is this effect, you can use the following approximate formula:
$${E_{em} \over E_{obs}} = \sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}$$
where $E_{em}$ is the energy of a cosmic ray in the frame where it is emitted by the supernova, $E_{obs}$ is the energy of the cosmic ray as seen here in Earth's frame, $v$ is the speed of the supernova w.r.t. the Earth and $c$ is the speed of light.
For a relative velocity $v=-300$ km/s, you get $E_{obs} = 1.001 E_{em}$.
This formula will work with highly relativistic particles (where $p \gg mc$) and when the supernova is moving mostly radially with respect to the Earth. (For transverse velocity the formula is different, see Redshift)
Edit:
If the cosmic rays come from an very distant source like a quasar, the cosmological redshift comes into play. In that case
$${E_{em} \over E_{obs}} = 1+z$$
Where $z$ is the redshift of the source. Since $z>0$, this effect can only reduce the energy of the cosmic ray, not increase it.
It can be interpreted as the result of the expansion of the universe, that moves very far objects away from us at superluminal speed. For the most distant quasars we know today, $z \approx 7$.
Best Answer
As I discuss in https://physics.stackexchange.com/a/415248/59023 and https://physics.stackexchange.com/a/708183/59023, a sunspot is a pressure-balance structure. The total pressure in a collisionally mediated plasma includes the thermal and magnetic pressures. Since the regions involved in generating sunspots arise from the enhanced magnetic fields of flux ropes rising to the photospheric surface, we know that the magnetic pressure is high. We assume the thermal pressure is given by: $$ P = n \ k_{B} \ T \tag{0} $$ where $n$ is the number density [number per unit volume], $k_{B}$ is the Boltzmann constant, and $T$ is the temperature.
Since this is a collisionally mediated system, the density inside and outside these structures will roughly remain the same but the temperature can vary. To remain in pressure balance (i.e., total pressure outside equals total pressure inside), the temperature must then drop to accomodate the higher magnetic pressure.