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2006-11-19 01:04:29 · 8 answers · asked by Nerzz O 1 in Science & Mathematics Astronomy & Space

8 answers

that seems imposible the weight would always change.....

2006-11-19 01:05:28 · answer #1 · answered by @ubreY 3 · 0 1

The "weight" of the Earth would depend on what gravitational field you wanted to weigh it in, which is a bit of a problem in defining your terms.

Supposing you mean the "mass" of the earth, there are two ways of doing it.

The Astronomer Royal, Nevil Maskelyne, did it in 1774 by measuring the deviation between the gravitational vertical and the astronomical vertical around the Scottish mountain Schiehallion. The mountain is relatively isolated and regularly shaped, and the results give the mass ratio of the mountain to the whole Earth. He was only about 20 per cent out.

The physicist Henry Cavendish did it in 1798 by directly measuring the gravitational force between two large masses in the laboratory, using a torsion balance. The results give the mass of the Earth quite simply, provided its radius is known.

2006-11-20 01:18:17 · answer #2 · answered by bh8153 7 · 0 0

Weight must be defined in the context of under what circumstances the object to be weighed exists.

For example, when you step on a scales you are measuring the force by which the earth is pulling you towards itself. This force is called gravity.

Now what force are you intending the earth to be pulled by?

What I mean is, are you interested in what the force is that the sun's gravity is pulling at the earth? Or the moon? or another planet? Or star?

These all have different answers.

Or are you interested in the hypothetical what does the earth weigh if it were on a scale on the earth itself?

Again, that would be another answer.

Be clear and an answer can be calculated.

2006-11-19 03:57:29 · answer #3 · answered by Carl 3 · 0 1

Do you want to find the mass of the earth?

Ok, I have an inaccurate method of doing it. I'll explain why it's inaccurate later.


"Two objects will attract each other proportional to their masses and inversely proportional to the square of distance between them. If the mass of one body is designated as M, the mass of the other as m, and the distance between them is r, then the force of attraction F between the two bodies is:
F = G*M*m/r^2

where G is the universal gravitational constant.

G = 6.67*10-11 N-m^2/kg^2. The units of G can be stated as Newton meter-squared per kilogram-squared or Newton square meter per square kilogram."

Ok, find an object's mass by inertia. Using a spring for measuring force, find the force it takes to move an object at constant speed in a place with as little friction as possible.
Then, in the same place, use a bigger force, and keep the force constant as you drag the object.

Ok, when you do that, you need to measure its speed. If you have a cool laser that measure distance, then cool. have someone write what it says exactly every 2 seconds or something like that. Or you can videotape it and a clock next to it.

F= M * A
F= Overall Force
M = Mass
A = acceleration or increase in speed.

The first time you did it, you found out how much F it took to move the object at constant speed, with no acceleration. This is how much friction there is, because if there wasn't friction, then it would have gone at a constant speed all by itself.

(Force of trial 2 - Force of trial 1) = Mass * Acceleration of trial 2

Mass = (Force of trial 2 - Force of trial 1) /Acceleration of trial 2


Now you know the mass.

Now put the object under the spring that measures force, and lift the object.

Now you know the force of attraction between the object and the Earth.

Remember:
F = G*M*m/r^2

F = Force of attraction (You know it)
G = 6.67*10-11 N-m^2/kg^2 (A constant, so you know it.)
M = Mass of object 1 (You know it)
m = mass of object 2 (Mass of Earth, we want it)
r = distance between the two. (Upps, how do you get this?)


Now, some parts of the Earth are closer to you, and some are farther away. Lets assume that the earth is totally even and that it is the size of a ping pong ball.

How many feet meters, whatever, are you above sea level?

Use that, and the fact that Sea level is ~ 21,000,000 ft or
6.38 x 10^6 meters from the core. If you live near the equator, add another 3,000 meters or 10,000 ft.
If you live close, to the equator, like Mexico, add 1,500 meters or 5,000 ft, because the Earth's sealevel changes all over it.

Your distance, r, is sealevel + distance above sealevel
or
sealevel - distance under sealevel.

Now plug it all in!

F = G*M*m/r^2

F = Force of attraction (You know it)
G = 6.67*10-11 N-m^2/kg^2 (A constant, so you know it.)
M = Mass of object 1 (You know it)
m = mass of object 2 (Mass of Earth, we want it)
r = distance between the two. (You know it


F = G*M*m/r^2

m = (F*r^2) / (G * M)

Wow. The way I told you to do it is really inaccurate, and is technically based on the Earth's Mass.

We used the Earth's mass to calculate the Earth's mass.

But that's the best we can do.

2006-11-19 01:18:33 · answer #4 · answered by husam 4 · 0 1

You have to understand the difference between "weight" and "mass". Mass is the quantity of matter, "stuff" if you will, in an object. Weight is the measure of an objects response to a gravitational field, specifically, the response "left over" after it is responding fully. So, since the earth is fully responding to the gravitational fields acting upon it, and is in free fall, the earth is "weightless" just as astronauts in free orbit around the earth are.

2006-11-19 02:09:21 · answer #5 · answered by JIMBO 4 · 0 1

I fond the answer on the link

2006-11-19 01:09:26 · answer #6 · answered by maussy 7 · 0 0

well there is a formula given by newton
F=(G*m*M)R*R
G = gravitational constant
m = mass of one body
M = second body
R = distance btw them
if you know F,G,R and M then you can get 'm'

2006-11-19 01:10:36 · answer #7 · answered by new_einstein 2 · 0 1

its impossible, unless you use some great science and math

2006-11-19 01:05:45 · answer #8 · answered by dirtbiker82004 2 · 0 1

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