I know the following: 1” is equivalent to 0.0254 m, the radius (R) of the Earth is 6,370,000 m, the surface area of a sphere is 4πR2, and the density of water is 1000 kg m-3. (You may approximate π as 3.14).
Also, the earths atmosphere (aka enough water to cover the surface of the earth with 1") is 3.75 x 10^6 gallons
Thank you!
Also, the earths atmosphere (aka enough water to cover the surface of the earth with 1") is 3.75 x 10^6 gallons
Thank you!
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So, first you calculate the surface area of the earth. Then, volume is the surface area multiplied by the height (0.0254m). Once you have the volume, you use the density of water to convert it into a mass.
It will go like this:
Surface Area: 4πR^2 = 4π(6370000m)^2 = 5.1*10^14 m^2
Volume with 1": Area*Height = (5.1*10^14 m^2)(0.0254m) = 1.3*10^13 m^3
Mass: density*volume = (1000kg/m^3)(1.3*10^13 m^3) = 1.3*10^16 kg
I don't know why you need to know the gallons of water in the atmosphere. That isn't the amount needed to cover the earth in 1" of water. A gallon is 0.00379m^3, so (0.00379m^3/gal)(3.75*10^6 gal) = 14,212.5 m^3.
It will go like this:
Surface Area: 4πR^2 = 4π(6370000m)^2 = 5.1*10^14 m^2
Volume with 1": Area*Height = (5.1*10^14 m^2)(0.0254m) = 1.3*10^13 m^3
Mass: density*volume = (1000kg/m^3)(1.3*10^13 m^3) = 1.3*10^16 kg
I don't know why you need to know the gallons of water in the atmosphere. That isn't the amount needed to cover the earth in 1" of water. A gallon is 0.00379m^3, so (0.00379m^3/gal)(3.75*10^6 gal) = 14,212.5 m^3.