The weight of Earth

Originally posted by Palynka
Don't forget that since it is a sum of vector forces then many cancel out.

With respect to the forces we can perceive, it seems to be zero because the Earth is in free fall w.r.t. the sun, which is the next logical reference.

Note that you need a reference to calculate that vector. Since we have no idea where the center point of the universe is, it makes sense to choose the sun as the next reference.

I didn't forget that they may cancel out. However, the distribution of matter in the universe has been observed to be highly non-uniform, so I stand by my original statement that I think it unlikely that the vector sum of the forces would be zero. The arbitrariness of the choice of origin won't affect this.

Also, "apparent weight" is not the same thing as "weight". Gravity doesn't disappear because we're in free fall, as you know.

Originally posted by PBE6
Where did the reference to the Earth come from? It's not in the definition:

http://dictionary.reference.com/search?q=weight&x=0&y=0
http://en.wikipedia.org/wiki/Weight

"Weight" is simply a measurement of the gravitational forces acting on an object. So the "weight" of the Earth is really the vector sum of gravitational forces acting on it due to all other bodies in the universe, which (I would think) is unlikely to be zero.

I stand corrected. However the definition is not infact what is used in science. In science we measure the gravitational force exerted in releation to a given object. If you do not include a point of reference then objects on one side of the earth would have to have negative weight compared to objects on the other side as the force of gravity acting on them results in acceleration in opposite directions, thus the forces acting on them must be in opposite directions.

When I say "I weigh x Newtons" there is infact an implied "In the direction of the centre of the earth" as Newtons are vectors and thus directional.

This means that to measure the weight of the earth (and in fact any object) (by the dictionary definition) we need to measure its current velocity in relation to the universe (if such a concept is meaningfull) and combine that with its mass to work out what force is acting on it. (of course we might have to take into account any other known forces other than gravity that may be acting).

Originally posted by twhitehead
I stand corrected. However the definition is not infact what is used in science. In science we measure the gravitational force exerted in releation to a given object. If you do not include a point of reference then objects on one side of the earth would have to have negative weight compared to objects on the other side as the force of gravity acting on th ght have to take into account any other known forces other than gravity that may be acting).

Your post is very confusing. The point of reference is arbitrary and does not change the magnitude of the gravitational force, only its direction. I'm not arguing about the direction, I'm arguing that the vector sum is unlikely to be zero.

Also, it is not necessary to measure the velocity of the Earth in relation to the rest of the universe in order to calculate the gravitational attraction between the Earth and the rest of the universe, nor do we need to account for any other forces. Newton gave us a handy-dandy formula for calculating gravitational attraction all by itself.

To that end, I've made some quick calculations using some simplifying assumptions. I got the masses of the Sun and all the planets in the solar system, and calculated an "average" distance from the Sun by taking the mean of the aphelion and the perihelion (all data provided by Wikipedia). Then I assumed that all the planets lay in a straight line - this turned out to be a sketchy assumption based on the current state of the solar system as seen here:

http://www.fourmilab.ch/solar/

But it ends up not mattering much, since the Sun, the Earth, and Jupiter all do lie on a straight line (more or less), and they turn out to be the only important bodies for this calculation.

If positive is towards the Sun, the vector sum of all the gravitational attractions between the Earth and the rest of the bodies in the solar system turns out to be 3.33E+22 N. If you only take into account the Sun, the Earth, and Jupiter, it turns out to be 3.33+22 N again. I also checked the effect of Alpha Centauri, which is about 4.5 light years away, but it only contributes 4.66E+11 N which doesn't change the answer (no matter which direction it pulls in).

If you divide the force by the mass of the Earth, you get 5.58E-3 m/s^2, which is very small, especially compared to the standard acceleration due to gravity at the Earth's surface of 9.81 m/s^2. Fine measurements (especially in orbit) would probably have to take this into account, but for everyday purposes we can neglect it without much error.

Originally posted by twhitehead
It is an incoherent question. Weight is a measure of the gravitational force acting on a body on the earths surface. The earth is not a body on the earths surface and thus to talk about its weight (without an expanded definition - which you did not specify) is incoherent.
We could use an expanded definition of weight to mean a the gravitational force on ...[text shortened]... definition of weight and the centrifucal force due to the motion of the object is included too.)

🙂 I was thinking, "Wouldn't the weight of earth depend on gravity?"
Kelly

Just lay a bathroom scale on the ground face down. That'll tell you. I guess you'd have to subtact the weight of the scale.

Originally posted by Sam The Sham
Just lay a bathroom scale on the ground face down. That'll tell you. I guess you'd have to subtact the weight of the scale.

I'd think you'd need to add in the weight of the scale because it's upside down.

I think that you have to just read the scale upside down.

Originally posted by PBE6
But it ends up not mattering much, since the Sun, the Earth, and Jupiter all do lie on a straight line (more or less), and they turn out to be the only important bodies for this calculation.

Thats nonsense. They do not lie on a straight line.

I think that the easiest way to find out the gravitational force acting on a body is to know the bodies acceleration and mass and work it out from there. The earths acceleration is almost entirely due to the sun, so I think we can quite easily ignore all the planets - and the rest of the universe too.

