Question

what is the value of g at center of earth?

Answer 1

0 since at the centre of earth , the force is acting equally in all directions thus cancelling every force

Answer 2

how equally?
Earth is not perfectly spherical

Answer 3

Is not perfectly spherical, but still its 0.

Answer 4

its infinite

Answer 5

shape of earth doesn't matter. if your at the CENTER, then logically the forces would seem to cancel out. I believe Gravitons pull rather than push on objects according to there mass. So you would have some weight because all the mass would not fit in the exact place of cancellation...so you would basically weigh the same .just a small fraction less. Just my thoughts. JJ

Answer 6

First of all you need to define "the center of the earth" as you said it was not perfectly spherical. Is the center the center of mass of earth, or center of volume of earth. (it won't matter much unless you don't want the answer 0). At the center of mass of earth the mass of earth would be distributed equally in all directions thus the force of gravity would be 0.

at the center of volume of earth:
the 109 highest mountains are about at the same place
http://en.wikipedia.org/wiki/List_of_highest_mountains
so an assumption can be made that these mountains contain the offset mass while rest of mass cancels out.
now we need the volume of the mountains, assuming the mountains come down at a 35 degree angle and using an arbitrary height of 8000m we can find the radius of the base of the mountain.

tan 35 degrees * h= r
=> 5601m, I will use \[5.6* 10^{3}\]
find volume of mountain \[V= 1/3 \pi r ^{2}h\]
which comes to \[2.6 * 10^{11}m^{3}\] multiply by 109 mountains you get \[2.86*10^{13}m^{3}\]
now we have volume of earth that is exerting the force but need mass
so mass=density*volume
taking density of 5540kg/m3 *volume we get \[1.59*10^{17}kg\]
so we now have mass which is separated by the center by a distance of 6371km.
now we can calculate acceleration due to gravity g.
\[g=GM/r ^{2}\] \[g=6.67*10^{-11} * 1.9*10^{17}/(6.371*10^{6})^{2}\]
which comes to \[3.12 * 10^{-7}m/s^2\] ( 0.000000312) which is basically 0.

You can continue to make it more complex by adding effects of tides as water moves around the earth g would change slightly but all in all will be too close to 0 to have any real effect.

Answer 7

if we dig a tunnel from one point of the earth to the diametrically opposite point and put a small object inside it it will perform an shm with mean position at the center of the earth
and as on mean position of shm the velocity is constant with the accelaration remains zero so g at the mean position is zero

Answer 8

Aditya96 we have derived the expression of shm through a tunnel which is dug into the earth passing through the center of the earth by already assuming that force is zero at the
center of earth so it is not a proof,however proof can be achieved by using relativity theory

Answer 9

PS: I've used the symbol 'a' instead of 'g' as a is the standard expression for acceleration.

We know the gravitational force is given by the equation:
(G*M*m)/R^2 , where G is the gravitational constant, M=mass of earth, m= mass of body which is pulled , R= radius of earth.
Now force= m*a (a=Acceleration of the body)
Therefore, m*a= (G*M.*m)/R^2
Or, a=G*M/R^2

Now, considering the density (ρ) of Earth to be almost uniform,
M=ρ*(4/3(π*R^3)) [ since 4/3(π*R^3) represents the volume of the sphere or earth)]

Therefore, a= (G*ρ*4/3(π*R^3))/R^2 [replacing M]
Or, a=G*ρ*4/3*π*R

But, R=0 at the centre of earth
therefore acceleration due to gravity, a= 0

Hence Proved. Ciao.
:)

Answer 10

@TELIHILAL: Well, the accepted shape of Earth is an oblate spheroid, that is slightly bulging at the equator . Now, both the sphere and the oblate spheroid are central curves, plus both are symmetrical about the equator. So, for the mass element (shown as a ring) at a distance 'h' from the radius, there is a similar element below. The force from the above and below shall cancel each other.

Answer 11

Although it may seem to be infinite, the general equation change when it's "inside the earth". As you go deeper and deeper into the earth, the accelaration decreases. It will soon become 0.