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### Measurement: Raindrops, the Weight of the Earth

Date: 04/11/2001 at 19:23:58
From: Charles L. Moss
Subject: Raindrops

I think your web site is really cool. My question is this: since after
the Big Bang and the Earth was formed and it began to rain upon the
face of the Earth, I was wondering how many raindrops have fallen from
the sky to the earth.

Again, thanks for having this great Web site I stumbled on today.  You
are doing a good service for our kids and our country.

C. L. Moss

Date: 04/13/2001 at 03:31:30
From: Doctor Ian
Subject: Re: Raindrops

Hi Charles,

The problem with a question like this is that the assumptions that you
would need to make - about the age of the earth, and what kind of
weather existed billions or even millions of years ago, and which
continents have been above sea level at what times - are so loose that
any answer you would come up would be meaningless.  (It would be sort
of like if someone asked how many thoughts a person has in a normal
lifetime, and someone else answered 'Somewhere between 100 thousand
and 100 trillion'.)

But if you just wanted to get _some_ idea, here's what you might try.
Look in an almanac to get the average annual rainfall in every country
in the world, in inches.  Then, for each country, multiply the average
rainfall by the area to get the average volume of rain that falls on
the country in a given year (in units of square-mile-inches per year).
Then multiply that by your favorite value for the age of the earth.
That would give you total rainfall, in square-mile-inches.

Finally, convert that to liters, and multiply by 1000 to get
milliliters.  I believe that one raindrop is about 1 ml in volume, so

The radius of the earth is about 6350 km. So the surface area of the
earth is about 4 pi r^2, or 200 million square miles.  70% of that is
water, so 30% is land - about 60 million square miles.

There are 5280 feet in a mile, and 12 inches in a foot, so there are
63360 inches in a mile, so there are 63360^2 or about 4 billion square
inches in a square mile. That makes 60x10^6 * 4x10^9 = 240x10^15
square inches, or 2.4x10^17 cubic inches of water for each inch of
rainfall. We get about 40 inches of rain each year in Swarthmore -
let's pretend that's typical of the earth as a whole. Then we're
talking about 1x10^19 cubic inches of rainfall each year for the whole
earth. If you assume that it's been raining for a billion years, then
that's 1x10^28 cubic inches of rain, total.

How many raindrops is that?  Let's say that a raindrop is about 1 ml
of water. There are about 16 ml in a cubic inch, so we arrive at an
incredibly imprecise figure of 16x10^28 raindrops since the beginning
of time... which seems like an incredibly large number, until you
realize that there are about 10^24 molecules of sugar in one of those
packets you find in restaurants.

Does this help?  Write back if you'd like to talk about this some
more, or
if you have any other questions.

- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/

Date: 05/02/2001 at 10:01:43
From: Charles Moss
Subject: Re: Raindrops

Dr. Ian,

Thanks for the reply. I have one more math question. Has anyone ever
tried to calculate the weight of the Earth, first in lbs. and then in
tons?  We don't have a scale that big. How can something so heavy just
hang in the air on nothing?

C. L. Moss

Date: 05/02/2001 at 14:55:26
From: Doctor Ian
Subject: Re: Raindrops

Hi Charles,

There are all kinds of clever ways to weigh things. For example,
suppose you want to weigh an elephant, but you don't have a scale
large enough to do it. Here's one thing you can do: Put an empty boat
in the water. Draw a line on the side to show how far it sinks. Put
the elephant in the boat. Draw a new line. Remove the elephant, and
fill the boat with rocks, or some other heavy objects until it sinks
to the second line. Now weigh the individual objects, and add up their
weights. The sum will be the weight of the elephant!

The main idea is: If two things act as if they have the same weight,
then they have the same weight.

Long ago, a guy named Henry Cavendish came up with a way to determine
the absolute strength of the gravitational attraction between two
objects using a couple of lead balls and some very fine wire. In doing
so, he made it possible to weigh the earth.

How? Well, let's say that I know your weight is 150 lbs. That means
that the earth is exerting a force of 150 lbs on you, and you're
exerting the same amount of force on the earth. Now, I know that the
force of gravity is

m1 * m2
F = G --------
d^2

where m1 is your mass, m2 is the mass of the earth, d is the distance
between your center and the center of the earth (basically, the radius
of the earth), and G is the universal gravitational constant - the
number that Cavendish measured. Now, I know all of these values except
for m2, which means that I can solve for the mass of the earth:

F * d^2
------- = m2
G * m1

The value turns out to be about 6 x 10^24 kg.

A large part of science involves finding indirect ways to measure
things that we can't measure directly. For example, we can tell what
stars are made of, the temperature on Venus, and the speed with which
galaxies are moving, all by examining the light that we see coming
from the sky. We know the shape and composition of the earth in part
because of how the orbits of satellites are perturbed. And so on, and
so on.

This is the essence of modeling as measurement: If I can come up with
a formula for how quantity X should affect quantity Y, then I don't
have to measure X if I can measure Y instead.

I hope this helps.  Write back if you'd like to talk about this some
more, or if you have any other questions.

- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/

Associated Topics:
Middle School Measurement

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