Gravitation Force on the Earth's Centre

In summary, inside a solid sphere, the gravitational force is proportional to the distance from the center of the sphere. However, at the center of the Earth, the gravitational force is zero because the radius is zero.
  • #1
xCross
15
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I am having some problem understanding this.. By Newtons Gravitation Law we know:

F=Gm1m2/r^2

centre of the particle means r=0, am I right? if so, the equation says that the force will be infinite on the centre of the earth.. but we see that it has 0 force on the centre.. why such thing happened??
 
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  • #3
xCross said:
I am having some problem understanding this.. By Newtons Gravitation Law we know:

F=Gm1m2/r^2

centre of the particle means r=0, am I right? if so, the equation says that the force will be infinite on the centre of the earth.. but we see that it has 0 force on the centre.. why such thing happened??
That is the formula for the gravitational force of a point mass which can also be shown, by integrating over the volume of a sphere, to be the force of a spherical mass on any point outside the sphere.

Inside the sphere, if, say r< r0, where r0 is the radius of the sphere, it can be shown that the force for the mass outside distance r cancels. (The mass on "this" side pulls upward, on "that" side pulls downward and so cancel.) That is, only the mass for radius < r gives a non-zero total contribution to the force. Since mass (assuming constant density) is proportional to volume which is proportional to r3, the mass of a sphere of radius r inside a uniform sphere of radius r0 of mass M is Mr3/r03.

Putting that together with GmM/r2, the force on an object at distance r from the center of the sphere is Gm(Mr3/r03)/r2 and so is proportional to r, not 1/r2. Of course, at the center of the Earth r= 0 and so the gravitational force is 0. (Again, because of the symmetry of the sphere and the assumed constant density, the gravitational force from any "bit" of matter in the Earth is canceled by the force of a diametrically opposite "bit" of matter.)

Oh, and that does not mean that the PRESSURE on something at the center of the Earth due to the mass above it would be 0. That's a completely different calculation.
 
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1. What is the gravitational force on the Earth's center?

The gravitational force on the Earth's center, also known as the center of mass, is equal to the product of the Earth's mass and the acceleration due to gravity. This force is constantly pulling objects towards the Earth's center.

2. How is the gravitational force on the Earth's center calculated?

To calculate the gravitational force on the Earth's center, we use the equation F = G * (m1 * m2 / r^2), where G is the universal gravitational constant, m1 and m2 are the masses of the two objects (in this case, the Earth and an object on its surface), and r is the distance between the two objects.

3. What factors affect the gravitational force on the Earth's center?

The gravitational force on the Earth's center is affected by the mass of the Earth and the distance between an object and the Earth's center. The universal gravitational constant, G, is a constant value and does not change.

4. How does the gravitational force on the Earth's center affect objects on its surface?

The gravitational force on the Earth's center is responsible for keeping objects on the Earth's surface. This force pulls objects towards the center of the Earth, giving them weight and keeping them from floating off into space.

5. Can the gravitational force on the Earth's center change?

The gravitational force on the Earth's center can change if the mass of the Earth or the distance between an object and the Earth's center changes. However, these changes would be very small and would not significantly affect the overall force. The universal gravitational constant, G, remains constant.

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