How close could you get to the Sun and survive in a perfectly reflective spacesuit?

Question

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1. What the problem is really asking

You have a spacesuit that reflects every photon that hits it (i.e. it absorbs no electromagnetic radiation).
How close could a person wearing that suit get to the visible surface of the Sun and still stay alive for an hour?

In other words we must find the smallest distance from the Sun at which the net heat that reaches the astronaut is low enough that the body can survive (≈ 100 W of absorbed power for a resting adult).
Because a perfectly reflecting surface would absorb zero power, the only limitation comes from the fact that real materials are never 100 % reflective – there is always a tiny absorptivity α. We will therefore:

Write the solar radiative flux as a function of distance from the Sun.
Convert a tolerable absorbed power into a required maximum flux using the suit’s absorptivity and its effective area.
Solve for the distance that gives that flux.

If we truly had α = 0 the answer would be “arbitrarily close”, but we will keep a realistic tiny α to obtain a finite number and illustrate the physics.

2. Step‑by‑step solution

2.1 Solar radiative flux at a distance r

The Sun radiates as (approximately) a black‑body of temperature (T_{\odot}=5778\ {\rm K}).
Its total power (luminosity) is

[ L_{\odot}=4\pi R_{\odot}^{2}\sigma T_{\odot}^{4}=3.846\times10^{26}\ {\rm W}, ]

where

(R_{\odot}=6.96\times10^{8}\ {\rm m}) – solar radius,
(\sigma =5.670374419\times10^{-8}\ {\rm W\,m^{-2}\,K^{-4}}) – Stefan‑Boltzmann constant.

At a distance (r) from the Sun’s centre the irradiance (power per unit area) is the inverse‑square law:

[ \boxed{F(r)=\frac{L_{\odot}}{4\pi r^{2}} } \qquad\bigl[{\rm W\,m^{-2}}\bigr]. ]

For convenience we can also express it relative to the solar constant at Earth ((F_{\oplus}=1361\ {\rm W\,m^{-2}})):

[ F(r)=F_{\oplus}\left(\frac{1\ {\rm AU}}{r}\right)^{2}, ] with (1\ {\rm AU}=1.496\times10^{11}\ {\rm m}).

2.2 How much absorbed power can a human tolerate?

A resting adult produces about (100\ {\rm W}) of metabolic heat.
To stay alive we must keep the additional absorbed power from the Sun at or below that value; otherwise the core temperature would rise too quickly.

So we set the tolerable absorbed power

[ P_{\text{tol}} = 100\ {\rm W}. ]

2.3 Relating absorbed power to the suit’s properties

Let

(A) = effective projected area of the astronaut+ suit that faces the Sun (≈ 2 m² for a standing person).
(\alpha) = absorptivity of the suit (fraction of incident radiation that is actually absorbed).
Perfect reflection would mean (\alpha=0).
In practice the best reflective coatings in the visible/IR have (\alpha) of order (10^{-6}) (reflectivity 99.9999 %).

The absorbed power is

[ P_{\text{abs}} = \alpha \, A \, F(r). ]

Setting (P_{\text{abs}} = P_{\text{tol}}) gives the maximum allowable flux:

[ F_{\max}= \frac{P_{\text{tol}}}{\alpha A}. \tag{1} ]

2.4 Choose a realistic absorptivity

Even the most advanced multilayer dielectric mirrors (used on solar sails) have (\alpha \sim 10^{-6})–(10^{-7}) over the whole solar spectrum.
We will use the optimistic value

[ \boxed{\alpha = 1\times10^{-6}} . ]

With (A = 2\ {\rm m^{2}}) and (P_{\text{tol}} = 100\ {\rm W}),

[ F_{\max}= \frac{100}{(1\times10^{-6})(2)} = 5.0\times10^{7}\ {\rm W\,m^{-2}}. \tag{2} ]

