Final Steps
At this junction in the lesson our goal to determine what the gravitational force would be on a point mass, **m**, if the point mass were to be placed (1) outside of a thin spherical shell, (2) inside a thin spherical shell, and then (3) inside a solid spherical mass. Our previous examples have paved the path for these investigations.
(1) First we will analyze the force that would be exerted on a point mass, **m**, if it were to placed outside of a thin spherical shell at a distance **x** from the shell's center. To do this, we can consider our uniform, spherical shell to be made up of thin rings symmetric to the x-axis. Each ring contributing to the shell's final total mass, **M**. The rings are going to increase in radius, from r = 0 to r = a, and then decrease back to 0 as we take our slices.
Complications arise from the number of variables that need to be related and then integrated. As you can see, the rings will have different radii and circumferences, have different values for angles θ and α, as well as different values of r and x. For those of you interested in this advanced solution, a derivation can be found in section 11-5 of your text (Tipler, __Physics for Scientists and Engineers__, 5th ed, pages 358-360).
Although the actual calculus for this derivation is beyond the scope of our lesson, remember that in general for a thin ring the gravitational attraction between it and an external point mass, **m**, is directed along the x-axis towards the center of the ring.
The conceptual understanding that we are adding together rings with forces all pointing towards the center of the shell will help us rationalize that the external point mass, **m**, "would see a massive shell" as if it were a concentrated point mass, **M**, located at the center of the shell and exerting a force
directed towards it center. This derivation can be revisited once we understand Gauss' Law and use its gravitational analogue to prove our result more simply.
If our point mass, **m**, were a distance **x** from a solid sphere, we would expect the same outcome since we can envision our solid sphere as being composed of tightly nested shells. What makes this result all the more **beautiful** is that it shows that the attractive force between the Earth and a softball arching through the air does not need to become a complicated calculation depending on local terrain. The gravitational force is between the mass of the softball, **m**, and the mass of the Earth, **
M**_{E}, as if it were a point mass at concentrated at the Earth's center. A truly amazing, simple result!
(2) To calculate the gravitational force on a point mass, **m**, placed inside a thin spherical shell at any point P, we will use properties of similar right circular cones along with Newton's Law of Universal Gravitation.
Through point P, draw two intersecting chords that terminate on opposite sides of our thin shell, crossing at the center of **m**. Our chords, with their equal apex angles, have created two similar right circular cones having base areas A_{1} and A_{2}. The mass segments represented by these areas are gravitationally attracted to our point mass, **m**. But what is the relationship between these areas?
Using the trigonometry ratio tangent, we can develop a relationship between the heights of each cone and the radii of their circular bases.
Gravitationally, these base areas each exert a force of attraction on our point mass, **m**.
where m_{1} represents the mass of area A_{1}, and m_{2} represents the mass of area A_{2}. If we let σ represent a uniform mass per unit area, kg/m^{2}, for a cross-sectional area of our shell, then m_{1} = σA_{1} and m_{2} = σA_{2}.
As you can see, the forces on **m** are exactly equal and would cancel. If we repeated this process for every surface segment, we would always discover that the forces are always balanced. That is, our point mass would feel NO gravitational force at point P, no matter where P is inside the shell.
(3) To calculate the gravitational force on a point mass, **m**, located at a radius **r** inside a solid uniform sphere of radius **R**, we will need to use our outcome for the gravitational force experienced inside a thin shell as well as our knowledge of density.
In the diagram given above, the point mass, **m**, is represented by the white dot. All of the sphere's mass above its location can be considered to consist of nested shells (shown in gray). From our earlier discussions, we know that the gravitational force on a point mass located anywhere inside a shell equals zero. Therefore we can discount their masses and only examine the mass for radii 0 <= r.
Using the fact that the sphere has a uniform density, ρ, in kg/m^{3}, we can express the mass of the interior sphere in terms of the total mass of the entire sphere. This leads us to our final substitution showing that the gravitational force between the interior mass and our point mass is proportional to r's percentage of the original radius R. A graph summarizing the results for a solid sphere is given below. |