How parker solar probe survives death rays?

Question

What is the temperature due to death rays alone excluding plasma.

Clearly it's now possible for any material to survive that.

It's 20x closer than earth so radiation is 400x.

150million km / 7 million km = 20.1

1. So the temperature too is 20^2 times than on earth. Should temperature of space above earth be multiplied by 400 to get the temp there?

2. How to survive those death rays?

3. There is also water in the probe why it's not melting? How does conduction,convection and radiation's effects are nullified, how is heat being dissipated into space and no conduction outside is possible only inwards.

4.why is thermal temp not 400x than on earth i.e. 400x300kelvin=120,000kelvin but just 1200Celsius.

Answer 1

This article contains an excellent description of the design of the Parker Solar probe. Regarding the temperature on the sun-facing side of the probe, it says:

The corona through which Parker Solar Probe flies, for example, has an extremely high temperature but very low density ... compared to the visible surface of the Sun, the corona is less dense, so the spacecraft interacts with fewer hot particles and doesn’t receive as much heat. That means that while Parker Solar Probe will be traveling through space with temperatures of several million degrees, the surface of the heat shield that faces the Sun will only get heated to about 2,500 degrees Fahrenheit (about 1,400 degrees Celsius).

The probe's solar panels cannot be kept entirely behind its heat shield (otherwise they are not doing their job !) and so they need to be actively cooled. Regarding the use of water as a coolant, the article says:

The solar arrays have a surprisingly simple cooling system: a heated tank that keeps the coolant from freezing during launch, two radiators that will keep the coolant from freezing, aluminum fins to maximize the cooling surface, and pumps to circulate the coolant. The cooling system is powerful enough to cool an average sized living room, and will keep the solar arrays and instrumentation cool and functioning while in the heat of the Sun. The coolant used for the system? About a gallon (3.7 liters) of deionized water. While plenty of chemical coolants exist, the range of temperatures the spacecraft will be exposed to varies between 50 F (10 C) and 257 F (125 C). Very few liquids can handle those ranges like water. To keep the water from boiling at the higher end of the temperatures, it will be pressurized so the boiling point is over 257 F (125 C).

Answer 2

Consider that 400x the surface temperature of earth (~300K) is a terrifying 120000K, which is far far hotter than the radiative heat source (surface of the sun is below 6000K). That would break, not only the probe, but also the 2nd law of thermodynamics (assuming there is no other heating mechanisms than radiation). So quite clearly, no, you can't just multiply numbers like that and hope to get a sound result.

So what is the temperature? Let's do a back of the envelope calculation, supposing purely radiative transfer (well, we're in space, there's not so many alternatives I can think of). (disclaimer; i did that as I was typing, there might be some mistakes. Anyone is welcome to correct)

You are right in saying that if the probe is 20x closer, then the irradiated power is 400x greater. The power radiated by the sun towards a surface $S$ at distance $r$ is given by: $$P_{in}=\frac{S}{4\pi r^2}P_{sun}$$

Where $P_{sun}$ is the total radiative power of the sun. The term $\frac{S}{4\pi r^2}$ represents the "fraction of space covered by the surface" (if you've heard of solid angles, you might wanna think in those terms). So because of the $r^2$ down there, it's true that 20x cloxer = 400x more power.

To obtain the temperature of the probe at thermal equilibrium with the sun, assuming that the probe is a blackbody, we simply say that this Power received by the probe must be equal to the power radiated by the probe (0 net flux at equilibrium). Considering that the total surface of the probe is 2S (only half of the total surface is facing the sun), and using the Stefan-Boltzmann law, we get that the power radiated by the probe at temperature T is:

$$P_{out}=2S\sigma T^4$$

We can use the Stefan-Boltzmann law to express $P_{sun}$ in a similar way, and we equate $P_{in}=P_{out}$: $$\frac{S}{4\pi r^2}S_{sun}\sigma T_{sun}^4=2S\sigma T^4$$ $$\rightarrow T=(\frac{S_{sun}}{8\pi r^2})^{1/4}T_{sun}$$

Then we just plug the numbers. Since I have zero astronomical culture, I have to look up on google that $T_{sun}=5778K$, $r_sun=1.3e6km$ and I searched that the closest distance probe-sun was, $r=8.5e6 km$

and I get: $$T=1964K=1691celsius$$

Keep in mind that this is a very rough calculation, with quite a few assumptions. In particular, the probe is not a blackbody. In fact, if it reflects 50% of the incoming light, the temperature already drops down to 1400 degrees C. A heat shield will probably be an even better reflector than that, but I don't have the numbers. Anyway that gives you a ballpark number for the temperature of the heat shield (below 2000K for sure, and probably far below, depending on how reflective the shield is)

So, in brief:

1. No, the temperature doesn;t even get near 400x that of earth. Thankfully. As far as I know, no material would remain solid at such a temperature.

2. I'm not sure what a death ray is

3. Do you mean to ask why the water does not evaporate? I know nothing about the engineering design of the probe (and if it even has water cooling at all), but the above mentionned temperature is that of the exposed surface, the heat shield. the bulk of the probe will likely be nowhere near that temperature (that's the whole point of the shield). You can have a look at rob's link in the comments, it gives some more details on that. According to that link, the main systems, hidden behind the shield, remain below 30 celsius (because they're in the shadow of the shield and don't actually get directly heated by the sun)