25.8: Appendix 25A Derivation of the Orbit Equation

25A.1 Derivation of the Orbit Equation: Method 1

Start from Equation (25.3.11) in the form

\[d \theta=\frac{L}{\sqrt{2 \mu}} \frac{\left(1 / r^{2}\right)}{\left(E-\frac{L^{2}}{2 \mu r^{2}}+\frac{G m_{1} m_{2}}{r}\right)^{1 / 2}} d r \nonumber \]

What follows involves a good deal of hindsight, allowing selection of convenient substitutions in the math in order to get a clean result. First, note the many factors of the reciprocal of r. So, we’ll try the substitution \(u=1 / r, d u=-\left(1 / r^{2}\right) d r\), with the result

\[d \theta=-\frac{L}{\sqrt{2 \mu}} \frac{d u}{\left(E-\frac{L^{2}}{2 \mu} u^{2}+G m_{1} m_{2} u\right)^{1 / 2}} \nonumber \]

Experience in evaluating integrals suggests that we make the absolute value of the factor multiplying \(u^{2}\) inside the square root equal to unity. That is, multiplying numerator and denominator by \(\sqrt{2 \mu} / L\)

\[d \theta=-\frac{d u}{\left(2 \mu E / L^{2}-u^{2}+2\left(\mu G m_{1} m_{2} / L^{2}\right) u\right)^{1 / 2}} \nonumber \]

As both a check and a motivation for the next steps, note that the left side \(d \theta\) of Equation (25.A.3) is dimensionless, and so the right side must be. This means that the factor of \(\mu G m_{1} m_{2} / L^{2}\) in the square root must have the same dimensions as u , or \(length^{-1}\) so, define \(r_{0} \equiv L^{2} / \mu G m_{1} m_{2}\) This is of course the semilatus rectum as defined in Equation (25.3.12), and it’s no coincidence; this is part of the “hindsight” mentioned above. The differential equation then becomes

\[d \theta=-\frac{d u}{\left(2 \mu E / L^{2}-u^{2}+2 u / r_{0}\right)^{1 / 2}} \nonumber \]

We now rewrite the denominator in order to express it terms of the eccentricity.

\[\begin{aligned}
d \theta &=-\frac{d u}{\left(2 \mu E / L^{2}+1 / r_{0}^{2}-u^{2}+2 u / r_{0}-1 / r_{0}^{2}\right)^{1 / 2}} \\
&=-\frac{d u}{\left(2 \mu E / L^{2}+1 / r_{0}^{2}-\left(u-1 / r_{0}\right)^{2}\right)^{1 / 2}} \\
&=-\frac{r_{0} d u}{\left(2 \mu E r_{0}^{2} / L^{2}+1-\left(r_{0} u-1\right)^{2}\right)^{1 / 2}}
\end{aligned} \nonumber \]

We note that the combination of terms \(2 \mu E r_{0}^{2} / L^{2}+1\) is dimensionless, and is in fact equal to the square of the eccentricity \(\varepsilon\) as defined in Equation (25.3.13); more hindsight. The last expression in (25.A.5) is then

\[d \theta=-\frac{r_{0} d u}{\left(\varepsilon^{2}-\left(r_{0} u-1\right)^{2}\right)^{1 / 2}} \nonumber \]

From here, we’ll combine a few calculus steps, going immediately to the substitution \(r_{0} u-1=\varepsilon \cos \alpha, r_{0} d u=-\varepsilon \sin \alpha d \alpha\) with the final result that

\[d \theta=-\frac{-\varepsilon \sin \alpha d \alpha}{\left(\varepsilon^{2}-\varepsilon^{2} \cos ^{2} \alpha\right)^{1 / 2}}=d \alpha \nonumber \]

We now integrate Equation (25.A.7) with the very simple result that

\[\theta=\alpha+constant \nonumber \]

We have a choice in selecting the constant, and if we pick \(\theta=\alpha-\pi, \quad \alpha=\theta+\pi\) \(\cos \alpha=-\cos \theta\), the result is

\[r=\frac{1}{u}=\frac{r_{0}}{1-\varepsilon \cos \theta} \nonumber \]

which is our desired result, Equation (25.3.11). Note that if we chose the constant of integration to be zero, the result would be

\[r=\frac{1}{u}=\frac{r_{0}}{1+\varepsilon \cos \theta} \nonumber \]

which is the same trajectory reflected about the “vertical” axis in Figure 25.3, indeed the same as rotating by \(\pi\)

