Thread: Gravity at center of the earth

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This thread has me thinking. Is there any possible consequence from the fact that the core is essentially in a low gee environment? I can't think of any.

2. Originally Posted by Evan
This thread has me thinking. Is there any possible consequence from the fact that the core is essentially in a low gee environment?
Yes. It is a perfect analogy of my business role. I am under tremendous pressure but get no sense of any one direction. :-?

3. Originally Posted by Evan
This thread has me thinking. Is there any possible consequence from the fact that the core is essentially in a low gee environment? I can't think of any.
omypelt's info shows that the gravity at the surface of the core is about the same as at the surface of the earth itself. That is a consequence of the higher density of the core.

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Yes, it also produces a non-linear drop off through the mantle. What does that do to convection in the mantle?

5. Originally Posted by Evan
Yes, it also produces a non-linear drop off through the mantle. What does that do to convection in the mantle?
No, it stays fairly constant throughout the mantle, with small variations, down to the core-mantle boundary, where it is a little greater than at the surface of the earth.

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Constant is non-linear. Any ideas?

7. Originally Posted by Evan
Constant is non-linear. Any ideas?
Constant is linear.

But I was referring to the "drop off" anyway.

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Re: Other forces?

Originally Posted by George
Originally Posted by fossilnut
Physics isn't my strong point so forgive a naive question. Wouldn't gravity at the center of the Earth be equal in all directions only if the Earth itself was all of existence? What about the pulling of the Moon, Sun, and the rest of the Universe? The gravity acting on the Earth is not equal from all directions.
Your thinking is correct. It is the sum of all the forces which produces the final result.

However, the force you feel from distance objects are quite weak. [...]
Also, consider that the Earth is already freely falling under the influence of all those external gravitational forces. If you were at the center of mass of the Earth, you would be freely falling in exactly the same manner (assuming a perfectly uniform Earth). So you wouldn't feel them, except for the extremely weak tidal forces arising from the variation of the gravitational pull over the extent of your body.

9. If you drill a hole through center of the earth, and jumped in, how you would fall depends upon where the hole is. If it is near the equator, your motion in the direction the earth is spinning will pin you against the side in a hurry. You will slide down until your angular momentum is in equalibrium with the force of gravity, at which point you will kinda hang there in geosinc orbit inside the earth...

...if the hole is at one of the poles, you will fall, with your momentum carrying you up to the south pole...neglecting all the usually effects, like bouyancy...which are kinda important, so the dampening would kill the ride in a few thousand cycles.

10. I've understood most of the explanations regarding gravity inside earth, but now you have to clear one thing up for me.

When an object is in orbit around the sun, it is NOT accelerating towards the sun making its path of travel curved but "It is following a straight line in the gravitational potential field of the space around the sun"

Because if it was accelerating, there would have to be some unbalanced force acting upon it, right? But there isn't, right? So how can object's trajectory curve without it accelerating? Does it have something to do with the choice of reference frames in gravitational fields or some other einsteinishly complex idea? I didn't get the quoted part in previous paragraph.

This is a whole new thing for me. Please explain.

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Originally Posted by JMV
I've understood most of the explanations regarding gravity inside earth, but now you have to clear one thing up for me.

When an object is in orbit around the sun, it is NOT accelerating towards the sun making its path of travel curved but "It is following a straight line in the gravitational potential field of the space around the sun"

Because if it was accelerating, there would have to be some unbalanced force acting upon it, right? But there isn't, right? So how can object's trajectory curve without it accelerating? Does it have something to do with the choice of reference frames in gravitational fields or some other einsteinishly complex idea? I didn't get the quoted part in previous paragraph.

This is a whole new thing for me. Please explain.
The answer depends on whether you use the physics of Newton or of Einstein. To understand the motion of a planet around the sun, both are sufficient (there is a small difference in the behaviour of mercury between the two systems of physics). In the physics of Newton, it is accelerating towards the sun because of the force of gravitation. In the physics of Einstein, there is no force of gravitation. The presence of mass like the sun makes the space in its neighbourhood no longer obey the ordinary geometry of Euclid. In this geometry, if you travel in a straight line, you can come back to where you started this journey. Then the orbitting object is not accelerating, but moving in a straight trajectory through a curved space. Choice of reference frames is for convenience, the behaviour of the object does not depend on this choice.

Martin

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(In Newtonian Physics) The Earth is accelerating around the Sun, as acceleration is a vector quantiy and therefore has a direction and since the direction of the Earth's motion is changing it therefore has acceleration...

I think..

Edit: Crap someone beat me!

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Originally Posted by Zjm7891
(In Newtonian Physics) The Earth is accelerating around the Sun, as acceleration is a vector quantiy and therefore has a direction and since the direction of the Earth's motion is changing it therefore has acceleration...

I think..
Yes, the acceleration is in direction of the sun. This is perpendicular to direction of motion if the orbit is circular. If the orbit is elliptical, then it can be not exactly perpendicular. This is by Newtonian physics, by Einsteinian physics, then there is no acceleration, the effect is accomplished by bending of space near the sun. In Newtonian physics, to take a simple case of circular orbit, then distance d is fixed. If the orbit is in x-y plane with the sun at x=y=0 (because the sun is much larger than earth, the effect of earth's gravity on sun is minimal), then:

x(t) = d*cos(2*pi*t)
y(t) = d*sin(2*pi*t)

gives position of earth, where t is time in years (sine and cosine functions take inputs in radians). Velocity vector is given by first derivatives:

x'(t) = -2*pi*d*sin(2*pi*t)
y'(t) = 2*pi*d*cos(2*pi*t)

Acceleration vector is from second derivatives:

x''(t) = -4*pi*pi*d*cos(2*pi*t)
y''(t) = -4*pi*pi*d*sin(2*pi*t)

Components of acceleration vector are just negative (multiplied with 4*pi*pi) of components of position vector. Since the sun is at origin the acceleration vector points straight to sun. The acceleration is perpendicular to velocity (x''*x'+y''*y' = 0). This perpendicularity fails when the orbit is elliptical. This circumstance is much more complicated; I do not know how to express position as function in time, but can express time and distance from origin taking angle as independent variable in polar coordinate system. Also, it is possible to use three dimensions, but the orbit still exists in a plane in three-dimensional space. So a third dimension makes the maths more complicated, but adds nothing in way of concept.

Originally Posted by Zjm7891
Edit: Crap someone beat me!
I am not beating anyone!

Martin

14. I think I see it now. Thanks for your input.

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