I'm trying to determine how long a probe could survive on Venus' surface. Assuming its just a sphere of titanium with room temperature air inside, what would be the best method for determining the heat transfer between the atmosphere (superheated carbon dioxide) and the inside of the sphere? How long will it take to reach equivalent temperature?

  • 1
    $\begingroup$ In the first instance, the standard formulae for thermal conductivity. You will need to know the size and thickness of the insulator (the shell), and the heat capacity of your heatsink (the volume of the air inside the shell) and, assuming your probe actually does something, the heat it generates itself $\endgroup$
    – user20636
    Commented Oct 6, 2020 at 8:07
  • 2
    $\begingroup$ This seems more of a physics.SE question, it's about thermal dynamics not space. $\endgroup$
    – GdD
    Commented Oct 6, 2020 at 8:26
  • 1
  • 3
    $\begingroup$ Soviet landers had a layer of thermal insulation and heat absorbers. $\endgroup$
    – A. Rumlin
    Commented Oct 6, 2020 at 9:06
  • 1
    $\begingroup$ I’m voting to close this question because this is purely a physics (heat-transfer) problem $\endgroup$ Commented Oct 6, 2020 at 11:40

2 Answers 2


Summary: The time will depend on how the inside of the spacecraft is insulated, but if we assume that you are in contact with the metal shell of a spacecraft similar to the lunar module (and make a lot of approximations regarding convection in the Venusian atmosphere), you will get serious burns within 15 minutes. The assumptions I make break down as the temperature of the spacecraft approaches the outside temperature (convective transfer slows), but you can probably expect it to reach oven temperature (350 °F, 450 K) within two hours.

First we need to understand what kind of convective regime we are in. The atmosphere of Venus is mostly carbon dioxide (96.5% by volume?) in the supercritical phase, so I'm going to rely a lot on its properties.

Let's first compute the Rayleigh number, which tells us whether the natural convection will be turbulent.

$$\textsf{Ra} = \frac{\rho g \beta}{\eta \alpha} (T-T_\mathrm{env}) D^3$$

Here is the data I'm using.
Atmospheric density: $\rho = 65~\mathrm{kg/m^3}$ (source)
Acceleration of gravity: $g = 8.87~\mathrm{m/s^2}$ (source)
Temperature of environment: $T_\mathrm{env} = 737~\mathrm{K}$ (source)
Temperature of spacecraft surface: $T = 294~\mathrm{K}$
Film temperature (used in calculating the quantities below): $T_f = (T + T_\mathrm{env})/2 = 515.5~\mathrm{K}$
Thermal expansion coefficient at $515.5~\mathrm{K}$: $\beta = 0.00194~\mathrm{K}^{-1}$ (ideal gas law)
Atmospheric pressure: $p = 92~\mathrm{bar}$ (source)
Dynamic viscosity of CO2 at 515.5 K and 92 bar: $\eta = 2.50 \times 10^{-5}~\mathrm{N \cdot s/m^2}$ (source)
Thermal conductivity of CO2 at 100 bar and 450 K: $k = 0.03392~\mathrm{W/(m \cdot K)}$ (source)
Isobaric specific heat of CO2 at 500 K: $c_p = 1014~\mathrm{J/(kg \cdot K)}$ (source)
Thermal diffusivity of atmosphere: $\alpha = k/(\rho c_p) = 5.15 \times 10^{-7}~\mathrm{m^2/s}$
Characteristic size: $D = 6~\mathrm{m}$

I end up with $\mathsf{Ra} = 8.33\times 10^{15}$, which is more than sufficient for Rayleigh–Bénard convection cells to form.

I can't find relations for heat transfer coefficients for a sphere, so I'm just going to treat the spacecraft as a horizontal plate. Our Rayleigh number is also out of the range of validity for this equation (supposed to be $<3\times10^{10}$), but I'm going to ignore that too.

$$h = \frac{0.27 k}{D}\,\mathsf{Ra}^{1/4} \sim 1.5~\mathrm{W/(K \cdot m^2)}$$

How long this takes for this heat transfer coefficient to kill you will depend a lot on the interior structure of the spaceship and possibly convection inside the spaceship. You will survive a lot longer if you are in the center of the spaceship surrounded by foam insulation. I don't really want to try to address that, so let's just assume your body is in contact with the titanium shell. I'm also going to ignore the fact that the convection field outside the spacecraft will dissipate as it cools off (you will probably die before that effect is significant anyway). The specific heat of titanium also changes as it warms up, but let's just use a constant value.

Specific heat of Ti: $c = 0.52~\mathrm{J/(g \cdot K)}$
Spacecraft area: $A= 113~\mathrm{m}^2$
Spacecraft thickness: $L = 1.5~\mathrm{cm}$
Density of Ti: $L = 4.5~\mathrm{g/cm^3}$
Spacecraft mass: $AL\rho_\mathrm{Ti}$ = $2545~\mathrm{kg}$ (similar to dry mass of the lunar module ascent stage)

The characteristic time for Newton's law of cooling is: $$\tau = \frac{mc}{hA} \approx 130.1~\mathrm{minutes}$$

The temperature of the spacecraft shell as a function of time is: $$T(t) = T_\mathrm{env} + \left( T(0) - T_\mathrm{env} \right) \mathrm{e}^{-t/\tau}$$

The time at which the spacecraft shell reaches a temperature $T$ in kelvin is: $$t_T = -\tau \ln \left( \frac{737-T}{443}\right)$$

Burns develop on skin within a second for temperatures above 343 K, so you will start getting severe burns on any portion of your body in contact with the spacecraft shell within 15 minutes.