25A.2 Derivation of the Orbit Equation: Method 2

The derivation of Equation (25.A.9) in the form

\[u=\frac{1}{r_{0}}(1-\varepsilon \cos \theta) \nonumber \]

suggests that the equation of motion for the one-body problem might be manipulated to obtain a simple differential equation. That is, start from

\[\begin{aligned}
\overrightarrow{\mathbf{F}} &=\mu \overrightarrow{\mathbf{a}} \\
-G \frac{m_{1} m_{2}}{r^{2}} \hat{\mathbf{r}} &=\mu\left(\frac{d^{2} r}{d t^{2}}-r\left(\frac{d \theta}{d t}\right)^{2}\right) \hat{\mathbf{r}}
\end{aligned} \nonumber \]

Setting the components equal, using the constant of motion \(L=\mu r^{2}(d \theta / d t)\) and rearranging, Equation (25.A.12) becomes

\[\mu \frac{d^{2} r}{d t^{2}}=\frac{L^{2}}{\mu r^{3}}-\frac{G m_{1} m_{2}}{r^{2}} \nonumber \]

We now use the same substitution u =1/r and change the independent variable from t to r , using the chain rule twice, since Equation (25.A.13) is a second-order equation. That is, the first time derivative is

\[\frac{d r}{d t}=\frac{d r}{d u} \frac{d u}{d t}=\frac{d r}{d u} \frac{d u}{d \theta} \frac{d \theta}{d t} \nonumber \]

From \(r=1 / u\) we have \(d r / d u=-1 / u^{2}\) Combining with \(d \theta / d t\) in terms of L and u, \begin{equation}d \theta / d t=L u^{2} / \mu\end{equation}, Equation (25.A.14) becomes

\[\frac{d r}{d t}=-\frac{1}{u^{2}} \frac{d u}{d \theta} \frac{L u^{2}}{\mu}=-\frac{d u}{d \theta} \frac{L}{\mu} \nonumber \]

a very tidy result, with the variable u appearing linearly. Taking the second derivative with respect to t ,

\[\frac{d^{2} r}{d t^{2}}=\frac{d}{d t}\left(\frac{d r}{d t}\right)=\frac{d}{d \theta}\left(\frac{d r}{d t}\right) \frac{d \theta}{d t} \nonumber \]

Now substitute Equation (25.A.15) into Equation (25.A.16) with the result that

\[\frac{d^{2} r}{d t^{2}}=-\frac{d^{2} u}{d \theta^{2}}\left(u^{2} \frac{L^{2}}{\mu^{2}}\right) \nonumber \]

Substituting into Equation (25.A.13), with \(r=1 / u\) yields

\[-\frac{d^{2} u}{d \theta^{2}} u^{2} \frac{L^{2}}{\mu}=\frac{L^{2}}{\mu} u^{3}-G m_{1} m_{2} u^{2} \nonumber \]

Canceling the common factor of \(u^{2}\) and rearranging, we arrive at

\[-\frac{d^{2} u}{d \theta^{2}}=u-\frac{\mu G m_{1} m_{2}}{L^{2}} \nonumber \]

Equation (25.A.19) is mathematically equivalent to the simple harmonic oscillator equation with an additional constant term. The solution consists of two parts: the angleindependent solution

\[u_{0}=\frac{\mu G m_{1} m_{2}}{L^{2}} \nonumber \]

and a sinusoidally varying term of the form

\[u_{\mathrm{H}}=A \cos \left(\theta-\theta_{0}\right) \nonumber \]

where A and \(\theta_{0}\) are constants determined by the form of the orbit. The expression in Equation (25.A.20) is the inhomogeneous solution and represents a circular orbit. The expression in Equation (25.A.21) is the homogeneous solution (as hinted by the subscript) and must have two independent constants. We can readily identify \(1 / u_{0}\) as the semilatus rectum \(r_{0}\), with the result that

\[\begin{array}{l}
u=u_{0}+u_{\mathrm{H}}=\frac{1}{r_{0}}\left(1+r_{0} A\left(\theta-\theta_{0}\right)\right) \Rightarrow \\
r=\frac{1}{u}=\frac{r_{0}}{1+r_{0} A\left(\theta-\theta_{0}\right)}
\end{array} \nonumber \]

Choosing the product \(r_{0} A\) to be the eccentricity \(\varepsilon\) and \(\theta_{0}=\pi\) (much as was done leading to Equation (25.A.9) above), Equation (25.A.9) is reproduced.