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# 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.