# Mid-latitude Mar's ancient covalent bonded ice?

1. Jun 7, 2010

### cph

Might ancient mid-latitude Mar's ice be covalently bonded? Would 1-2 billion year old ice have much greater density then, compared to earth ice? So drilling into Mars' ice, compared to drilling into earth ice, and measure relative difference in voltage/ampere as a proxy for relative density. Could one bring back a cube of Mars ice, without melting? Also wouldn't such guessed at Mars ice sink in water? Would it super cool your drink?

2. Jun 7, 2010

### alxm

No, because there's no stable covalent bond to be formed. The HOMO of water is anti-bonding.

Depending on pressure and temperature conditions, you can have different forms of ice; we already know of at least a dozen different phases of ice.
But I don't think that we're going to find anything new or unique in martian ice, because the conditions on mars aren't very exotic compared to what can be achived in the lab.

3. Jun 8, 2010

### alxm

Whoops, wrote HOMO, meant LUMO.

4. Aug 8, 2010

### cph

Re: Mars' orbital telescope and ice rock
Originally Posted by cph View Post

Might one place a telescope in Mars orbit, in order to obtain 2 mm resolution, looking for closeup geological views; such as mars' stromatolytic fossils, or mars' ice (rock) on the surface, in form of a smooth surface that seems a bit odd? Might billions year old Mars' ice be opaque to light, and take on the appearance of rock, but very light weight rock? So is there mars' ice rock on the surface? Do any of the Mars meteroites on earth have in part mars rock ice?

As noted 2 mm resolution (that of a field geologist) would be better. Also placing a Moon orbiter (rover-like) with a telescope, looking for Moon rock ice etc. For example, what appears to be a rock effacement, might continue into mottled appearance for near to reflected light. Thus a geological data base at 2mm resolution also for the Moon; equivalent to placing a geologist on the Moon.

5. Aug 9, 2010

### Staff: Mentor

2 mm at 200 km, that means sin(θ)=10-8. Using Rayleigh criterion

$$\sin \theta = 1.220 \frac {\lambda}{D}$$

and assuming visible light at 500 nm, you need lens with 61 meter diameter. Good luck.