2.5 Solve for the distance r that yields this flux

Set the solar flux (eq. 2) equal to the inverse‑square law:

[ F_{\oplus}\left(\frac{1\ {\rm AU}}{r}\right)^{2}=5.0\times10^{7}\ {\rm W\,m^{-2}}. ]

Insert numbers ((F_{\oplus}=1361\ {\rm W\,m^{-2}})):

[ \left(\frac{1\ {\rm AU}}{r}\right)^{2}= \frac{5.0\times10^{7}}{1361} \;\;\Longrightarrow\;\; \frac{1\ {\rm AU}}{r}= \sqrt{3.6748\times10^{4}} = 1.918\times10^{2}. ]

Hence

[ r = \frac{1\ {\rm AU}}{1.918\times10^{2}} = \frac{1.496\times10^{11}\ {\rm m}}{191.8} \approx 7.8\times10^{8}\ {\rm m}. ]

Recall that the solar radius is (R_{\odot}=6.96\times10^{8}\ {\rm m}).
The height above the photosphere is therefore

[ h = r - R_{\odot} \approx 7.8\times10^{8}\ {\rm m} - 6.96\times10^{8}\ {\rm m} \approx 8.4\times10^{7}\ {\rm m} = 8.4\times10^{4}\ {\rm km}. ]

2.6 Interpretation

With an ideal perfectly reflective suit ((\alpha=0)) the absorbed power would be zero at any distance, so theoretically you could get arbitrarily close to the Sun’s surface and still survive (ignoring mechanical stresses, radiation pressure, etc.).
In a realistic ultra‑reflective suit with (\alpha = 10^{-6}) you could approach to within ≈ 8 × 10⁴ km above the photosphere—roughly one‑tenth the Sun’s radius, or about one‑quarter the Earth–Moon distance—while keeping the extra heat load to ≤ 100 W.
If the absorptivity were ten times larger (α = 10⁻⁵) the safe altitude would increase by √10, to about 2.6 × 10⁵ km. Conversely, improving α by a factor of ten pushes the safe altitude down by √10.

3. Final answer

Perfectly reflective (α = 0): no radiative heating, so in principle you could get arbitrarily close to the Sun’s surface and still survive for an hour (subject to other non‑thermal hazards).
Practically achievable reflectivity (α ≈ 10⁻⁶): the maximum tolerable absorbed power (≈ 100 W) limits you to a distance of

[ \boxed{r \;\approx\; 7.8\times10^{8}\ {\rm m}\;( \;h\approx 8.4\times10^{4}\ {\rm km\ above\ the\ photosphere}\;)}. ]

4. Common mistakes

Mistake	Why it’s wrong	How to avoid it
Treating the solar flux as constant (using 1361 W m⁻² everywhere)	The Sun’s intensity follows an inverse‑square law; it grows dramatically near the Sun.	Write (F(r)=F_{\oplus}(1\ {\rm AU}/r)^{2}) before plugging numbers.
*Assuming a perfectly reflective surface absorbs nothing* without mentioning absorptivity**	No real material has α = 0; even the best mirrors absorb a tiny fraction, which matters because the flux is huge.	Introduce an explicit absorptivity α and keep it in the equations.
Neglecting the projected area	Power absorbed is proportional to the area facing the Sun. Using the wrong area leads to an order‑of‑magnitude error.	Use a realistic estimate (≈ 2 m² for a person) and keep the factor in the calculation.
Forgetting to compare absorbed power to a physiological limit	The question is about “survive for an hour”; you need a sensible limit (≈ 100 W) rather than an arbitrary temperature.	State the tolerable power (or temperature rise) and set (P_{\text{abs}}=P_{\text{tol}}).
Ignoring other hazards (radiation pressure, solar wind, tidal forces)	These become significant extremely close to the Sun, but they are secondary to the radiative heating problem asked.	Mention them qualitatively, but keep the quantitative analysis focused on radiative heat balance.

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