NASA's Parker Solar Probe Captures its First Images of Venus' Surface in Visible Light, Confirmed links to the new Geophysical Research Letter Parker Solar Probe Imaging of the Night Side of Venus.

The image below shows an image of Venus taken by Parker Solar Probe during a gravitational assist flyby on the left, and a topographical representation of the same surface on the right, with high altitudes shown as dark and low altitudes as bright shades, the idea being that the lowlands are hotter and so glow more brightly due to thermal radiation.

The modeled surface temperature of the night side here is 735 Kelvin.

While much of the light captured by the camera through its circa ~800 nm long wavelength cutoff can be called "near infrared" the authors estimate about 1/3 of the signal detected is below 750 nm, thus the first in visible light headlines.

Question: How brightly does Venus's hot surface glow at night? Could you see it? Could you see well enough to walk around? (assuming that you had dark-adapted eyes and a really nice suit!

Figure 1. (a) Wide-Field Imager for Parker Solar Probe-I (WISPR-I) image of the nightside of Venus from Venus gravity assist (VGA) 3, showing thermal emission from the surface on the disk and O2 nightglow emission at the limb.

Figure 1

(a) Wide-Field Imager for Parker Solar Probe-I (WISPR-I) image of the nightside of Venus from Venus gravity assist (VGA) 3, showing thermal emission from the surface on the disk and O2 nightglow emission at the limb. Black to white represents 0 DN s−1 to 40 DN s−1 with the scale saturated at 40 DN s−1. The image is contaminated by numerous roughly horizontal dust streaks, from material ablating off the Parker Solar Probe heat shield. (b) Topographical map from Magellan, using an inverse black and white scale to match the WISPR image, with bright regions being low elevation and dark regions being high elevation. (c) WISPR-I and -O images of Venus from VGA4. The same part of the Venusian surface is observed as in (a). Red numbers in all panels mark common features for ease of reference. A movie of the VGA4 images is available in the online article.

source: Wood et al. 2022 Parker Solar Probe Imaging of the Night Side of Venus

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    $\begingroup$ Can't find the source but I recall an article quoting a Soviet Venera engineer saying that the daytime lighting level was akin to dusk on Earth. $\endgroup$ Feb 10 at 0:19
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    $\begingroup$ Sorry, but I deal in deg F. Surface temp as stated to be 735 K would be about 860F. Just from experience steel at 900 - 1000 F appears to be very dull red. If that was the only light souce you couldn't see anything from it. I doubt if any material of venus surface would have any dramatically different emission properties than a piece of steel. So I would say you couldn't see any glow and you couldn't walk around in a safe manner based upon that light. $\endgroup$
    – tckosvic
    Feb 10 at 3:15
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    $\begingroup$ For example, the exhaust duct on a stationary power gas turbine engine glows dull red at 1000 F at night. Been there lots of times. $\endgroup$
    – tckosvic
    Feb 10 at 5:08
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    $\begingroup$ @BrendanLuke15: I recall reading something similar during the past couple of weeks, wjile reading material for this question. I think it might have been a Wikipedia article. $\endgroup$
    – Fred
    Feb 10 at 9:08
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    $\begingroup$ @BrendanLuke15 "... the idea being that the lowlands are hotter and so glow more brightly due to thermal radiation." $\endgroup$
    – uhoh
    Feb 10 at 19:14

2 Answers 2


Keep calm and apply the Planck Distribution Law for radiant heat transfer. Integrated between two different wavelengths and including an emissivity factor of 0.8 (reasonably good for most rocks), we render the intensity of radiation thusly:

$I=(0.8)(2\pi c^2h)\int_{\lambda_1}^{\lambda_2}\dfrac{\lambda^{-5}d\lambda}{\exp(ch/(\lambda kT))-1},$

where $c$ is the speed of light, $h$ is Planck's constant, $\lambda$ is tgw wavelength and $T$ is the temperature.

When we plug in $T=735\text{ K}$ and integrate through the visible range from $4×10^{-7}$ to $8×10^{-7}$ meters, we get only about $0.8 \text{ mW/m}^2$. This means the glow is equivalent to placing 25-watt light bulbs in a square array some tens of meters apart (how many "tens of meters" depends on how efficient they are at converting power to light).

Let us try converting this into lux, the standard unit for illumination. According to https://ambientweather.com/faqs/question/view/id/1452, the conversion factor for sunlight is $126.7 \text{ lux/(W/m²)}$, which brings us to roughly $0.1$ lux given the above radiant output of $0.8\text{ mW/m}^2$, comparable to the light of a full Moon.

The problem with this calculation, of course, is that the actual radiance from the rocks is heavily on the red end of the spectrum; more than half of the integrated visible radiance calculated from the Planck Distribution Law comes just from the wavelength range between $7.5×10^{-7}$ and $8.0×10^{-7}\text{ m}$. Therefore, light perception is less efficient than for the overall spectral range of sunlight. So we have to assume that the true perception will be much less than the full-Moon comparison noted above, making this a dim vista indeed.

So you're in the dark ! When the surface is said to glow at night, the glow is detected by sensitive instruments, as humans are as yet nowhere near capable of landing on the surface of Venus.

  • $\begingroup$ Hopefully for this question, including the case where all the light is red ... . $\endgroup$ Apr 24 at 15:30
  • $\begingroup$ From livescience.com/50678-visible-light.html : "At about 800⁰ C, the energy radiated by an object reaches the infrared. As the temperature increases, the energy moves into the visible spectrum and the object appears to have a reddish glow." The max. temp. on Venus is 482⁰ C, so how can this be visible to the human eye ? $\endgroup$
    – Cornelis
    Apr 24 at 21:26
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    $\begingroup$ Please read again. I am not saying it is visible. I am saying the luminosity is so low the setting will be dark. And the glow is not being detected by humans. $\endgroup$ Apr 24 at 21:29
  • $\begingroup$ "...and the glow is not being detected by human." can be supported and verified or just as likely disproven quantitatively. Just for example see Thresholds and noise limitations of colour vision in dim light (also here) to lesser extent What are the limits of human vision? The answer is out there, we don't need to remain in the dark. $\endgroup$
    – uhoh
    Apr 24 at 23:11
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    $\begingroup$ Answer(s) to How can I estimate the optical power that a single-color LED generates? can be convoluted with your $I(\lambda)$ $\endgroup$
    – uhoh
    Apr 24 at 23:29

No. You could not see by blackbody radiation in either IR or visible regardless of brightness.

To be able to “see” something depends on both brightness and contrast. There may be enough radiation (visible or infrared) but if there is no contrast between object and surround, the object cannot be discerned.

Low contrast is the problem in the “polar bear in a snow storm” scenario. The object (bear) and surround (snowscape) have the same reflectivity so the bear cannot be discerned by reflected light.

In a landscape illuminated by its own blackbody radiation, all objects with the same emissivity will have the same brightness and cannot be discerned from each other. An object in a forging furnace is initially black against the bright red interior. As its temperature equilibrates with the furnace, it almost disappears.

enter image description here

Some low level contrast is still present since the angular emissivity of real grey-body objects is not perfectly isotropic like an idealized black-body.



The image below is an Earth FLIR landscape by night time blackbody radiation. No foreground detail can be seen.

enter image description here

So on Venus at night, using FLIR or not, you would see a relatively featureless landscape under a slightly darker, featureless sky.

  • $\begingroup$ I think on Venus at night, you would see nothing at all, because there's no brightess at all, whatever how much contrast there may be. $\endgroup$
    – Cornelis
    May 8 at 10:36

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