# This is How to Calculate the Mass of our Earth in Gm and Total Number of Water Drops in it

## Earth

Though it looks like a very difficult job, it'll be really interesting if you go through these following calculations. These calculations are very simple and all the things which have been used here may recollect our school days where we may have studied about radius, volume and total surface area of the sphere. Our Earth is a great eliptical sphere and we need only the average density, radius of our earth and average depth of seawater to calculate both the mass of earth and total number of water drops in it.

Method for Calculating Mass of Earth in Gm

To calculate the mass of this great Earth we need only the following two things

2. Average Density of Earth

The radius of earth is about 6,378 Km. Normally the soil density is calculated as 2.6 gm/cc. But, the presence of larva (molten iron) in inner and outer core of the earth is giving rise to its overall density and its average density comes as 5.5 gm/cc. Now we can easily calculate the mass of our Earth as we have known both radius and average density of our earth.

In science, density is defined as mass per unit volume (i.e.) d (density) = m/v

Formula to find out the volume of earth or any sphere is 4/3*3.14*r^3

Before substituting the value for' r', we need to convert the radius of earth into cm which comes as 6.378*10^8 cm

Now we can calculate the total volume of earth as follows

4/3*3.14*6.378*10^8*6.378*10^8*6.378*10^8

=1.0862*10^27 cc

Now we can get the mass of earth by multiplying its volume and average density as follows

d=m/v
m=v*d

So, mass of the earth is = 1.0862*10^27 cc * 5.5 gm/cc

= 5.974*10^27 gm

Method for Calculating Total Number of Water Drops in our Earth

About 71% of earth surface has been covered by the seawater and it is estimated that average depth of sea on our earth surface is 4000 m (13,000 ft)

To calculate the total number of water drops in our great globe we need only the following things

1. Radius of Earth (6.378*10^8 cm)
2. Average depth of sea on our earth (4*10^5 cm)

First we need to calculate the total surface area of our earth. Formula to find out the total surface area of earth or any sphere is 4*3.14*r^2

Now we can substitute the value of ' r' to get the total surface area of the earth. But, only 71% of earth surface is covered by water. So, the total area of seawater on the earth is = 4*3.14*6.378*10^8*6.378*10^8*0.71 sq.cm

Average depth of sea is 4*10^5 cm

Now the total volume of seawater in our earth is = 4*3.14*6.378*10^8*6.378*10^8*4*10^5*0.71 cc

= 1.451*10^24 cc

Every 1 cc of water contains 20 drops of water. All the river and lake water in our earth consitute only 1% of total water in our earth. So, the total number of water drops in our earth including all the oceans, lakes and rivers is = 1.01*1.45*10^24*20

= 2.92*10^25

## More by this Author

ratnaveera 7 years ago from Cumbum Author

Vinayak 6 years ago

very good concept of calculation..

Thanks....!

vinayezone 6 years ago from Banglore

very good concept of calculation..

Thanks....!

ratnaveera 6 years ago from Cumbum Author

Dear vinayezone, Thanks a lot for coming here and your nice comments. VERY HAPPY NEW YEAR 2011

Shahid Bukhari 5 years ago from My Awareness in Being.

Very interesting ... no wonder, Mathematics was a Sport, in the medieval Arabian Concept !

Anyway, your calculations, cannot be Exact, because, of three factors, thus, Theoretical. These are :

1. Averages, involved in all your calculated assumptions.

2. No Mathematical Numbers Exist, to describe, except by hypothetical means, with raising the Power of simple, into theoretical numbers.

3. The factor of Gravity, which determines the Weight, of any object on earth ... has not been taken into account ... Thus, these are Calculations of the Dead Weights, and Volumes " of Solid, Liquid and Gaseous Masses of the earth,

Hope when you grow up, you will look into these questions.

ratnaveera 5 years ago from Cumbum Author

Dear Shahid Bukhari, Thanks for coming here and your valuable comments and points. Yes! I agree with your third point. But, In the above article I have used exact measurements as far as possible. I hope the above two answers could have more than 90% accuracy. Thanks again!

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