Originally posted by twhitehead
Thats nonsense. They do not lie on a straight line.

In order to make calculation he "Then I assumed that all the planets lay in a straight line". Of course, he can make any assumption in order to make his calculations.

We all know, he does, you do, and I do, that this is not the general positions of the planets in our solar system. But sometimes it is easier to make calculation simplifications.

Originally posted by FabianFnas
In order to make calculation he "Then I assumed that all the planets lay in a straight line". Of course, he can make any assumption in order to make his calculations.

We all know, he does, you do, and I do, that this is not the general positions of the planets in our solar system. But sometimes it is easier to make calculation simplifications.

What you and he are implying is that the planest lie in approximately a straight line. That is not even remotely true. It wont make the calculation simpler, it will make the calculation totally wrong.

I suspect that he did in fact think that the sun, the earth and Jupiter were in approximately a straight line - a mistaken impression you might get if you look at a textbook which doesn't bother to put the orbits of the planets in the diagram.

If you want to assume totally untrue things for the sake of a simple calculation then why not assume that the sun and Jupiter dont exist and that the earth is in fact a one kg lead weight. See much simpler!


What is interesting is that because the earths orbit is eliptical, the force of the suns gravity on the earth varies, so the earths weight is different at different times of year!

Originally posted by twhitehead
What you and he are implying is that the planest lie in approximately a straight line. That is not even remotely true. It wont make the calculation simpler, it will make the calculation totally wrong.

I suspect that he did in fact think that the sun, the earth and Jupiter were in approximately a straight line - a mistaken impression you might get if yo ...[text shortened]... suns gravity on the earth varies, so the earths weight is different at different times of year!

I quote him once again: "Then I assumed that all the planets lay in a straight line". He didn't say 'I believe that all the planets lay in a straight line'. That's a huge difference. In mathematical calculations we often do that. He calculate the worst case possible and drew conclutions out of that. Our conversation here is way ot of topic.

Originally posted by PBE6
I didn't forget that they may cancel out. However, the distribution of matter in the universe has been observed to be highly non-uniform, so I stand by my original statement that I think it unlikely that the vector sum of the forces would be zero. The arbitrariness of the choice of origin won't affect this.

Also, "apparent weight" is not the same thing as "weight". Gravity doesn't disappear because we're in free fall, as you know.

Ok, I'm out of my league here, but isn't the sum vector of forces relative to the Sun zero, if the Earth is in free fall?

The thing is that you can only measure relative forces as there is no way to account for the rest of the universe properly. Hence the need to pick an arbitrary reference point.

I would say that the probability that the vector sum of forces is zero is not just unlikely, it is of mass zero. Earth would need to be in the universe's center of gravity for that to be true. What I'm saying is that there is no way to know where this center of gravity is, because we are measuring only relative forces (there's no way to account for all bodies), so it is meaningless to conjecture about it. We need then to pick a reference point for the weight of the Earth and I think the Sun is an intuitive choice in this case.

Originally posted by FabianFnas
I quote him once again: "Then I assumed that all the planets lay in a straight line". He didn't say 'I believe that all the planets lay in a straight line'. That's a huge difference. In mathematical calculations we often do that. He calculate the worst case possible and drew conclutions out of that. Our conversation here is way ot of topic.

If he knew that the planets orbit the sun at different frequencies then he would know that for the vast majority of the time they are not even close to being in a straight line. The assumption remains ridiculous.
It is not even a 'worst case'. What is 'worse' or 'better' anyway when it comes to weight?
I still think he foolishly thought that the planets are in approximately a straight line.

Originally posted by twhitehead
Thats nonsense. They do not lie on a straight line.

I think that the easiest way to find out the gravitational force acting on a body is to know the bodies acceleration and mass and work it out from there. The earths acceleration is almost entirely due to the sun, so I think we can quite easily ignore all the planets - and the rest of the universe too.

Two things:

1) Check the link. It shows a representation of the solar system as it stands now. The Sun, Earth, and Jupiter lay more or less in a straight line - straight enough for a "back-of-the-envelope" calculation. It turns out that it doesn't matter where the rest of the planets are, because the answer is the same to a few decimal places (at least 2) regardless.

2) If you make some quick calculations yourself, you'll see that Jupiter has a small but noticeable effect on the answer. If you leave it out, the answer changes by about 5%.

Originally posted by Palynka
Ok, I'm out of my league here, but isn't the sum vector of forces relative to the Sun zero, if the Earth is in free fall?

No. When a body is moving with constant velocity then there is no force acting on it. When the body is accelerating then there is a force acting on it.
A free falling body in a gravitational field has the force of gravity acting on it and it is accelerating.
An apple that is falling is accelerating towards the earth -due to the force of gravity
The earth is accelerating towards the sun - due to the force of gravity.
The resultant motion of the earth in a near uniform elipse around the sun tells us that the main gravitational force acting on it is entirely due to the suns mass acting on the earths mass. Thus simply by knowing the masses of the two bodies and the path of the earths orbit we can work out the gravitational pull being exerted.
In fact I suspect we can do the calculation with even less data than that (and actually work out the mass of the earth from there).