The Solar Probe Plus fact-sheet states that the craft will approach to the distance of 9 solar radii to the surface of the Sun (Approx 6.26e6 km) and its heat shields must withstand 1644K of heat.
I wonder how did they arrive at this number? Based on that, how can I calculate the max temperature an object has to withstand at 3.68e6 km (about 5 solar radii)?
Of course I know that at some point simple formula wouldn't work as we'll enter corona.
Answer
The energy per unit area radiated by an object at a temperature $T$ is given by the Stefan-Boltzmann law:
$$ J = \varepsilon\sigma T^4 \tag{1} $$
where $\sigma$ is the Stefan-Boltzmann constant and $\varepsilon$ is the emissivity. A spaceship in a vacuum can only lose heat by radiation, so it will heat up until the energy loss given by equation (1) is equal to the rate of energy absorption from the Sun. So if we calculate the energy flux from the Sun and plug it into equation (1) we can solve for the temperature.
However there's an easier way to do the calculation. Suppose the temperature of the Sun's surface is $T_s$ and it radiates some energy flux $J_s$ given by (the emissivity of the Sun is close to one):
$$ J_s = \sigma T_s^4 $$
If we go out to a distance of $n$ solar radii then the area goes up as $r^2$ so the energy flux per unit area is $J_n = J_s/n^2$ so:
$$ \frac{J_s}{n^2} = \sigma T_n^4 $$
where $T_n$ is the temperature at our distance of $n$ solar radii. Substituting for $J_s$ gives:
$$ \frac{\sigma T_s^4}{n^2} = \sigma T_n^4 $$
which rearranges to:
$$ T_n = \frac{T_s}{\sqrt{n}} $$
So at 9 solar radii we get:
$$ T = \frac{T_s}{3} $$
and since the temperature of the surface of the Sun is 5778K we get:
$$ T = 1926\,\text{K} $$
This is higher than the figure your article mentions, though only by 17%. Presumably the difference is down to the emissivity of the Solar probe.
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