Question

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.

Short answer

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

Step-by-step solution

Definition of Angular momentum.

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 given data:

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 $\theta$.

Find the angular momentum of the comet.

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

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

Find the value of  :

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 $\frac{v_1{r}_1}{r_2\sin \theta }$.