Does the Parker Solar Probe experience radiation pressure being so close to the sun? Since the Parker Solar Probe does close fly-bys around the sun, does it experience any significant radiation pressure from photons emitted by the sun?
Is this push, significant enough to require course corrections or does the heat shield itself reduce it to a point that it can be ignored?
 A: This is taken from: https://space.stackexchange.com/a/57575/12508
On it's latest perihelion pass, Parker Solar Probe (PSP) passed within ~13.3 $R_{s}$ of the Sun and had a maximum speed relative to the Sun of ~160 km/s.  The particle number density topped out at ~8000 $cm^{-3}$.  The solar wind speed relative to the spacecraft topped out around ~600 km/s (excluding a few peaks up to ~800 km/s for very short intervals).  All of this adds up to a ram pressure of ~2.41 $\mu$Pa (even if I used the 800 km/s peaks, it would increase this by less than a factor of 2).  The solar wind near Earth typically has number densities around 10 $cm^{-3}$ and speeds around 400 km/s, or ram pressures of ~0.0013 $\mu$Pa, or about 3200 times smaller than at ~13.3 $R_{s}$.
The Sun's luminosity is ~$3.828 \times 10^{26}$ W, so at one astronomical unit (AU) (~214.94 $R_{s}$) the power per unit area is ~1361 W $m^{-2}$.  At ~13.3 $R_{s}$ the solar power per unit area is ~355,500 W $m^{-2}$, or an increase by a factor of ~261.
PSP's heat shield has a radius of ~4 ft (~1.2192 m).  If we approximate it as a circular disk, then its area would be ~4.6698 $m^{2}$.  The force from radiation pressure is given by:
$$
F_{rp} = \frac{ L_{s} }{ 4 \ \pi \ r^{2} \ c } \ C_{r} \ A_{abs} \tag{0}
$$
where $L_{s}$ is the solar luminosity given above, $r$ is the radial distance from the Sun's center, $c$ is the speed of light in vacuum, $C_{r}$ is the coefficient of reflectivity, and $A_{abs}$ is the absorbing area.
If we divide both sides by $A_{abs}$ we get a solar radiation pressure, $P_{rp}$, exerted on whichever area is of interest.  In this case, it's the heat shield.  Throwing in the numbers for $r$ ~ 13.3 $R_{s}$, we get $P_{rp}$ ~ 1185.8 $\mu$Pa, or nearly 500 times that of the ram pressure given above.
The reaction wheels are necessary to make sure the heat shield always points at the Sun and keeps all necessary equipment in shadow.  This is absolutely essential as several instruments in shadow were not designed to handle the solar radiation fluxes that will be experienced at it's closest approach of ~9.5 $R_{s}$.  In fact, the thermal conductivity of many of the parts in shadow are so high that were the whip antenna at the tail of the magnetometer boom to be exposed to sunlight, it would ablate in a few 10s of seconds and start to over heat the magnetometer sensors.  This would also drastically change the spacecraft's moment of inertia, which would throw the automated attitude corrections into a state of confusion.  There are lots of checks to try and mitigate scenarios like this one but the worst case is that the attitude keeps turning the reaction wheels under the assumption the spacecraft is whole.  This will be insufficient and the spacecraft will continue to rotate such that more and more of the side is exposed to sunlight.  The end result would be that the heat shield would come flying out on the other side of perihelion alone, the rest of the spacecraft bus would be completely ablated.
