Gravitational Torque on Axisymmetric Mass Distribution

Consider an axisymmetric mass distribution. Let us set up a Cartesian coordinate system whose origin corresponds to the center of mass of the distribution, and whose $z$-axis corresponds to the symmetry axis. The fact that the origin corresponds to the center of mass implies that

$$\int_V x\,\rho\,dV= \int_V y\,\rho\,dV = \int_V z\,\rho\,dV=0,$$ (1.383)

where the integrals are over the volume, $V$, of the distribution, and $\rho$ is the distribution's mass density. (See Section 1.4.2.) By symmetry, the Cartesian axes are principal axes of rotation, which implies that all of the products of inertia are zero. (See Section 1.7.2.) In other words,

$$\int_V x\,y\,\rho\,dV = \int_V y\,z\,\rho\,dV=\int_V z\,x\,\rho\,dV=0.$$ (1.384)

Finally, the principal moments of inertia of the distribution are

$$I_\parallel = I_{zz} = \int_V (x^2+y^2)\,\rho\,dV,$$ (1.385)

$$I_\perp = I_{xx}=I_{yy} = \int_V (x^2+z^2)\,\rho\,dV = \int_V(y^2+z^2)\,\rho\,dV.$$ (1.386)

Suppose that the mass distribution lies in the gravitational field of a distant object. Let ${\mit\Phi}(x,y,z)$ be the gravitational potential generated by the distant object. The gravitational torque, about the center of mass, that the distant object exerts on the mass distribution is

$$\mbox{\boldmath$\tau$} =-\int_V {\bf r}\times \nabla{\mit\Phi}\,\rho\,dV.$$ (1.387)

[See Section 1.4.5 and Equation (1.269).] However, we expect ${\mit\Phi}(x,y,z)$ to only vary slightly across the mass distribution, assuming that the distance of the distant object from the origin is much larger than the dimensions of the mass distribution. Thus, we can Taylor expand ${\mit\Phi}(x,y,z)$ about the origin to give

$${\mit\Phi}(x,y,z) \simeq {\mit\Phi}_x\,x + {\mit\Phi}_y\,y+{\mit\... ..._{zz}\,z^2 + {\mit\Phi}_{xy}\,x\,y+{\mit\Phi}_{yz}\,y\,z+{\mit\Phi}_{zx}\,z\,x.$$ (1.388)

Here, ${\mit\Phi}_x$ denotes $\partial{\mit\Phi}/\partial x$ evaluated at the origin, whereas ${\mit\Phi}_{xy}$ denotes $\partial^2{\mit\Phi}/\partial x\,\partial y$ evaluated at the origin, et cetera. Note that we have set the value of ${\mit\Phi}$ at the origin to zero, as we are free to do, given that gravitational potential (like gravitational potential energy) is undefined to an arbitrary additive constant. The previous six equations can be combined to give

$$\tau_x = - (I_\parallel-I_\perp)\,{\mit\Phi}_{yz},$$ (1.389)

$$\tau_y = (I_\parallel-I_\perp)\,{\mit\Phi}_{zx},$$ (1.390)

$$\tau_z =0.$$ (1.391)

We conclude that the distant object exerts a torque on the mass distribution whose direction lies in the distribution's equatorial plane. Note, however, that the distant object is incapable of exerting a torque on a spherically symmetric mass distribution (i.e., a distribution for which $I_\parallel=I_\perp$).