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Problems for Newton's Theorem

Problems for Newton's Theorem

Problems for Newton's Theorem

Problems for Newton's Theorem

Problems for Newton's Theorem

Problems for Newton's Theorem

Problem : If the Universal Law of Gravitation was a 1/r3 force instead of a 1/r2 force, would it still be possible to treat the mass of a sphere as concentrated at its center?

No. As we have seen in Newton's Shell Theory, the proof of Newton's Shell Theorem relies on a the potential going as 1/r, which corresponds to a 1/r2 force.

Problem : Show that the gravitational force is independent of the path taken by evaluating some line integrals explicitly. Take a 10 kilogram mass at (1, 0) and calculate the work done to move a one kilogram mass from (2, 0) to the origin. Take one path as being directly along the x-axis and the other taking a quarter-circle path from (2, 0) to (1, 1), and then a straight path from (1, 1) to (1, 0).

The work done is the integral .dr, and the first integral is given by:

W = dx = = –10G = - 5G    

In the second path the circular path contributes nothing to the integral because the distance between the masses remains constant. The line integral from (1, 1) to (1, 0) is exactly the same calculation as above, except now along the y-direction instead of the x-direction. Thus the two paths yield the same result.

Problem :

A gravitating beam of length 10 meters.
What is the gravitational potential energy of a 10 kilogram mass which is 10 meters away from a 100 kilogram beam of length 10 meters as shown in the figure?

Consider dividing the beam into small pieces of mass dm. If the mass of the beam is evenly distributed along its length, then dm = , where dx is a small length of the beam. The gravitational potential can be found by summing over all the small pieces dm:

U = - 10G    

However, we can observe that l = and then using the relationship between dm and dx from above we can write this as an integral:

= - 10G = - 100Gln(x + ) = - 100G(2.78–1.82) = - 96.2G    

I evaluated the integral simply by looking it up on a table.

Problem : Determine the magnitude and direction of the force for the set-up described in the previous problem.

The direction of the force is, just by an argument from symmetry, just on the line directly connecting the mass and the center of the beam. To find the magnitude of the force, we must do the problem more generally and then take a spatial derivative. In fact, let us do the problem in exactly the same way, except with the distance between the mass and the center of the beam as a variable, y. We can write the same expression:

U = - 10G = - 100Gln(x + ) = - 100G(ln(5 + )–ln(- 5 + ))    

The force is given by: F = - . It is possible to calculate this using the chain rule:

F = - 100G    

To find the force at any point we must substitute in our value for y. In this case we have y = 10 , so F = - 100G× = -71.6G

Problem : Show that the gravitational force on a point P inside a spherical shell is zero by the following method: 1) pick a point P; 2) draw two lines through P which intersect the edges of a circle representing the spherical shell--in fact, in three dimensions these lines will makes two cones with their base as pieces of the shell; 3)show that the force from each piece of mass at the ends of the cones cancels.

Call one mass A and the other B. Drop perpendicular bases of the cones onto one of the lines drawn through P. Call these perpendiculars A' and B'. Call the distance between P and A, a, and the distance between P and B, b. Then the ratio of the areas A' to B' is a2/b2. Now because the chord between A and B meets the circle at equal angles at its ends, the angle between A and A' is the same as the angle between B and B'. This means that the ratio of the areas A to B is also a2/b2. The amount of mass contained in the cone base is proportional to the area, so the ratio of the mass pieces is a2/b2. But the gravitational force decreases as 1/r2. The ratio of the distances is b/a, so the ratio of the force is 1. Thus the magnitude of the forces exerted by the conic base sections is equal, but opposite in direction. We can divide up the whole spherical shell into a series of cones in this way, and the same conclusion will hold for each. Thus the force on a mass inside the sphere cancels out.