### One of a continuing series

Isaac Newton is credited with the first scientific explanation of gravity. His powerhouse was his mathematical expertise, and his analyses were comprehensive. To begin with, Newton described the attraction between two bodies due to gravity. The attractive force is proportional to the product of the masses of the two objects, and inversely proportional to the square of the distance between them. But what does “between them” mean. For idealized objects, such as homogeneous spheres, the solution was straight forward. The distance of significance was the distance between their mass centers, their mass centers also being their geometrical centers.

The attraction between two homogeneous, spherical objects is inversely proportional to the square of the distance between their centers.

The implication of the foregoing is that a large sphere, such as the one pictured above, can be viewed, as far as the computation of the gravitational attraction goes, as a very small, even point-like sphere of the same mass and located at the center of the large sphere.

Newton went further. Suppose all the mass of the large sphere were concentrated into a spherical shell.

The answer is still the same. The spherical shell acts, gravitationally, as a point-like sphere at the center.

Now for the Quiz Question. Newton went ever further. What is the gravitational attraction if you are inside the spherical shell?

Provide your answer as a comment below.

## Update and answer

Two answers have been submitted, so I am closing down the competition. Here is a discussion:

Greg writes:

Zero. All the attractions due to all the “particles” in the ring cancel, as you’re being pulled in all directions at once.

That is correct. It’s the same answer Issac Newton came up with 300 years ago. Great minds travel in like circles.

Prasad writes:

At the surface it would be the gravity of the whole object, as we go towards the center of the sphere the gravity decreases, eventually at the center the gravity is zero. This would be the same for even a shell as with a solid sphere.

Prasad seems to be saying the gravitational attraction would vary within the hollow shell. Greg’s answer is correct. Everywhere inside the shell there would be no net force due to the gravitational attraction of the shell.

Here’s a bonus question: Demonstrate a simple proof of this week’s answer.

Zero. All the attractions due to all the “particles” in the ring cancel, as you’re being pulled in all directions at once.

At the surface it would be the gravity of the whole object, as we go towards the center of the sphere the gravity decreases, eventually at the center the gravity is zero. This would be the same for even a shell as with a solid sphere.

Let’s say we are in a mine 1 mile below the surface of the earth. Would the gravity be zero there? It will definitely be a fraction less than on the surface but not zero.

For the spherical shell with uniform mass, yes the forces will cancel out inside the sphere.

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