Marineris: The Invisible Valley? Yes It Is


Last week, I posed a question about Valles Marineris (The Mariner’s Valley) on Mars.  Being 200km wide, I asked you if the cliff faces of Marineris would be visible from the centre of the valley, or would they be so far away they would be over the horizon.  According to the current poll, about 66.6% think yes, and 33.3% think no.

Credit: NASA (Viking 1)

Well, let’s find out.  The distance to the horizon d is

 d = \sqrt{2Rh + h^2}

R is the radius of Mars (3,396 km) and h is the height of the observer in the canyon (6 feet, say).  Plugging in the numbers gives the distance as about 3.5 km, way less than the canyon’s width.  Even on Earth, the horizon distance is only about 5 km, so a Marineris-sized canyon would be invisible to us here too.  In fact, if you wanted to just be able to make out the canyon face (assuming an infinitely clear day of course), the planet you were standing on would have to be about 2 million kilometres in radius! That’s about 4 times the radius of the Sun! As you can see from the scale model (left), that is very, very big.  Indeed, it’d be very unlikely that a terrestrial planet of that size could ever exist in the Universe (not least because gathering up that much rocky material is very difficult).  So Marineris truly is an invisible valley – and at 7 km deep (i.e. 7 El Capitans high), quite a challenge for any would-be extraterrestrial climbers.

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4 thoughts on “Marineris: The Invisible Valley? Yes It Is

  1. Um. Shouldn’t the relevant value for h in your formula be 7 km? Yeah, the *base* of the cliffs (assuming a rectangular profile) will be over the horizon, but you might still be able to see the *top*. In fact, I get 218km, so, from the centre (100km), easily visible.

    Of course, that doesn’t settle it: for a non-rectangular profile, your horizon could be almost any suitably convex slope on the valley side.

    1. Duncan, the calculation you gave at the top is from the observer to the bottom of the cliff. Changing R-> R+cliffheight us the wrong way round. The above correspondant is correct about the top of the cliff. It is symmetrical can someone at the top of a 7km cliff see the bottom of the valley. In this case you don’t change R: that is just asking the question can someone on a 7km cliff see someone in the centre of the valley raised up 7 km and with the valley filled in. You can also ask what is the lowest height of the cliff that you can see. Here you can set d=100km and solve for h. I will leave this to you as I want to get back to watching the rugby.

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