# Could a probe fly through the Sun's transition zone between chromosphere and corona?

I wonder if a cryogenically cooled probe could fly very fast tangentially pass the Sun and through its transition zone where the temperature drops to below 6 000 K, from millions in the corona. What maximum speed would a spacecraft achieve from the Sun's gravity and how long would it be within, say, one Solar radius of the surface?

Is the plasma in the lower corona too dense even for an ablative shield? The spacecraft could be shaped like a sphere and tumble to spread the heat. It should still be able to detect temperature, density, mass concentrations and magnetic field.

• There is another very strong heat source, the solar radiation in the visible and IR range. The heat load due to radiation may be much bigger than by the plasma with high temperature but very low density.
– Uwe
Commented Sep 28, 2017 at 17:13
• Why don't you add some information about the distances you are thinking - how far is the transition zone from the center of the Sun? Of course you still have to pass through the corona twice, it's not like you can bypass it.
– uhoh
Commented Sep 28, 2017 at 17:51
• – Uwe
Commented Sep 28, 2017 at 20:09
• A tangential pass of the Sun is not a possible orbit. A highly elliptical orbit like the Parker probe space.stackexchange.com/questions/17498 is possible. A hyperbolic orbit requires much more speed, but even a ray of light can't pass the Sun tangentially, see en.wikipedia.org/wiki/…
– Uwe
Commented Sep 28, 2017 at 20:24
• @Uwe Everything that orbits the Sun necessarily moves tangentially to it at times. I meant it as opposed to plunging through it orthogonally. Commented Oct 4, 2017 at 9:33

The sun, modelled as a 5800K black body, releases heat at 64MW per square meter (Stefan-Boltzmann const * T^4). Wikipedia describes the transition region as "tens to hundreds of kilometers thick", so this hypothetical spacecraft is essentially so close to the sun that one side is directly exposed to 64MW/m^2$, as you can pretty much ignore circular geometry. If we coat the sunlit side with the shiniest material feasible (optical solar reflector with a visible light absorbtance of 0.07) and paint the shaded side black for an infrared emmittance of 0.9, we get via some slightly fudged maths$ T^4 = \frac{solar\_intensity * absorbtance}{SB const * emmittance} \$ an equilibrium temperature of 5443K, due to those tricksy 4th power terms.