What's your weight at the center of the earth???

the weight will be zero for sure. "the effect of gravity is inversely proportional to distance" is true only from the surface of the earth.it is the other way inside the earth bcos the mass responsible for gravity decreases and is zero at the centre.

At the geometric centre of mass of the Earth you would be weightless. This is because the local gravitational field would be equally balanced in all directions. As the Earth (and you within it) are in orbit around the sun, the gravitational force pulling you towards the sun is equally balanced by the centripedal force of your orbital motion so, aside from minor perturbations, you essentially experience weightlessness in the sun's orbit also. If it weren't for the extreme heat at the Earth's core, it might be quite relaxing...

In 1687 Newton published his work on the universal law of gravity in his Mathematical Principles of Natural Philosophy. Newton’s law of gravitation states that: every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. If the particles have masses m1 and m2 and are separated by a distance r (from their centers of gravity), the magnitude of this gravitational force is: F = -G \frac{m1 * m2}{r^2} where: F is the magnitude of the (repulsive) gravitational force between the two point masses G is the gravitational constant m1 is the mass of the first point mass m2 is the mass of the second point mass r is the distance between the two point masses Assume: m1 = Your weight, m2=earth mass (m2 ~ r^3) F ~ G*m1*r^3/r^2 ~ G*m*r for all r >= 0 and r <= Radius_of_Earth F(r=0) => m2=0 => F=0 !!!

The huge ball of iron is stationary, because nothing has caused it to move in the first place. A person falling from the surface to the centre would have a velocity as it reached the equilibrium position (the centre). The core is not moving, because it is already at the equilibrium position. Also, the pull of gravity does not only act in one direction. It acts towards the centre of mass of an object. So, if you are at this centre of mass, the resultant force is zero (you would either not move, or continue moving at your current speed)

at the excat center of earth u will have no weight because all the (earth)mass around u will be pulling u to all sides equaly. If u drill a tunnel from one side of earth to other side thru the center and jump into it, in theory u'll oscilate between the two sides forever ( a simple harmonic motion)

if ever a hole was dug from 1 pole to other then any object thrown would oscillate from one pole to other ,u will never be crushed as the pressure would not be as high as it was before the hole was dug.......

These answers sound good. But oscillation? If you would oscillate what is stopping the huge ball of iron oscillating within the soft inner mantle? Gravity pulls in one direction. Whilst some oscillation would occur you would be pulled fairly quickly to the mid point and be stuck there. You would be crushed to an infinitesimally small size long before you got there. In that situation weight would have no function and the little blob of you would become part of the larger mass.

weigth would be zero but mass would,nt be b/c W=mg so * g* could be zero but mass will be constant i think we will not be injured

Everyone here is wrong, or at least not completely correct. 1. You would be weightless THE ENTIRE time that you were falling as you are -by definition- in free fall. 2. At the center of the earth the mass of the entire earth would be equally distributed around you to a sum effect of nil. But momentum (not acceleration) would be unaffected. 3. Momentum would indeed factor in and you would oscillate repeatedly until you finally came to rest at the center.