Earth's Gravity

Liokh
I was wondering what kind of force is the force of gravity ? Is there a subcategory of that specific force ? (Got any internet documentation about it, too ? It would be nice.)
I got another question: in the air, are there particle of metals (negligeable when studying the air's components)?

Thanks in advance for feeding my sudden thirst of knowledge :tongue:

(By the way is this the best forum to post these questions ?)

Norman
This really has nothing to do with Quantum Physics- unless you are asking about Quantum Gravity... but let us assume you are not. This should be in the Classical Physics or General Physics section.

I guess I don't understand what you mean by "what kind of force is the force of gravity"... please be more specific

Here is some basic information:

Gravity is one of the 4 fundamental forces of nature, which in addition to gravity includes the strong, weak and electromagnetic forces. Newton's formulation of gravity tells us that gravity is a $\frac{1}{r^2}$ type force. That means, for instance, if we have two spherical massive bodies, the force of gravity between them falls off as one over the distance between their center of masses squared. The gravitational force is also a vector.
Scienceworld is a good site usually. Here is a link to the Newtonian Gravity section:
http://scienceworld.wolfram.com/physics/Gravity.html

General Relativity describes gravity in a very different way but I have only a rudimentary understanding of it.
hope this helps.

Mentor
Liokh said:
I got another question: in the air, are there particle of metals (negligeable when studying the air's components)?
There are all sorts of suspended particles in air. Not much in the way of metal though.

orsanyuksek
Gravity itself is one of the four fundemantal forces. As a force, it has both magnitutude and direction.

timeformation
Does this mean that if you charge a particle, which is a 1/3 spin, then it could counteract gravity, and then if you give both the eweak and strong forces a jolt, you could overengage gravity, so that it becomes less than the electromotive force?

Redfox
Liokh said:
I was wondering what kind of force is the force of gravity ?

We don't know yet.

However it may turn out to be one of the moving mechanisms too.

Judging by the way sattellites behave it may be that gravity force has a multiple nature, i.e. it could be that the vector of the gravity force would not be described as a straight line.

Imagine a globe with a lot of lines sticking out of it, just like hair. Now spin the globe in all directions at one time (I know it's a bit hard to imagine). The lines will curve forming something alike numerous discending orbits, curving around it but finally leading to the globe itself. This is what the vector force of gravity may turn out to be. A spinning, discending to the mass all-directional spiral-like vector force.

In other words gravity field may have torsion nature.

Planets obviously stay in their orbits because the outward spinning around pull compensates for the discending factor of the gravity force.

If gravitons exist then they probably form a part of gravitational mass to mass exchange occurring in the above spin-like-spiral fashion or they may simply spin about masses causing gravitational pull. Partly the spinning approach may be supported by the idea that quarks, forming particles with masses are also supposed to have a kind of spin.

However until the exchange particle is identified, above will remain a fruit of imagination.

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Liokh
Thanks for the explanations !
It made me think of more questions. Sorry it's more lengthy this time :tongue:

If the Earth is attracted by the sun, why aren't we ?
Or are we, having a mass lighter in the day at noon than during the night at midnight?
And somewhat along Redfox's hairy vector, ( I see it clear with the head-hair example) it would mean that some days, it's easier to walk east if the sun is west of you, would it ?
Is there a way to predict(measure) the path(force/vector) of those simili-waves ?
Lastly, the gravitons you speak of would make that hair(line) tangible. Is it regroupable ? I suppose it would have some kind of nuclear power due to it's electronegativity interaction, would it ? Would that be palpable if regrouped ?

Mentor
We are attracted to the sun, but since we're in orbit around the sun, we don't feel much of a change in weight. You would weigh very slightly more at 6:00 am and pm than at noon or midnight, essentially due to tidal effects.

Gold Member
Liokh said:
Thanks for the explanations !
If the Earth is attracted by the sun, why aren't we ?
We are, just like the Earth is. Which is why we and the Earth move together, none the wiser.

(Consider what would happen if we weren't attracted to the Sun. The Earth would move in a circle around the Sun, but we would tend to go straight (tangent to the Earth's orbit). The people on side of the Earth nearest the Sun would have their faces plowed into the ground as the Earth went in a circle while they went straight. And the people on the side of the Earth away from the Sun would almost fly off, because the Earth would move in a circle while they went straight.)

Norman
DaveC426913 said:
We are, just like the Earth is. Which is why we and the Earth move together, none the wiser.

(Consider what would happen if we weren't attracted to the Sun. The Earth would move in a circle around the Sun, but we would tend to go straight (tangent to the Earth's orbit). The people on side of the Earth nearest the Sun would have their faces plowed into the ground as the Earth went in a circle while they went straight. And the people on the side of the Earth away from the Sun would almost fly off, because the Earth would move in a circle while they went straight.)

HUH? Am I missing something here... Order of magnitude calculation for centripetal force on a 100kg person:
orbital velocity $\approx 10^4 \frac{m}{s}$
orbital radius $\approx 10^{11} m$
which means the force due to our circular motion around the sun (since the Earth's radius in much smaller than the earth-sun distance) I will assume that the force on either side of the Earth will be approx. the same.

$$F = \frac{m v^2}{r} \approx \frac{10^2 (10^4)^2}{10^{11}} = 10^{-1} N \ll 10^3 N \approx mg$$

So you see we really wouldn't even notice the centripetal acceleration, let alone be thrown off the Earth by it.

WeeDie