5.12: Gravitational Potential of any Massive Body

You might just want to look at Chapter 2 of Classical Mechanics (Moments of Inertia) before proceeding further with this chapter.

In figure $\text{VIII.26}$ I draw a massive body whose centre of mass is $\text{C}$, and an external point $\text{P}$ at a distance $R$ from $\text{C}$. I draw a set of $\text{C}xyz$ axes, such that $\text{P}$ is on the $z$-axis, the coordinates of $\text{P}$ being $(0, 0, z)$. I indicate an element $δm$ of mass, distant $r$ from $\text{C}$ and $l$ from $\text{P}$. I’ll suppose that the density at $δm$ is $ρ$ and the volume of the mass element is $δτ$, so that $δm = ρδτ$.

$\text{FIGURE V.26}$

The potential at $\text{P}$ is

$$ψ = -G \int \frac{dm}{l} = -G \int \frac{ρdτ}{l}. \label{5.12.1} \tag{5.12.1}$$

But $l^2 = R^2 + r^2 - 2Rr \cos 2 θ$,

so $$ψ = -G \left[ \frac{1}{R} \int ρ dτ + \frac{1}{R^2} \int ρ r \cos θ d τ + \frac{1}{R^3} \int ρ r^2 P_2 (\cos θ) dτ + \frac{1}{R^4} \int ρ r^3 P_3 (\cos θ) d τ ... \right]. \label{5.12.2} \tag{5.12.2}$$

The integral is to be taken over the entire body, so that $∫ ρdτ = M$, where $M$ is the mass of the body. Also $∫ ρr \cos θd τ = \int z dm$, which is zero, since $\text{C}$ is the centre of mass. The third term is

$$\frac{1}{2R^3} \int ρ r^2 (3 \cos^2 θ - 1) dτ = \frac{1}{2R^3} \int ρ r^2 (2-3\sin^2 θ ) dτ . \label{5.12.3} \tag{5.12.3}$$

Now

$$\int 2 ρ r^2 d τ = \int 2r^2 d m = \int \left[ (y^2 + z^2) + (z^2 + x^2) + (x^2 + y^2) \right] dm = A + B + C$$

where $A$, $B$ and $C$ are the second moments of inertia with respect to the axes $\text{C}x$, $\text{C}y$, $\text{C}z$ respectively. But $A + B + C$ is invariant with respect to rotation of axes, so it is also equal to $A_0 + B_0 + C_0$, where $A_0, \ B_0, \ C_0$ are the principal moments of inertia.

Lastly, $\int ρ r^2 \sin^2 θ dτ$ is equal to $C$, the moment of inertia with respect to the axis $\text{C}z$.

Thus, if $R$ is sufficiently larger than $r$ so that we can neglect terms of order $(r/R)^3$ and higher, we obtain

$$ψ = - \frac{GM (2MR^2 + A_0 + B_0 + C_0 -3C)}{2R^3}. \label{5.12.4} \tag{5.12.4}$$

In the special case of an oblate symmetric top, in which $A_0 = B_0 < C_0$, and the line $\text{CP}$ makes an angle $γ$ with the principal axis, we have

$$C = A_0 + (C_0 - A_0) \cos^2 γ = A_0 + (C_0 - A_0) Z^2/R^2, \label{5.12.5} \tag{5.12.5}$$

so that $$ψ = -\frac{G}{R} \left[ M + \frac{C_0 - A_0}{2R^2} \left( 1 - \frac{3Z^2}{R^2} \right) \right]. \label{5.12.6} \tag{5.12.6}$$

Now consider a uniform oblate spheroid of polar and equatorial diameters $2c$ and $2a$ respectively. It is easy to show that

$$C_0 = \frac{2}{5} Ma^2. \label{5.12.7} \tag{5.12.7}$$

Exercise $\PageIndex{1}$

Confirm Equation $\ref{5.12.7}$.

It is slightly less easy to show (Exercise: Show it.) that

$$A_0 = \frac{1}{5} M \left( a^2 + c^2 \right) . \label{5.12.8} \tag{5.12.8}$$

For a symmetric top, the integrals of the odd polynomials of Equation $\ref{5.12.2}$ are zero, and the potential is generally written in the form

$$ψ = - \frac{GM}{R} \left[ 1 + \left( \frac{a}{R} \right)^2 J_2 P_2 (\cos γ) + \left( \frac{a}{R} \right) J_4 P_4 (\cos γ) ... \right] \label{5.12.9} \tag{5.12.9}$$

Here $γ$ is the angle between $\text{CP}$ and the principal axis. For a uniform oblate spheroid, $J_2 = \frac{C_0 - A_0}{Mc^2}$. This result will be useful in a later chapter when we discuss precession.