Measurement: Raindrops, the Weight of the EarthDate: 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 that would be your answer. 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/ |
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