Q:

If one were to travel closer to the center of the earth, then would there be less gravity?

- Stephen (age 16)

- Stephen (age 16)

A:

Yes. In the middle, the gravitational field from the Earth is zero. If it weren't, which way would it point?

As you leave the middle, the field grows approximately proportional to the distance from the middle, until you reach the surface. That's where it's biggest. As you go out into space, the field falls off as the square of the distance from the middle.

Mike W.

As you leave the middle, the field grows approximately proportional to the distance from the middle, until you reach the surface. That's where it's biggest. As you go out into space, the field falls off as the square of the distance from the middle.

Mike W.

*(published on 10/22/2007)*

Q:

How can be the gravitational force can object zero because according to formula of it, the distance from the centres will object zero so it must object infinity?

- Razin Shaikh (age 13)

Navsari, Gujarat, India

- Razin Shaikh (age 13)

Navsari, Gujarat, India

A:

You're thinking of a formula that says that the gravitational field falls off as the inverse of the square of the distance from an object. That formula only applies to point-like objects. When you have an extended object, you need to consider two complications:

1. The distances to different parts of the object are different.

2. The directions to different parts of the object are different, so the fields contributed don't all point the same way.

Calculating the net force then requires some effort. It was to solve this problem that Newton invented integral calculus.

For spherical shells, it turns out that the field outside the shell looks just the same as if all the mass were right at the middle. However, inside the shell it gives a field of zero. Right at the middle of the earth, you're inside all the shells, so the field is zero.

Of course, as we argued before, you know by symmetry that that has to be the answer.

Mike W.

1. The distances to different parts of the object are different.

2. The directions to different parts of the object are different, so the fields contributed don't all point the same way.

Calculating the net force then requires some effort. It was to solve this problem that Newton invented integral calculus.

For spherical shells, it turns out that the field outside the shell looks just the same as if all the mass were right at the middle. However, inside the shell it gives a field of zero. Right at the middle of the earth, you're inside all the shells, so the field is zero.

Of course, as we argued before, you know by symmetry that that has to be the answer.

Mike W.

*(published on 10/03/2012)*

Q:

In the above question which you answered, you told that the gravity is zero but that zero is approximately zero or exactly zero?

- Razin Shaikh (age 13)

Navsari, Gujarat, India

- Razin Shaikh (age 13)

Navsari, Gujarat, India

A:

If the earth had perfect spherical symmetry, the gravitational field from it at its center would be exactly zero. That's also true for some other highly symmetrical shapes. For unsymmetrical shapes, it's not even clear what point you should call the center. There's always some point there in the middle where the field is zero.

Mike W.

Mike W.

*(published on 10/13/2012)*

Q:

As you told in the follow up:1, that as we go inside the earth's crust, we go inside the shells and when we are at the centre of the earth we are in the innermost shell so there is no mass and so gravitational force of attraction is zero. But when we go at the centre the distance is also zero. Then according to the formula it would be zero divided by zero which is undetermined. Can you please explain me about that?

- Razin Shaikh (age 13)

Navsari, Gujarat, India

- Razin Shaikh (age 13)

Navsari, Gujarat, India

A:

The field at some point comes only from the spherical part closer to the center than that point. For uniform density, the mass of that part is proportional to the cube of the distance from the center. The force law gives a force proportional to the mass divided by the square of the distance to the center. So the distance cubed over the distance squared gives a force proportional to the distance. It smoothly goes to zero as the distance goes to zero. Trying to calculate that answer by taking 0/0 of course gives something undefined, so you have to use procedures like the one above, or the symmetry argument, to find out what the actual answer is.

Mike W.

Mike W.

*(published on 10/14/2012)*