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

Question

If you fell through (an imaginary tunnel) from one pole to the other or from one side to the other,

how would your weight profile be as you fall from the surface towards the center of the earth?

The weight of an object increases or decreases from the center of gravity. The effect of gravity is inversely proportional to the distance from the centre. so if you were to fall to the center of the earth, how much would you weigh? the radius of the earth at the center is zero so the force on you is infinite??? will you be crushed by gravity?? any clarifications?

Answer 1

Your weight would be proportional to the distant from the center of the earth. So it would it act like a spring (Hookes law) and oscillate, where you would fall through and pass the center of the earth and assuming the tunnel went through to the other side you would end up on the other side minus the air drag. So with air drag consideration, you would be a damped oscillator, falling through with increasing speed that would "whip" you through the center and almost exactly the same distance on the other side, and then you would fall back through again from the other side. Because of air drag you would not make it quite as far each time you passed the center, so you would eventually end up wiggling around the center of the earth at a weight of zero.

Answer 2

Actually the acccln due to gravity is proportional to the distance from the center inside a solid sphere such as the Earth. And it is inversely proportional to the SQUARE of the distance from the center of the sphere, outside the sphere. And ur weight is your mass multiplied by the gravitational accln. So when u move towards the center INSIDE THE EARTH ur weight will decrease and it'll b 0 at the center.

Answer 3

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)

Answer 4

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 !!!

Answer 5

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.

Answer 6

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.

Answer 7

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.

Answer 8

Your weight at the centre of the Earth would be zero.

Your mass would be unchanged.

Your velocity if you had fallen in would (negelcting air resistance of course) be the escape velocity of Earth. You would shoot right out to the other side.

Once you are below the Earth's surface you are effectively only pulled down by the mass in a sphere of your radius from the centre of Earth.

Answer 9

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...

Answer 10

When you reach the center of earth ( provided you are not crusehed in the matter inste you are just surrounded by it) you will weigh Zero. But remember your mass will still be same.
The weight is zero because at the center you are attractd by all particles arround you equally(earth being uniform).