diff --git a/images/jwst-trajectory.gif b/images/jwst-trajectory.gif new file mode 100644 index 00000000..7b20241f Binary files /dev/null and b/images/jwst-trajectory.gif differ diff --git a/interplanetary-maneuvers/interplanetary-transfer-phasing.md b/interplanetary-maneuvers/interplanetary-transfer-phasing.md index 64c26b62..4fdf1fb1 100644 --- a/interplanetary-maneuvers/interplanetary-transfer-phasing.md +++ b/interplanetary-maneuvers/interplanetary-transfer-phasing.md @@ -277,10 +277,10 @@ The wait times are shown in {numref}`tab:heliocentric-hohmann-wait-times`. The t | N | $t_{\text{wait}}$ (years) | |---|-----------------------| -| 0 | {glue:text}`heliocentric-hohmann-t_wait_0` | -| 1 | {glue:text}`heliocentric-hohmann-t_wait_1` | -| 2 | {glue:text}`heliocentric-hohmann-t_wait_2` | -| 3 | {glue:text}`heliocentric-hohmann-t_wait_3` | +| 0 | {glue:text}`heliocentric-hohmann-t_wait_0:.4f` | +| 1 | {glue:text}`heliocentric-hohmann-t_wait_1:.4f` | +| 2 | {glue:text}`heliocentric-hohmann-t_wait_2:.4f` | +| 3 | {glue:text}`heliocentric-hohmann-t_wait_3:.4f` | ::: Clearly, the total mission time is dominated by the transfer time. This is because the synodic period of Venus relative to Neptune is quite small, at only {glue:text}`heliocentric-hohmann-T_syn:.2f` Earth years. Since Venus whips around the Sun, relative to Neptune, the same phase angle occurs relatively often. diff --git a/the-n-body-problem/circular-restricted-three-body-problem.md b/the-n-body-problem/circular-restricted-three-body-problem.md index 0c54141e..cb05fb9c 100644 --- a/the-n-body-problem/circular-restricted-three-body-problem.md +++ b/the-n-body-problem/circular-restricted-three-body-problem.md @@ -255,9 +255,9 @@ The characteristic length is the circular orbit radius, $r_{12}$. Using this, we :::{math} :label: eq:non-dim-r-vectors-cr3bp \begin{aligned} - \vector{\rho} &= \frac{\vector{r}}{r_{12}} = x^_\uvec{\imath} + y^_\uvec{\jmath} + z^_\uvec{k} \\ - \vector{\sigma} &= \frac{\vector{r}_1}{r_{12}} = \left(x^_ + \pi_2\right)\uvec{\imath} + y^_\uvec{\jmath} + z^_\uvec{k} \\ - \vector{\psi} &= \frac{\vector{r}_2}{r_{12}} = \left(x^* - 1 + \pi_2\right)\uvec{\imath} + y^_\uvec{\jmath} + z^_\uvec{k} + \vector{\rho} &= \frac{\vector{r}}{r_{12}} = x^*\uvec{\imath} + y^*\uvec{\jmath} + z^*\uvec{k} \\ + \vector{\sigma} &= \frac{\vector{r}_1}{r_{12}} = \left(x^* + \pi_2\right)\uvec{\imath} + y^*\uvec{\jmath} + z^*\uvec{k} \\ + \vector{\psi} &= \frac{\vector{r}_2}{r_{12}} = \left(x^* - 1 + \pi_2\right)\uvec{\imath} + y^*\uvec{\jmath} + z^*uvec{k} \end{aligned} ::: @@ -281,7 +281,7 @@ where $\tau = t/t_C$. Making the terms on the right hand side of Eq. {eq}`eq:fiv :::{math} :label: eq:non-dim-five-term-accel-cr3bp -\ddot{\vector{\rho}} = \left(\ddot{x}^* - 2\dot{y}^* - x^_\right)\uvec{\imath} + \left(\ddot{y}^_ + 2\dot{x}^* - y^_\right)\uvec{\jmath} + \ddot{z}^_\uvec{k} +\ddot{\vector{\rho}} = \left(\ddot{x}^* - 2\dot{y}^* - x^*\right)\uvec{\imath} + \left(\ddot{y}^* + 2\dot{x}^* - y^_\right)\uvec{\jmath} + \ddot{z}^*\uvec{k} ::: Now we have the non-dimensional inertial acceleration, we need to make Eq. {eq}`eq:vector-eom-cr3bp`, the equation of motion, non-dimensional. After a bunch of algebra, not shown here, we end up with: diff --git a/the-n-body-problem/lagrange-points.md b/the-n-body-problem/lagrange-points.md index 5fcf930e..03a42245 100644 --- a/the-n-body-problem/lagrange-points.md +++ b/the-n-body-problem/lagrange-points.md @@ -207,7 +207,7 @@ ax.plot([0, 1], [0, -1], lw=1.0, color="silver", ls="--", label="$m_2$") ax.set_xticks(np.arange(0, 1.1, 0.1)) ax.set_yticks(np.arange(-1.5, 1.75, 0.25)) ax.grid() -ax.set_xlabel("$\pi_2$") +ax.set_xlabel(r"$\pi_2$") ax.set_ylabel("$x^*$") ax.annotate("$m_2$", xy=(0.55, 0.5), ha="center", va="bottom") ax.annotate("$m_1$", xy=(0.55, -0.5), ha="center", va="bottom") @@ -228,7 +228,7 @@ On this figure, the $x$ axis is $\pi_2$ and the $y$ axis is $x^*$. For a given v The solutions for $x^*$ for the collinear Lagrange points lie on the S-curve shape. For a given value of $\pi_2$, we can see there are 3 solutions of the function, corresponding to the three collinear Lagrange points for that system. By convention, the Lagrange points are numbered such that $L_1$ lies between $m_1$ and $m_2$, $L_2$ lies to the right of $m_2$, and $L_3$ lies to the left of $m_1$. Thus, we can see on the figure that the upper part of the S-curve is the solution for $L_2$. Below $x^{*} = 1.0$, the solution is for $L_1$, since that lies between $m_1$ and $m_2$. Finally, below $x^{*} = -1.0$, the solution is for $L_3$. - + For the Earth-Moon system, the value of $\pi_2$ is approximately 0.012. @@ -499,7 +498,7 @@ The Trojan and the Greek asteroids are clusters of asteroids that have collected The collinear Lagrange points, $L_1$, $L_2$, and $L_3$ are all **saddle points** in {numref}`fig:pseudo-potential-energy-cr3bp`, meaning that the function increases when going in one axis, but decreases going in the other axis. This means that the three collinear Lagrange points are **unstable** and an object placed at one of those points, if perturbed, will diverge from the position. -Nonetheless, these are quite useful points for observation of the solar system. Several satellites have been placed at the $L_1$ point of the Earth-Sun system for solar observation, and the James Webb Space Telescope is planned to launch to the $L_2$ of the Earth-Sun system sometime ~this year~ in 2022. +Nonetheless, these are quite useful points for observation of the solar system. Several satellites have been placed at the $L_1$ point of the Earth-Sun system for solar observation, and the James Webb Space Telescope (JWST) is located at the $L_2$ point in the Earth-Sun system specifically to avoid sunlight interefering with observations. These satellites orbit around the unstable Lagrange points in a [Lissajous orbit](https://en.wikipedia.org/wiki/Lissajous_orbit). This type of orbit requires a very small amount of propulsion onboard the satellite to keep position, but the orbit can last for a very long time with only a little fuel. One example is the [Wilkinson Microwave Anisotropy Probe](https://en.wikipedia.org/wiki/Wilkinson_Microwave_Anisotropy_Probe) (WMAP) which was sent to the $L_2$ point in the Earth-Sun system to study the [Cosmic microwave background](https://en.wikipedia.org/wiki/Cosmic_microwave_background). The trajectory of WMAP is shown in {numref}`fig:wmap-trajectory`. @@ -509,4 +508,12 @@ These satellites orbit around