The highly elliptical orbit of Halley's comet is shown in the representations. When the comet is closest to the Sun, at the location specified by the position vector $\vec{r}_{1}$ (“perihelion”), it is $8.77$ x $10^{10}$m from the Sun, and its speed is $5.46$ x $10^4$ m/s. When the comet is at the location specified by the position vector $\vec{r}_{2}$, its speed is $1.32$ x $10^{4}$ m/s. At that location the distance between the comet and the Sun is $1.19$ x $10^{12}$ m, and the angle $\theta$ is $17.81^{\circ}$. The mass of the comet is estimated to be $2.2$ x $10^{14}$ kg. Calculate the translational (orbital) angular momentum of the comet, relative to the Sun, at both locations.

At $\vec{r}_{1}$ comet is 8.77$x$10^{10}$m from the Sun.

The comets speed at $\vec{r}_{1}$ is $5.46$ x $10^4$ m/s.

At $\vec{r}_{2}$ the comets speed is $1.32$ x $10^{4}$ m/s.

The distance between the comet and the Sun at $\vec{r}_{2}$ is $1.19$ x $10^{12}$ m.

Angle $\theta$ in representation is $17.81^{\circ}$

The mass of the comet is estimated to be $2.2$ x $10^{14}$ kg.

Calculate the translational (orbital) angular momentum of the comet, relative to the Sun, at both locations.

No other interactions the rest of the solar system.

Assume main interaction is with the sun.

$\left|\vec{L}_{trans}\right| = \left|\vec{r}_A\right|\left|\vec{p}\right|\sin \theta$

Direction: At both locations, the direction of the translational angular momentum of the comet is in the -z direction (into the computer); determined by using the right-hand rule.

Given this information we know at location 1 the translational angular momentum of the comet relative to the sun will be:

$\left|\vec{L}_{trans,Sun}\right| = \left|\vec{r}_A\right|\left|\vec{p}\right|\sin \theta$

We don't know the momentum but we do know the mass and velocity of the comet at $\vec{r}_{1}$ so our equation becomes:

$\left|\vec{L}_{trans,Sun}\right| = \left|\vec{r}_A\right|\left|\vec{v}\right|\left|m\right|\sin \theta$

Substituting in for the known variables we get:

$\mid\vec{L}_{trans,Sun}\mid$ = $(8.77$ x $10^{10}m)(2.2$ x $10^{14}kg)(5.46$ x $10^4m/s)sin 90^{\circ}$

Solving for $\mid\vec{L}_{trans,Sun}\mid$ we get:

$= 1.1$ x $10^{30}$ $kg \cdot m^2/s$

In vector form $\vec{L}_{trans,Sun}$ is:

$\vec{L}_{trans,Sun}$ = $\langle{0, 0, -1.1 x 10^30}\rangle$ $kg \cdot m^2/s$

The same step by step process is used to solve for $\vec{L}_{trans,Sun}$ at location 2:

$\mid\vec{L}_{trans,Sun}\mid$ = $(1.19$ x $10^{12}m)(2.2$ x $10^{14}kg)(1.32$ x $10^4m/s)sin 17.81^{\circ}$

$= 1.1$ x $10^{30}$ $kg \cdot m^2/s$

$\vec{L}_{trans,Sun}$ = $\langle{0, 0, -1.1 x 10^30}\rangle$ $kg \cdot m^2/s$

Even in the highly elliptical orbit, the comet's translational angular momentum is constant throughout the orbit, despite the fact that it's position, its momentum, and the angle between them change continuously implying that angular momentum is a conserved quantity.