# 5 Developing a General Solution@

Info

Developing a General Solution: Write down the solution to the 2-body problem in terms of $$r$$ and $$\theta$$ using the previously developed properties of the system.

Description

See above.

Author

Matt Werner

## 5.1 We’re finally ready to solve the 2-body problem! Well, kind of…@

A solution of the radius $$r = r(\theta)$$ is achievable, but we fall just short of knowing $$\theta(t).$$ The grim reality of the problem is that the angle $$\theta$$ has to be calculated numerically.

The results are summarized below.

Important

The radius $$r$$ is a known function of the angle $$\theta$$ and is parameterized by the system’s characterization ($$\mu = \mathrm{const.}$$) and initial conditions ($$h = \mathrm{const.}$$ and $$A = \mathrm{const.}$$).

(5.1.1)@$r = \frac{h^2}{\mu + A\cos\theta}$
• We know exactly what path the reduced mass must trace out in the orbital plane.

The angle $$\theta$$ is NOT a known function of time. It must be obtained numerically and depends on the system’s characterization and initial conditions as well as time.

• We don’t know exactly where the reduced mass is in its trajectory at any given time (but we will be able to get “close enough”).

## 5.2 Finding the radius as a function of the angle@

Fact: The distance between the two bodies measured with respect to the angle subtended from the Laplace vector in the orbital plane is given by (5.1.1).

Proof using the Laplace vector

The Laplace vector can be used to obtain an expression for the radius $$r$$ as a function of the angle $$\theta$$ very directly.

Since $$\theta$$ is measured from the $$x$$ axis, which is aligned with the Laplace vector $$\mathbf{A}$$, we have

$\begin{split}\mathbf{A} \cdot \mathbf{r} &= \left(\dot{\mathbf{r}} \times \mathbf{h} - \frac{\mu}{r}\mathbf{r}\right) \cdot \mathbf{r} \\ &= (\dot{\mathbf{r}} \times \mathbf{h}) \cdot \mathbf{r} - \left(\frac{\mu}{r}\mathbf{r}\right) \cdot \mathbf{r} \\ &= \mathbf{r} \cdot (\dot{\mathbf{r}} \times \mathbf{h}) - \frac{\mu}{r} (\mathbf{r} \cdot \mathbf{r}) \\ &= \mathbf{h} \cdot (\mathbf{r} \times \dot{\mathbf{r}}) - \mu r \\ &= \mathbf{h} \cdot \mathbf{h} - \mu r \\ &= h^2 - \mu r \\ &= A r \cos\theta. \quad\square\end{split}$

Alternatively, knowing (4.2.3) and combining it with (4.2.4) yields the radius $$r$$ even more simply.

$\begin{split}A\cos\theta &= r^3\dot{\theta}^2 - \mu \\ &= \frac{h^2}{r} - \mu. \quad\square\end{split}$

Proof using the radial dynamics

The $$\ddot{r}$$ equation of motion (4.2.2) combined with the conservation of angular momentum (4.2.3) is

$\begin{split}\ddot{r} &= r\dot{\theta}^2 - \frac{\mu}{r^2} \\ &= \frac{h^2}{r^3} - \frac{\mu}{r^2}.\end{split}$

We can convert the derivatives in time $$t$$ to those in the angle $$\theta$$ to understand the geometry of the trajectory in polar coordinates. Using the conservation of angular momentum (4.2.3) repeatedly, we see

$\begin{split}\frac{d^2 r}{d\theta^2} &= \frac{d}{d\theta}\left(\frac{dr}{d\theta}\right) \\ &= \frac{d}{d\theta}\left(\frac{\dot{r}}{\dot{\theta}}\right) \\ &= \frac{d}{d\theta}\left(\frac{r^2\dot{r}}{h}\right) \\ &= \frac{1}{\dot{\theta}}\frac{d}{dt}\left(\frac{r^2\dot{r}}{h}\right) \\ &= \frac{r^2}{h^2}(2r\dot{r}^2 + r^2 \ddot{r}) \\ &= \frac{r^2}{h^2}\left(2r\left(\frac{h}{r^2}\frac{dr}{d\theta}\right)^{\!2} + r^2 \left(\frac{h^2}{r^3} - \frac{\mu}{r^2}\right)\right) \\ &= \frac{r^2}{h^2}\left(2\frac{h^2}{r^3}\left(\frac{dr}{d\theta}\right)^{\!2} + \frac{h^2}{r} - \mu\right) \\ &= \frac{2}{r}\left(\frac{dr}{d\theta}\right)^{\!2} - \frac{\mu r^2}{h^2} + r.\end{split}$

