Irodov Problem 1.217

a) Imagine that the shell is made of a large number of very tiny particle point masses such that . The total interaction energy of the shell is the energy required to separate all these point masses of the spherical shell against their mutual attractive forces into infinity. Let E(dmi,dmj) be work done by gravity when masses dmi and dmj are pulled apart to infinity - this is the interaction energy of two particles dmi and dmj. The the total interaction energy of the entire spherical shell is given by the sum of the interaction energies of all pairs of particles E(dm1,dm2) + E(dm1,dm3) + E(dm2,dm3) + E(dm1,dm4) + E(dm2,dm4) + E(dm3,dm4) + ...

The gravitational field strength at any point outside the spherical shell at a distance r from its center can be determined using Gauss law of gravity and choosing a spherical Gaussian surface of radius r and that is concentric with the spherical shell as shown in the figure. So we have,


Now we can determine the gravitational potential at any given point on the spherical shell as,







The potential energy of any point mass dm in the spherical shell will be . This gravitaional potential energy is actually the work done by gravity in pulling this point mass dm away from the rest of the sphere to infinity.

In other words, = E(dm1,dm2) + E(dm1,dm3)+ E(dm1,dm4).... or,








Now since E(dm1,dm2) = E(dm2,dm1) in the sum we end up adding the gravitational interaction energy of each pair of point masses dmi,dmj , E(dmi,dmj), twice. In other words the total energy of interaction of all the particles in the spherical shell is given by,









b) The basic idea behind the solution is identical to that in part a) execpt that now we have consider the entire volume of the sphere. The gravitational potential at a point that is r distance away from the center of the sphere was computed in the solution to problem 1.217 as,





As in case of part a), the total gravitational energy of interaction is given by,

Irodov Problem 1.216

Pressure is the derivative of component of the force acting in the normal direction to a surface with respect to the surface area (dF/dA). In this problem the gravitational force
due to the mass in the sphere causes the pressure at each layer of the sphere. The closer a certain layer to the center, the higher the pressure since the column of the mass pushing this layer is larger.

In the solution to problem 1.214 we already derived the gravitational field strength in a uniform sphere at a location that is r distance from the center as,





Here p is the density of the sphere. Since the gravitational field strength is the force per unit mass , the force acting on an infinitesimally small mass dm located at a distance r from the center is given by gdm. Consider an infinitesimally small cylindrical column of area dA located at a distance r from the center and height dr (as shown in the figure). The total mass of this infinitesimally small elemement is given by dm = pdAdr and so the gravitaional force acting on it is given by,