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.

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

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:

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

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

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