This equation still proves to be formidable in obtaining a solution. However, a final, clever change of variables will yield a useful relation by letting $$\eta = 1/r$$. In this case,

$\begin{split}\frac{d^2\eta}{d\theta^2} &= \frac{d}{d\theta}\left(\frac{d\eta}{d\theta}\right) \\ &= \frac{d}{d\theta}\left(-\frac{1}{r^2}\frac{dr}{d\theta}\right) \\ &= -\frac{1}{r^2}\frac{d^2r}{d\theta^2} + \frac{2}{r^3}\left(\frac{dr}{d\theta}\right)^{\!2} \\ &= -\frac{1}{r^2}\left(\frac{2}{r}\left(\frac{dr}{d\theta}\right)^{\!2} - \frac{\mu r^2}{h^2} + r\right) + \frac{2}{r^3}\left(\frac{dr}{d\theta}\right)^{\!2} \\ &= \frac{\mu}{h^2} - \eta.\end{split}$

This is a linear oscillator with constant forcing, where both $$\mu$$ and $$h$$ are (positive) constants. We can therefore immediately write the general solution as

$\eta = \frac{\mu}{h^2} + C \cos(\theta - \phi),$

where both $$C$$ and $$\phi$$ are constants of integration. Note that $$C$$ represents an amplitude and, as such, satisfies $$C \geqslant 0$$. The orbital radius ($$r = 1/\eta$$) is then

$r = \frac{h^2}{\mu + (C/h^2) \cos(\theta - \phi)}.$

Since $$\mathbf{A}$$ already provides a constant direction in the orbital plane from which $$\theta$$ is measured, we do not wish to shift the angle $$\theta$$ to be measured with respect to anything else (yet). Thus, without loss of generality in this coordinate frame, we can set $$\phi = 0.$$

To determine the constant $$C$$, we can calculate $$\dot{r}$$ exactly and then use the conservation of angular momentum (4.2.3) and the Laplace vector (4.2.4) to compare expressions.

$\begin{split}\dot{r} &= \frac{C\sin\theta}{(\mu + (C/h^2) \cos\theta)^2}\dot{\theta} \\ &= \left(\frac{r^2}{h^4}C\sin\theta\right)\left(\frac{h}{r^2}\right) \\ &= \frac{C}{h^3}\sin\theta \\ &= \frac{A}{h}\sin\theta\end{split}$

From this, we obtain that the integration constant $$C$$ is given in terms of known quantities as $$C = A h^2.$$

The final form of $$r$$ is (5.1.1). $$\quad\square$$

### 5.2.1 Rewriting the radius in terms of the eccentricity@

The radius $$r$$ can be more simplified if we define the new variable $$e = A/\mu$$ called the eccentricity. Doing this allows us to write the radius (5.1.1) as

(5.2.1)@$r = \frac{h^2/\mu}{1 + e\cos\theta}$
• This reduces the amount of parameters needed to study the trajectory from two ($$\mu$$ and $$A$$) to one ($$e$$)
• (There were only two parameters because we didn’t actually need $$h$$ to determine the trajectory’s behavior)

### 5.2.2 Plotting the trajectory@

Because the trajectory in the orbital plane is simply

$\begin{split}\mathbf{r} = \begin{pmatrix}r\cos\theta \\ r\sin\theta \\ 0\end{pmatrix} = \frac{h^2}{\mu} \begin{pmatrix}\frac{\cos\theta}{1 + e\cos\theta} \\ \frac{\sin\theta}{1 + e\cos\theta} \\ 0\end{pmatrix},\end{split}$

we can easily visualize what the trajectory will do as the parameter $$e$$ is varied.

• We do not need to vary the angular momentum $$h$$ in this case because the behavior is simply scaled by $$h^2/\mu.$$ Fig. 5.2.1 Variation of trajectories in the orbital plane; each curve is labelled with its corresponding eccentricity value $$e.$$ Closed trajectories (red) are ellipses and open trajectories (green) are hyperbolas. The trajectory separating the two cases is shown in black.@

The above plot can shortly be summarized by the following table.

