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.