# physics

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Assuming that 68.1% of the Earth's surface is covered with water at average depth of 1.21 mi. ,estimate the mass of the water on Earth. One mile is 1.609 km and the radius
of the earth is 6.37 x 10^6 m. Answer in units of kg

• physics -

The volume in cubic meters is the product of the ocean area (in m^2) and the average depth (in meters).

For the ocean area, take 68.1% of 4*pi*R^2.

For the mass of the water, multiply the volume by 1000 kg/m^3. Actually you should use a higher density of 1027 kg/m^3, corresponding to ocean salt water, since that is what most of it is.

Calculating these numbers yourself will improve your learning experience

• physics -

I multiplied 68.1 X 5.099043638x10^14=3.472448717x10^16 to get the ocean area.
Then to get the average depth i took 1.21mix1609m=1946.89.
Then i multiplied the ocean area and volume i got to equal 6.760475683x10^19.
To get the mass, I took that answer and multiplied it to 1000 kg/m^3. I got 6.760475683x10^22. If i put that in significant figures it woud just be 6.76 x 10^22 kg right? was my method correct?