Table 5.2.1 Limits of the angle $$\theta$$@

Class

Condition

Limits

Closed

$$e < 1$$

$$|\theta| < \pi$$

Open

$$e \geqslant 1$$

$$|\theta| < \cos^{-1}(-e^{-1})$$

• Closed trajectories are also called elliptic

• Open trajectories are also called hyperbolic (unless $$e \equiv 1$$, then the trajectory is called parabolic).

## 5.3 Finding the time as a function of the angle@

Knowing $$r = r(\theta)$$, we would most like to know the angle $$\theta = \theta(t)$$. This would fully complete the parameterization as then $$\mathbf{r} = \mathbf{r}(x(t), y(t), 0)$$ would be a known function for all time.

However, we’re not so fortunate. Instead, we can obtain $$t = t(\theta),$$ but inverting this function $$t$$ is difficult.

Fact: The time since the reduced mass crossed the $$x$$ axis ($$\theta = 0$$) is

(5.3.1)@$t = \frac{h^3}{\mu^2}\left(\frac{2}{(1 - e^2)^{3/2}}\,\tan^{-1}\!\left(\sqrt{\frac{1 - e}{1 + e}}\tan\frac{\theta}{2}\right) - \frac{e\sin\theta}{(1 - e^2)(1 + e\cos\theta)}\right)$

where $$\theta$$ is limited according to the parameter $$e.$$

• This result is written for $$e < 1,$$ but it is still valid when $$e \geqslant 1$$ (where equality is taken in the sense of a limit).

Proof using the angular momentum

The conservation of angular momentum states $$r^2\dot{\theta} = h,$$ but the orbital radius $$r$$ is now a known function of only $$\theta.$$ Thus, we can write

$\begin{split}\dot{\theta} &= \frac{h}{r^2} \\ &= \frac{(\mu + A\cos\theta)^2}{h^3},\end{split}$

which provides a separable equation for $$\theta$$. Naively integrating obtains

(5.3.2)@$\int_{\theta_0}^\theta \frac{d\theta'}{(\mu + A\cos\theta')^2} = \frac{1}{h^3}\int_{t_0}^t dt',$

Taking $$t_0 = \theta_0 = 0$$ yields the above result. $$\quad\square$$

### 5.3.1 The orbital period@

Looking closer at (5.3.1) shows that the time $$t$$ is odd in the angle $$\theta.$$ (That is, $$t(-\theta) = -t(\theta).$$) This is useful because we can (more) easily calculate the period ($$T$$) for a closed trajectory. (An open trajectory can’t have a period since it simply doesn’t repeat itself.) The period is

$T = 2\pi \frac{h^3/\mu^2}{(1 - e^2)^{3/2}}, \qquad (e < 1).$

Proof

Assume $$e < 1$$ (so that a period exists). Using the previous proof, calculate

$\begin{split}T &= \int_0^T dt \\ &= \int_0^{2\pi} \frac{h^3}{\mu^2} \frac{d\theta}{(1 + e\cos\theta)^2} \\ &= \int_0^\pi 2\frac{h^3}{\mu^2} \frac{d\theta}{(1 + e\cos\theta)^2} \\ &= 2\pi\frac{h^3/\mu^2}{(1 - e^2)^{3/2}}.\end{split}$

Of course, the time has already been calculated by (5.3.1), so plugging in limits works just as well – the result comes out directly. $$\quad\square$$

The (normalized) orbital period is plotted below as a function of the eccentricity $$e.$$ Fig. 5.3.1 The orbital period $$T$$ (scaled to a constant) of varying closed orbit trajectories characterized by eccentricity. The period lengthens as the orbit becomes more eccentric, approaching infinity when parabolic.@

• A circular orbit has (constant) angular velocity $$h^3/\mu^2.$$

• The period is valid over one cycle of the orbit, but $$t(\theta = 0) = 0$$ and $$|\theta| < \pi.$$

• We say the time over an orbit occurs between $$-T/2 < t < T/2.$$

### 5.3.2 Plotting the time of flight@

Using modern computational tools makes it “easy” to visualize what a typical flight path looks like for any given eccentricity $$e$$ as seen below.

• We do not need to know what $$h^3/\mu^2$$ is since this is simply a constant that scales the actual behavior we’re interested in. Fig. 5.3.2 Orbital position in equal timesteps. The timesteps were chosen such that points fall on the $$x$$ axis and $$(0,1)$$.@

• The circular orbit ($$e = 0$$) has equally spaced time steps (as we expected)

• The motion is slowest at the furthest distances and fastest at the closest distances.