PhysicsLAB Resource Lesson
Advanced Gravitational Forces

The Law of Universal Gravitation states that every object in the universe attracts every other object in the universe with a force that has a magnitude which is directly proportional to the product of their masses and inversely proportional to the distance between their centers squared.
 
 
where
 
  • G is the gravitational constant, 6.67 x 10-11 Nm2/kg2
  • M1 is the mass of the first body in kg
  • M2 is the mass of the second body in kg
  • R is the distance from the center of M1 to the center of M2
 
In this lesson we will look at some special cases of universal gravitation whose calculations require the use of more sophisticated math and sometimes, integral calculus.
 
 
Force between a point mass and an extended mass
 
Our first example involves calculating the gravitational force between a point mass M and an extended rod of mass m, length L, and mass per unit length,.
 
To begin, divide the rod into a finite (countable) number of segments of mass  each located at a distance x from M.
 
 
Each of these segments will contribute a gravitational force of attraction.
 
 
If we take the limit as  approaches zero, then our expression for F becomes
 
 
Before we can integrate we must express  in terms of .
 
 
Substituting and integrating gives us
 
 
where
 
 
 
Force between a point mass and two equidistant particles
 
Our next investigation will be to calculate the gravitational force exerted by two equal, equidistant massive particles, M on a point mass, m, located at point P.
 
 
 
In setting up this diagram, we place the x- and y-axes to reflect the problem's symmetry.
 
 
Since the y-components of these attractive forces will cancel,
 
 
we will sum up the two x-components to determine our answer.
 
 
Using the formula for universal gravitation and the distances labeled in the above diagram
 
 
 
we can calculate our equation for net Fx.
 
 
Notice that this net gravitational force on our point mass, m,
 
  • points to the origin, or the center of the 2M system.
 
Our result will allow us to calculate the gravitation force at any position P along the x-axis. In particular, notice that
 
  • when x = 0, the net force will become zero; that is, the point mass, m, would feel no gravitational force.
 
 
 
Force on a point mass located along the axis of a thin ring
 
Suppose we now allow the two masses in the previous example to be rotated about the x-axis to create a thin ring in the yz-plane having a radius of a and a mass of 2M. To accomplish this let each mass "create" half of the ring. Our point mass, m, is now located on the axis of our thin ring. How will this continuous distribution of mass affect our previous result?
 
Now, imagine that the ring, like the bar in example #1 is made up of little segments . If we pair up segments based on the requirement that they are on opposite sides of any selected diameter, then we can calculate the force the ring exerts on the point mass, m.
 
 
Each pair of  segments will be the same distance r from m. No matter where the mass segment is located on the ring, only the force components pointing along the x-axis, towards the center of the ring at the origin, will contribute towards the force of gravitational attraction between the ring and the point mass. This symmetry will allow us to calculate the gravitational force on the mass by the ring.
 
As before, using the formula for universal gravitation and the distances diagram shown above, note that:
 
 
Adding up both x-components that point to the center of the ring will yield the following expression for net Fcenter. Note that we must also pass to the limit to convert our summation into the desired integral.
 
 
Before we can integrate we must express dM in terms of the arc length, s, of the ring. This arc length for the entire ring will equal the ring's circumference.
 
 
This substitution along with our original supposition that "each mass, M" was stretched out to form half of our ring will allow us to set up the limits of the integral and evaluate it.
 
 
As we complete our work also notice that G, m, a, x, and are all constants and can be removed from the integral.
 
 
Our last step is to find an expression for M and replace it.
 
 
Notice that this is the EXACT same expression that we calculated for the net gravitational force in the previous example involving two equidistant particles. Stretching the masses, M, out into a ring did NOT affect the magnitude of the gravitational force on the point mass.
 
 
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, ME, 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 A1 and A2. 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 m1 represents the mass of area A1, and m2 represents the mass of area A2. If we let σ represent a uniform mass per unit area, kg/m2, for a cross-sectional area of our shell, then m1 = σA1 and m2 = σA2.
 
 
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/m3, 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.
 
 




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