Is the acceleration due to gravity at the centre of the earth zero??

Question

everyone says g=0 but i say NO
because ,
by universal law of gravitation,

g=G Mm\R*R
WHERE,G is universal gravitational constant
M IS MASS OF EARTH
m is mass of a body say "a"
R is radius of earth
since G and M are constant,
so g is inversely proportional to R*R
but R=0 at the centre of earth ,
so g cannot be 0
please tell me whether i am correct

Answer 1

I understand you reasoning, but it is flawed.

The law of universal gravitation treats the masses being studied as points (point masses). When you are ‘inside’ one of the masses, you have to treat it a bit differently. Without going into all of the mathematical explanation (which I can if you want me to), in this sort of situation, one must only consider the mass which is ‘beneath’ you. If you are below the surface of the Earth, not all of the Earth’s mass is beneath you anymore; some is ‘above’ you. If you actually work it out, all of the mass which is now above you cancels itself out due to the symmetry of the sphere, but the mass below you acts normally as a point mass still.

An analogy I like to compare this to is an onion. As you travel below the surface of the Earth, the mass which is now located above you is like a layer of the onion which can be ‘pealed’ away, so to speak, without affecting the gravitational pull below you.

If you are at the center of the Earth, there is no mass beneath you any more, all of it is above you and it all is canceling itself out in terms of its gravitational pull on you in all directions. Since M = 0 now, the gravitational acceleration = 0.

Answer 2

It is in fact zero, and the reason is interesting: it has to do with a bit of mathematics called Gauss's law. Suppose a well that reaches to the center of the earth. At any point in the well, the measure of the gravitational force is given by the mass of the earth BELOW the point in question: all the mass above that point, whether on the same side of the earth, the opposite side, or somewhere in between, cancels out. So, as you descend the well, the force of gravity decreases nearly linearly with increasing depth, reaching zero at the center. (It would be exactly linear if the earth's density were constant with depth, but it isn't.) Gauss's law says that the total flux from any source equals the total flux passing through any surface surrounding the source. This is usually applied to electrostatics, but it applies to gravity as well.

Answer 3

The simple formula you cite applies to point masses. To apply it inside the earth, you would need to sum all the force vectors between you and each particle of earth. But see the reference for the simplification of that, where someone else has already done the volume integral for you. With the usual assumptions of a spherically symmetric earth, the net gravitational force on you due to the earth is zero. You still feel the gravitational attraction of all other masses in the universe. That's primarily the sun, and to a lesser extent, the moon. You would feel weightless and would remain stationary even if you didn't hold on, since you would be orbiting the sun just the same as the earth is orbiting the sun. Because of that assumption of spherical symmetry, the gravitational force you experience from each particle on earth is counterbalanced exactly by the force from the symmetrically opposite particle.

Answer 4

mrjeffry3 is correct for an earth isolated in its own universe. However the real earth is not isolated. While g due to the mass of earth is 0 at the center there are other bodies such as the sun and moon in the vicinity. The actual value of g at the center would be the sum of g's for all these bodies. It would very small but not negligible.

Answer 5

The question and thus the answer is relative. Presuming that a person could be at the exact center of earth and the the earth was completely homogenous, then the Accel. of gravity in reference to earth could be zero. The problem is that the earth is not the only body in the universe that affects gravity upon an object.

Answer 6

The force of gravity is a particular form of energy. In that it performs work, it must have a continual source of energy able to be expended as its basis (like gasoline being expended in an automobile is the energy source making the machine move). Stephen Hawking in “A Brief History of Time,” page 92, paragraph 3, states; “Like light, gravitational waves carry energy away from the objects that emit them.” We find that the formation of gravitational waves is an energy expenditure of a particular kind. This value is found in the physics trilogy, which is: E = mc2, m = E/c2, and c2 = E/m. The last is that for a field of gravity, or one of physical time - these two concepts are the same.

The energy in this particular equation is that of the heat energy found within a solar mass. As all energy must exhibit its presence with associated with a mass, the manner in which this is done within our planet, moon, and sun is by an increase of a field of gravity. These waves from on one side of our planet, pass through its center, and exit out the other side. In this manner the waves attract the heat energy within our planet towards it center as well as all mass. Were this not so, then our planet would be almost the same temperature throughout. Our planet expends 0.00444 kg./sec to keep things in place. Our sun expends 665 lbs./sec to keep the planets where they are. In the very innermost location of our planet, as our sun, there must exist a very small location where all these waves pass through, tearing in every possible direction. Here, there would be, not zero gravity, but all gravity.

The mass of our planet keeping us of the US in place is found in the Indian Ocean, and we of the US keep waters in that location where they should be. Because the force is universal all about the center of our planet, our planet, as other solar bodies, form into a sphere.

http://360.yahoo.com/noddarc Were you to view the blog, then click on "list view", the post entitled "An Experiment You Can Help With" gives information on an experiment to prove what has been written.

Answer 7

Yes. It is 0 because the value of g decreases as we go under the surface of the earth.
g= G* mass of earth/square of radius of earth
Now,the value of square of radius of earth is also considered as the distance of centre of the earth from its surface. At the centre of the earth this distance is 0. So, the value of g is 0 at that place.

Answer 8

Ideally for a uniform sphere yes, if you could actually mesure it, no. The earth isnt uniform density, different areas of the earth have slightly different g's because of varying crust thickness

Answer 9

While dealing with such problems we first find the equation for very small change and then by integrating it we find value for greater change.So your mathematical approach is wrong.

Answer 10

g''=g{1-h/R]
g''=gravity at depth h
g=gravity of earth
R=radius of earth
so in center h=Rmeans g''=0...ok (effect of depth)