Expert: Kevin Johnson - 8/22/2005

Question
Kevin, my knowledge level on this is almost minimal.

The thought has come up about the volume of water that would have to have fallen during the biblical great flood.

Considering the surface area of the earth. Then what the surface area would be at the elevation of the peak of Mr. Arafat. Namely, finding the space between. How much water (rain) would have to have fallen to fill that void?

In 40 days, what would be the average rainfall per hour?

What would the weight of the water have been at peak flood?

How long would it have taken for said water to evaporate into the atmosphere?

This is a point of curiosity. Not an earth shaking question. (Unless the weight would have compressed the earth.) Or, the rate of fall would have been a solid sheet of water.

Thanks, Jim.  

Answer
Hello Jim Martens,

The earth has a radius of approx. R=6374 km.
Its surface area is thus A= 4*Pi*R² = 5.1 *10^14 m².

Mr. Arafat (the late President of Palestine) was not very tall, but Mt. Ararat (the volcano mountain in Northeastern Turkey) is 5165 m high.
This makes the water volume necessary to flood it to its peak V= 2.6 * 10^18 m³. (Or 2.5 million million million cubic meters of water)
This water weighs 2.6 * 10^18 tons.

Spread out to 40 days the average rainfall would be 5165m / 40 d = 129 000 mm per day or 5375 mm per hour. Imagine standing under a waterfall.

Evaporated it would saturate the whole atmosphere plus the stratosphere with 105% humidity. In other words: Clouds would fill the atmosphere from the ground on upwards plus the stratosphere (where there are usually no clouds - "above the weather"). This would lead to the earth freezing under the clouds since no sunlight reaches the ground. The consequences would be harsher than the imagined "nuclear winter" after a global thermonuclear war.

I hope this answer was helpful for you.

Peace,
Kevin.