A certain comet of mass $\mathit{m}$at its closest approach to the Sun is observed to be at a distance${\mathit{r}}_{\mathbf{1}}$ from the center of the Sun, moving with speed ${\mathit{v}}_{\mathbf{1}}$ (Figure 11.92). At a later time the comet is observed to be at a distance from the center of the Sun, and the angle between $\overrightarrow{{\mathbf{r}}_{\mathbf{2}}}$ and the velocity vector is measured to be $\mathit{\theta }$. What is ${\mathit{v}}_{\mathbf{2}}\mathbf{?}$Explain briefly.

Angular momentum (rarely, moment of momentum or rotational momentum) is the rotational analog of linear momentum. It is an important quantity in physics because it is a conserved quantity—the total angular momentum of a closed system remains constant.

Angular momentum is an important property of a rotating object and is expressed as the product of the moment of inertia and angular velocity.

The mass of the comet is $m$.

The initial distance of the comet from the center of the Sun is $r_1$.

The initial speed of the comet is $v_1$.

The final distance of the comet from the center of the Sun is $r_2$.

The final speed of the comet is $v_2$.

Angle between the radius vector ${\overrightarrow{r}}_2$and velocity vector $v$is $\theta$.

The figure shows the orbit of a comet of mass $m$around the Sun.

From the figure, the component of the radius vector ${\overrightarrow{r}}_2$in the direction perpendicular to the velocity vector ${\overrightarrow{v}}_2$is,

$r_{2,y}=r_2\sin \theta$

Angle between the radius vector ${\overrightarrow{r}}_1$ and velocity vector ${\overrightarrow{v}}_1$is $90°$.

The initial angular momentum of the comet is,

$$\begin{gathered} {\overrightarrow{L}}_i={\overrightarrow{r}}_1\times {\overrightarrow{p}}_1 \\ ={\overrightarrow{r}}_1\times m{\overrightarrow{v}}_1 \\ =m{v}_1{r}_1\sin 90° \\ =m{v}_1{r}_1 \end{gathered}$$

Angle between the radius vector ${\overrightarrow{r}}_2\sin \theta$and velocity vector ${\overrightarrow{v}}_2$is$90°$

The final angular momentum of the comet is

$$\begin{gathered} {\overrightarrow{L}}_f={\overrightarrow{r}}_2\times {\overrightarrow{p}}_2 \\ ={\overrightarrow{r}}_2\sin \theta \times m{\overrightarrow{v}}_2 \\ =\left(m{v}_2{r}_2\sin \theta \right)\sin 90° \\ =m{v}_2{r}_2\sin \theta \end{gathered}$$

Let us consider the comet plus the Sun as a system. The net external torque acting on the system is zero, so the angular momentum of the system is conserved.

$\begin{array}{rcl}{\overrightarrow{\tau }}_{net}& =& \frac{d\overrightarrow{L}}{dt}\\ 0& =& \frac{d\overrightarrow{L}}{dt}\\ & & \end{array}$

${\overrightarrow{L}}_f={\overrightarrow{L}}_i$

….. (1)

From the equation (1), you get the speed $v_2$as,

$\begin{array}{rcl}{L}_f& =& {\overrightarrow{L}}_i\\ m{v}_2{r}_2\sin \theta & =& m{v}_1{r}_1\\ v_2& =& \frac{v_1{r}_1}{r_2\sin \theta }\\ & & \end{array}$

Hence, the value of speed $v_2$is $\frac{v_1{r}_1}{r_2\sin \theta }$.