Edit: I'm also ignoring wind. Winds at the surface of Venus are supposed be pretty still ($<2~\mathrm{m/s}$). But if there are high winds, expect to roast more quickly.

  • $\begingroup$ Wow, so the heat transfer between a gas and a solid surface can be at least estimated simply from the conductivity of the gas itself, plus $\mathsf{Ra}$ and some $D$. $$h =\frac{0.27 k}{D}\,\mathsf{Ra}^{1/4}$$ What is $D$ and why 6 meters? This gets the coveted +n! upvote award btw $\endgroup$
    – uhoh
    Commented Oct 7, 2020 at 22:36
  • 1
    $\begingroup$ @uhoh It started out being roughly the size of my office, but it's also roughly the size of the lunar module. (7.04 m * 4.22 m * 9.4 m)^(1/3) = 6.5 m $\endgroup$ Commented Oct 7, 2020 at 22:41
  • 2
    $\begingroup$ Ah, I see, $\frac{\mathsf{Ra}^{1/4}}{D} \sim D^{-1/4}$ is almost independent of size; it's not a big effect $\endgroup$
    – uhoh
    Commented Oct 7, 2020 at 23:10
  • 1
    $\begingroup$ @uhoh That's an interesting point. I guess the strongest contributor is the conductivity $k$, which contributes $k^{3/4}$. $\endgroup$ Commented Oct 8, 2020 at 14:17

The question seems primarily interested in the rate of heat transfer from Venus to the spherical cow spacecraft via conduction and convection types of transfer. That includes both inelastic collisions of atmospheric molecules with the sphere and also adhesion of hot particulates and/or droplets of any aerosol if that happens.

The high density of the atmosphere means that atomic collisions will happen about 160x more frequently than in Earth's atmosphere; 100x more due to density and 1.6x more due to speed (flux ~ areal density x velocity).

This kind of calculation is central to engineering any aerocraft or lander that may go to bodies with atmospheres. In addition to hot ones like Venus there are cold ones like Titan; either way there's heat transfer from the craft to the environment and it's very important to have a realistic estimate.

That being said, I don't know!

However, the Stefan–Boltzmann constant $\sigma$ is 5.67E-8 W/m2/K4 It's the only constant besides pi that I can remember four digits of off the top of my head, because it's got 5, 6, 7, 8

and the Stefan–Boltzmann law says that a sphere of radius $R$ will receive a radiant power of something of the order of

$$4\pi R^2 \sigma T^4$$

A sphere with a volume of 1 cubic meter has a diameter of 1.24 meters. That's about the same size as a one ton spherical cow.

It has an area of about 4.84 square meters. Thinking of the Venusian surface and atmosphere as a blackbody cavity, there will be 73 kilowatts of radiant power incident on our cow sphere!

If it's fairly highly reflective over most of its area, maybe one can get that down to 10 kW (a surface-averaged reflectivity of thermal IR of 85%). It's going to have to be smooth, shiny metal, to do that, and the metal will have to withstand Venus' chemical environment.

If our sphere were indeed a spherical cow with the heat capacity of water needing 4200 Joules for each kg to be raised by one degree, then our spacecraft would warm one degree every 7 minutes, or 100 degrees per every 11.7 hours.

Make all of that 7 times shorter if the spacecraft is dark/absorbing for thermal IR.

While not the central theme of the question, the other answers that will hopefully be posted can be compared to this.

  • 1
    $\begingroup$ @Uwe I've addressed exactly that within the question: "That being said, I don't know!" and "While not the central theme of your question, the other answers that will hopefully be posted can be compared to this." We're both looking forward to better answers! $\endgroup$
    – uhoh
    Commented Oct 6, 2020 at 10:20
  • 1
    $\begingroup$ @Cornelisinspace Thanks. hmm... well then a 5 mm thick titanium wall is 22.5 kg/m^2 and it takes 540 Joules to raise 1 kg by 1 Kelvin, or 12 kJ/m^2. The raw IR flux is 16.7 kJ/sec (kW) so if the shell is dark it will heat about 1.3 K/sec! at first. Of course it will slow down because it's an exponential and the rate that the shell radiates back becomes significant (the rate is really $\sigma (T_{venus}^4 - T_{sphere}^4)$ so it will equilibrate most of the way in about 15 minutes if dark, and a few hours if shiny. The air inside does nothing to slow it down, that's why I added water. $\endgroup$
    – uhoh
    Commented Oct 7, 2020 at 14:41
  • 2
    $\begingroup$ @Cornelisinspace I learned this the hard way in shop class a million years ago. We made chisels from hexagonal steel stock first with a furnace, anvil and hammer, then using a grinding wheel to make it sharp. You could hold one end in your hands while grinding the other end until it got so hot it turned blue from oxidation. Years later I tried to grind one end of an aluminum rod holding the other end in my hand, screamed and almost burned myself, conductivity of a metal like aluminum is far higher than steel. Ti is not very high, but this is only 5mm vs 20 cm. $\endgroup$
    – uhoh
    Commented Oct 7, 2020 at 15:47
  • 1
    $\begingroup$ @Cornelisinspace looks like the area 4.84 m^2 is correct but I left off the 4 in the expression; feel free edit & fix, thanks! $\endgroup$
    – uhoh
    Commented Oct 18, 2020 at 15:37
  • 1
    $\begingroup$ @Cornelisinspace "...then our spacecraft would warm one degree every 7 minutes... Make all of that 7 times shorter if the spacecraft is dark/absorbing for thermal IR." The idea being that they would have tried to make it mostly reflective so it would fry (or boil) too quickly. $\endgroup$
    – uhoh
    Commented Oct 19, 2020 at 9:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.