the unstable Lagrange points in a [Lissajous orbit The trajectory of the [Wilkinson Microwave Anisotropy Probe](https://en.wikipedia.org/wiki/Wilkinson_Microwave_Anisotropy_Probe) (WMAP) as viewed from Earth. Note the distance in the bottom of the animation, showing the satellite as approximately 1.5 million km from the earth. [Phoenix7777](https://commons.wikimedia.org/wiki/File:Animation_of_Wilkinson_Microwave_Anisotropy_Probe_trajectory_-_Viewd_from_Earth.gif), [CC BY-SA 4.0](https://creativecommons.org/licenses/by-sa/4.0), via Wikimedia Commons. ::: +Another example is the JWST, mentioned previously. JWST has a simpler [halo orbit](https://en.wikipedia.org/wiki/Halo_orbit) around $L_2$. The orbit of JWST is shown in {numref}`fig:jwst-trajectory`. + +:::{figure} ../images/jwst-trajectory.gif +:name: fig:jwst-trajectory + +The trajectory of the [James Webb Space Telescope](https://en.wikipedia.org/wiki/James_Webb_Space_Telescope) (JWST) as viewed from above the ecliptic plane with Earth fixed. Note again the distance in the bottom of the animation, showing the satellite as approximately 1.5 million km from the earth. [Phoenix7777](https://commons.wikimedia.org/wiki/File:Animation_of_James_Webb_Space_Telescope_trajectory_-_Polar_view.gif), [CC BY-SA 4.0](https://creativecommons.org/licenses/by-sa/4.0), via Wikimedia Commons. +::: + $L_1$ and $L_2$ in the Earth-Sun system are about 1.5 million km towards the Sun and away from the Sun, starting at the Earth, respectively. $L_3$ lies on the other side of the Sun, and has long been the predicted location of a hidden planet, since it could not be observed from Earth prior to the advent of satellite observation. Now, of course, we know there is no planet at that location. diff --git a/the-orbit-equation/hyperbolic-trajectories.md b/the-orbit-equation/hyperbolic-trajectories.md index abd65240..c1049d53 100644 --- a/the-orbit-equation/hyperbolic-trajectories.md +++ b/the-orbit-equation/hyperbolic-trajectories.md @@ -26,9 +26,9 @@ From the orbit equation, Eq. {eq}`eq:scalar-orbit-equation`, we see that the den As the true anomaly approaches $\nu_{\infty}$, $r$ approaches infinity. $\nu_{\infty}$ is restricted to be between 90° and 180°. -For $-\nu_{\infty} < \nu < \nu_{\infty}$, the trajectory of $m_2$ follows the occupied or real trajectory shown on the left in {numref}`fig:hyperbolic-trajectory-animation`. For $\nu_{\infty} < \nu < \left({360}^{\circ} - \nu_{\infty}\right)$, $m_2$ would occupy the virtual trajectory on the figure below. This trajectory would require a repulsive gravitational force for a mass to actually follow it, so it is only a mathematical result. +For $-\nu_{\infty} < \nu < \nu_{\infty}$, the trajectory of $m_2$ follows the occupied or real trajectory. For $\nu_{\infty} < \nu < \left({360}^{\circ} - \nu_{\infty}\right)$, $m_2$ would occupy the virtual trajectory. This trajectory would require a repulsive gravitational force for a mass to actually follow it, so it is only a mathematical result. -```{code-cell} ipython3 + Periapsis lies on the apse line, as usual, of the occupied trajectory. Interestingly, apoapsis lies on the virtual trajectory. Halfway between periapsis and apoapsis lies the center of a Cartesian coordinate system.