In this question about the amount of sulfuric acid in the atmosphere of Venus it is calculated that the amount of water in it's atmosphere is $9.6 \times 10^{15} \text{ kg }$ H$_2$O.

In one of the answers to that question it is calculated that there is a total of $12 \times 10^{13} \text{ kg }$ H$_2$SO$_4$ in the clouds of Venus.
Assuming that at least 75% by weight of the droplets is sulfuric acid, this means that no more than $3 \times 10^{13} \text{ kg }$ H$_2$O is in those clouds.

So almost all of the water must be outside the clouds !

The figure on page 14 of the dissertation of Yeon Joo Lee shows that beneath the cloud layers there is a thin haze region which has a thousandfold less mass loading than the clouds above it, so the amount of water there is negligible.

So almost all of the water must be above the clouds !

Is this true or must there be some miscalculations ?

Is there any publication that can confirm the question ?

  • 1
    $\begingroup$ Could the water not simply be vapour rather than clouds of droplets? $\endgroup$ – Steve Linton May 26 '18 at 18:33
  • $\begingroup$ @SteveLinton Yes, could be, i don't know. So i've changed the question again ! $\endgroup$ – Cornelisinspace May 26 '18 at 19:14

No !

The conclusion that the amount of water in the haze region is negligible is false !

Figure 4 of the article The Recent Evolution of Climate on Venus from Bullock and Grinspoon shows the abundances of H$_2$SO$_4$ and H$_2$O in the cloud region of the atmosphere of Venus.

Fig. 4 Bullock & Grinspoon 2001 Venus aerosols

The dashed line for H$_2$O there shows that above the H$_2$SO$_4$ production region there is less than 4 ppm H$_2$O, gradually growing to above 26 ppm below the lower cloud boundary.

So there's much less water above than below the clouds whereas the conclusion in the question that most of the water in the atmosphere is outside the clouds looks justified.

So almost all of the water in the atmosphere of Venus must be below the clouds !

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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