This was just released today by NASA. It is quite a humbling image from the Mars rover Curiosity, though it’s not quite the same impact as the Blue Marble image from Apollo 8, but historically significant nonetheless.
This view of the twilight sky and Martian horizon taken by NASA’s Curiosity Mars rover includes Earth as the brightest point of light in the night sky. Earth is a little left of center in the image, and our moon is just below Earth.
Researchers used the left eye camera of Curiosity’s Mast Camera (Mastcam) to capture this scene about 80 minutes after sunset on the 529th Martian day, or sol, of the rover’s work on Mars (Jan. 31, 2014). The image has been processed to remove effects of cosmic rays.
A human observer with normal vision, if standing on Mars, could easily see Earth and the moon as two distinct, bright “evening stars.”
NASA’s Jet Propulsion Laboratory, a division of the California Institute of Technology, Pasadena, manages the Mars Science Laboratory Project for NASA’s Science Mission Directorate, Washington. JPL designed and built the project’s Curiosity rover. Malin Space Science Systems, San Diego, built and operates the rover’s Mastcam.
More information about Curiosity is online at http://www.nasa.gov/msl and http://mars.jpl.nasa.gov/msl/.
Click for a full resolution image:
Source: http://www.jpl.nasa.gov/spaceimages/details.php?id=PIA17936
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Gunga Din –
I am sure somone can tell you the exact relative brightness of Earth from Mars vs Venus from Earth, but here would be my off the top of my head thoughts:
The minimum brightness of Venus when it is on the far side of the Sun is magnitude –3.82.
The maximum brightness of Venus when it is lit up as a crescent is magnitude -4.89.
The Earth from Mars at maximum elongation*** (47.4°), the Earth and Moon would shine at apparent magnitudes −2.5 and +0.9, respectively.
Astrobob says, “If you could be there in person, the home planet would appear as a pale blue “star” shining at magnitude -1, a little fainter than Sirius, the brightest star in the skies of both planets.”
Ah, -1 is pretty bright.
Here is an equation: (m2 – m1) = 2.5log10(b1/b2)
– – – m1 and m2 are the two brightnesses
– – – (b1/b2) is the apparent brightness of one versus the other
Solving for log10(b1/b2) gives (m2 – m1)/2.5.
So from the numbers above (for the Curiosity photo) we can figure as follows:
For Venus min magnitude:
m2 = -1 [Earth]
m1 = -3.82 [Venus minimum]
log10(b1/b2) = (-1 – (-3.82))/2.5 = 2.82/2.5 = 1.128
The antilog10 of 1.128 = 3.0895
For Venus max magnitude:
m2 = -1 [Earth]
m1 = -4.89 [Venus minimum]
log10(b1/b2) = (-1 – (-4.889))/2.5 = 3.89/2.5 = 1.556
The antilog10 of 1.556 = 4.7398
So Earth in the Curiosity image is
3.0895 times brighter than Venus from Earth when Venus is at minimum magnitude.
4.7398 times brighter than Venus from Earth when Venus is at maximum magnitude.
That is, if I did the math right. (I am rusty and old now…LOL)
You should be able to do the same for the Moon from Mars vs Venus from Earth, if you google the magnitude of the Moon from Mars.
The -1 magnitude does explain to some extent why the image shows Earth and the Moon and VERY barely shows other stars. If Sirius was in the photo it would be slightly brighter, but most stars would be fainter. I would think that a longer exposure would bring out the other stars.
I was expecting Earth to be about the same magnitude as Venus from Earth, meaning like a bright star, but it is MUCH brighter. I learned something. Even if I might have messed up the exact numbers.
Thank you, Steve Garcia, for that fine answer! Much appreciated. Now, if there were only an easy way to let Gunga Din know… .
Steve Garcia says:
February 9, 2014 at 12:28 pm
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Thank you, Steve.
(And thank you, Janice, for making sure I say it.)
“(And thank you, Janice, for making sure I say it.)”
Should be:
“(And thank you, Janice, for making sure I saw it.)”