what is the value of acceleration due to gravity at the centre of the earth? on what factors it depends?

Question

g=GM/Rpow2 . at the centre R=0. what value of M should be chosen at the centre.

Acc to formula g is evalauted to 0 or infinity. Zero g indicates every body moving through the centre of earth will come to rest at the centre as g=0, but it is against to Newton's gravity train concent, which says every body will make a harmonic motion if dropped from earth's surface to move through a duct passing through earth's centre. and the body wil come to other end of the earth's surface

Answer 1

To find this out you cannot use the simple GMm/r^2. Instead you have to consider gravity as a vector field and integrate over the volume of the Earth in all directions from the point you are considering.

This is, of course, tricky. However, Gauss' theorem comes to the rescue (look it up on Wikipedia if you want to know more). If we assumes that the mass of the Earth is uniformly distributed, Gauss' theorem shows that the net field at point R from the cetnre of the Earth is simply that due to the mass contained within a sphere of that radius - the rest (the mass above you) cancels out.

Answer 2

the gravitational force at the centre of the earth is 0 because on seeing the formula , we can make out that at the centre of the earth the effective mass of the earth becomes 0. also the considering the shell theorem by newton ,at the centre of the earth , the entire mass of the earth will behave like a shell an dthus the force inside will be zero.

Answer 3

the gravity inside the earth is zero.

if you want to know why it observes harmonic oscillations consider this proof (we assume that the earth is of uniform density ρ)

Mass of Earth (M) = ρ(4/3)πR^3 where R is the radius of the earth

and we consider the mass effective of the earth

Mass effective (M') = ρ(4/3)πr^3 where r is the distance from the center of the earth

so m/M = r^3/R^3
and m = (r/R)^3 * M

by another calculation, all the outer shells' force equals zero within the earth

so F = -GmM'/r^2 (within the earth, or R = r at the surface)

substituting M' = (r/R)^3 * M we get

F = -GmMr/R^3

with F = mr'' where r'' is the double derivative with respect to time in x we get

mr'' = -(GmM/R^3)r
r'' + (GM/R^3)r = 0
r'' + (ω^2)r = 0, ω = sqrt(GM/R^3)

this differential equation has a solution of
r = Acosωt, where A is a constant which can be found by boundary conditions

thus we have an oscillatory motion with angular velocity ω
and period of 2π/ω

Answer 4

at the centre, g=o.
As g=o, a body at the centre of the earth will not be attracted towards the centre. this is because gravitational force is directly proportional to g. This is why the body continues in its motion and reaches the other end of the earth, hence exhibiting simple harmonic motion.
At the centre of the earth the value of the M will be 0.

Answer 5

The formula for g from the center of the Earth to the surface is gr/R (assuming constant density, which is not really the case, but it's workable), where g = acceleration due to gravity at the surface, r = distance from the center, and R = radius of the Earth, or, if you prefer, g =-GMr/R^3.

Roughly speaking, M = ρ(4/3)πr^3, so -GM/r^2 becomes
-ρ(4/3)πGr^3/r^2, or -ρ(4/3)πGr.

Answer 6

Zero is fine and dandy. But as experimental sciences go, we need to validate the data. And nobody has been to the center of the earth. Ha Ha Ha.

Hey, what if it were actually measured and the result is not zero? Hmmm, we will have to rethink everything something like the concept of friction, which gave results that are quite counter-intuitive from what was previously known.

Answer 7

At the center of the earth, the acceleration due to gravity is zero because the mass of the earth is surrounding you, and the force all cancels out.

Answer 8

that formula is a special case... for a point on the surface or outside it...
i suppose you must have heard of gauss's theorem...
and if not then read any good book on physics you will get the answer to your question...
still i will give you a hint... line of thought...
actually gravitational force depends on the mass enclosed inside an imaginary surface to be more precise gaussian surface...
so at the center of earth...
mass enclosed will be 0...
hence, F = 0.... that implies g = 0.

Answer 9

When a stone falls in a hole to the center of earth, the gravity acceleration it 'feels' decreases to zero proportional to the distance to the center.

Th

Answer 10

You can dedude the force by formula G*Mass(body1)*Mass(body2)/(square of distance between them).
Earth is not a perfect sphere so if you are at center. assume rings of elements outside you. For each ring, divide them into subelements and compute the force that they exert on you. Force is a vector so resultant forces exerted on you, is a vector addition of each of these forces. Depending upon whether the ring is symmetrical or not, you will get either get zero force exerted on you or not.
But theoretically you can assume that for each element on earth there is symmetrical element so at center of earth you will receive zero force and so zero acceleration.