How large are the various forces and torques that the Parker Solar Probe will experience during its closest flyby of the Sun?

Question

The Parker Solar Probe will experience forces and torques as it passes within about 8.5 solar radii of the Sun's surface that even a General Products hull will not protect it from.

How large will the tidal acceleration be along the 3 meter length of the spacecraft at perihelion? Will it be detectable in any way?

Will there be any (real or effective) torques that will tend to rotate the spacecraft during it's closest flybys? What about effects due to the oblates of the sun?

Will any of these be a new record - or have Cassini or Juno experienced greater effects during their closes approaches to Saturn and Jupiter already?

above: Parker Solar Probe cropped, from here.

above: screen shot from NASA's Solar Probe Plus Fact Sheet note: the link seems to be dead now, but there is plenty of related information available at http://solarprobe.jhuapl.edu/index.php#spacecraft

above: Illustration of Parker Solar Probe's orbit, from here.

Answer 1

The largest torque affecting the Parker Solar Probe during its closest approach is caused by the solar wind blasting against the heat shield (Thermal Protection System, a.k.a TPS).

The TPS is like an umbrella held facing into a strong wind. The wind whips against it, and tries to force it around away from the wind. The spacecraft has to "fight" against this force to maintain it's proper attitude with the TPS pointed right at the sun and right into the solar wind.

To "fight" against the solar wind, the primary mechanism is the spacecraft's reaction wheels. They are mounted at angles to each other (there are four) and they are spun up or slowed down to create the counter force to the solar wind acting on the TPS. The spacecraft subsystem that controls the wheels is the Guidance and Control (G&C) system. Fifty times a second (50 Hz) the inertial measurement unit provides change-in-attitude readings and the G&C system calculates the reaction wheel response required to counter act the change and bring the spacecraft back to the correct attitude with the TPS pointing at the sun.

In addition to controlling attitude with the reaction wheels, the spacecraft thrusters can also be used, but this is not normally done because it degrades the science measurements. However, as the reaction wheels counter the torques from the solar wind, they wind up and spin faster and faster. In order to bring the speed back down, the spacecraft periodically performs a "monmentum dump" by firing the thrusters to counter-act the wheel's momentum and allow them to slow down to a manageable rate.

Answer 2

The other answer suggests that ram pressure is the dominant effect on the spacecraft. I messed this up in my first thoughts on the problem too, but it's actually radiation pressure from sunlight.

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.

In regards to the other answer given before mine, 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.