only the distance between centres of body and earth are changing , but the mass of the earth is not changing. And also having the same centre of gravity.
2006-06-19 01:53:56 · 7 answers · asked by sara_swathi m 1 in Science & Mathematics ➔ Physics
The source for gravity is not concentrated at one point. The mass is the source of gravity. At the center of the earth, you are surrounded by mass, equally distributed in all directions. Every particle is pulling at you from each direction with the same force, so the gravity is zero at the earth's center. If you are somewhere within the earth, let's say halfway, more mass will be on one side (towards the center) than on the other ... the result is a reduced gravity force, pointed towards the center. At the surface of the earth, all mass is on one side, so you have the full effect there.
2006-06-19 02:00:24 · answer #1 · answered by dragolt 3 · 1⤊ 0⤋
The gravitational pull is due to attraction of two masses.
The earth is a sphere of huge mass of radius R.
Suppose a mass is on the surface of earth. If we imagine a horizontal plane surface below the mass, on one side of the plane there is the huge earth and on the other side is the mass. The attraction of the mass is toward the center of earth.
If we consider a mass at the inner side of the Earth and if we, again consider a horizontal plane surface below the mass, now there is a portion of earth is on one side of the plane and another small portion is on the other side of this plane. Thus the mass is pulled towards the center by the large portion and also pulled in opposite direction by the small portion of earth. The net force is small compared to the case when the mass was on the surface of earth.
It can be seen by the above consideration the force on the center will be zero.
2006-06-19 03:50:07 · answer #2 · answered by Pearlsawme 7 · 0⤊ 0⤋
If I understand your question, the answer is that if you view the earth as a series of concentric spheres, the NET gravitational field contribution from all of the mass contained in spheres with greater radius than the observer cancels out, but the net gravitational field from all of the mass contained in spheres with radius less than the observer does not. Calculus shows this quite readily. Recall that the gravitational field is inversely proportional to the square of the separation distance. If one takes a small volume element dV, with thickness dr, width r* (d-theta), height r* (d-phi), integrating all of the gravitational field contributions from r > r of the observer will yield zero net field.
This has been demonstrated by analogy to the electrostatic force, which is also an inverse-square force. There is no net electric field inside a charged sphere for the same reason.
Therefore, the acceleration due to gravity will decrease as one approaches the center of the earth, as more and more of the earth's mass would be contained in concentric spheres with radius greater than the observer and therefore would not contribute to g.
2006-06-19 02:34:22 · answer #3 · answered by volume_watcher 3 · 0⤊ 0⤋
Like Robin says. The "center of gravity" is not a physical point, but an average of the pull of gravity around the surface. The center of gravity changes (relative to you) as you descend into the Earth's mass.
2006-06-19 02:01:38 · answer #4 · answered by thylawyer 7 · 0⤊ 0⤋
The equation is f = G * m / r^2 and as you flow in the direction of the middle of Earth the linked fee of "m" decreases considering the fact that there is way less mass between the factor closer to the middle than it truly is while the factor is on the exterior.
2016-10-31 03:05:12 · answer #5 · answered by Anonymous · 0⤊ 0⤋
Because the mass of the Earth above you is counteracting the pull of the mass below you.
2006-06-19 01:55:44 · answer #6 · answered by Robin the Electrocuted 5 · 0⤊ 0⤋
its true and this can be proved .
using
g'=g(1-d/r)(formula) hence proved
2006-06-19 02:03:28 · answer #7 · answered by cute girl 1 · 0⤊ 1⤋