If earth were completely smooth, excluding space for current water bodies, what weather should i expect?

I'm under the assumption that wind would be a lot stronger, but i have nothing to back that up.

  • 1
    $\begingroup$ How would the change between water and dry land go? Gentle slope onto a sandy beach? Also is there a moon? $\endgroup$ – Mormacil Mar 19 '17 at 16:37
  • $\begingroup$ Is there shrubbery? Trees? Wildlife? Vegetation? $\endgroup$ – Erin Thursby Mar 19 '17 at 16:43
  • $\begingroup$ Yes, the accepted answer of the older question specifically includes weather detail. $\endgroup$ – JDługosz Mar 19 '17 at 20:20

Down below the tip of Cape Horn, there is an open expanse of water which circumnavigates the globe with almost no obstructing land masses to disrupt the winds.
Referred to as The Roaring Forties, The Furious Fifties and the Screaming Sixties, the winds in this region are among the strongest and most consistent on the planet.

From that, I think we can assume that your version of earth, occupied by non-obtrusive, wind-friendly land masses would be a little more windy than our current world.

| improve this answer | |
  • $\begingroup$ I believe that your higher winds are caused by air currents falling from elevated regions and into your flat unobstructed region. Given you've described a path of least resistance in an otherwise resisting world I would expect its behavior to be rather different than a world where there is no unique path like this. I don't believe it can be generalised. $\endgroup$ – Lio Elbammalf Mar 19 '17 at 17:43
  • $\begingroup$ @LioElbammalf, I hadn't thought of it that way.. Thanks for the clarification. Does that leave us with no method of figuring out the wind speed on a perfectly smooth planet? $\endgroup$ – Henry Taylor Mar 19 '17 at 19:32
  • $\begingroup$ I think it does make it difficult to figure out. Wind is really just air of different densities (temperatures) mixing such that they would tend to a uniform density. On a uniform world I think it would be dominated by the difference in temperature between the poles and equator. I'm not quite sure how you would determine the strength of that wind though. $\endgroup$ – Lio Elbammalf Mar 19 '17 at 20:13

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