91Ó°ÊÓ

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 ${\stackrel{→}{r}}_2$and velocity vector $v$is $\mathrm{θ}$.

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

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

$r_{2,y}=r_2\sin \mathrm{θ}$

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

The initial angular momentum of the comet is,

$$\begin{gathered} {\stackrel{→}{L}}_i={\stackrel{→}{r}}_1×{\stackrel{→}{p}}_1 \\ ={\stackrel{→}{r}}_1×m{\stackrel{→}{v}}_1 \\ =m{v}_1{r}_1\sin 90° \\ =m{v}_1{r}_1 \end{gathered}$$

Angle between the radius vector ${\stackrel{→}{r}}_2\sin \mathrm{θ}$and velocity vector ${\stackrel{→}{v}}_2$is$90°$

The final angular momentum of the comet is

$$\begin{gathered} {\stackrel{→}{L}}_f={\stackrel{→}{r}}_2×{\stackrel{→}{p}}_2 \\ ={\stackrel{→}{r}}_2\sin \mathrm{θ}×m{\stackrel{→}{v}}_2 \\ =\left(m{v}_2{r}_2\sin \mathrm{θ}\right)\sin 90° \\ =m{v}_2{r}_2\sin \mathrm{θ} \end{gathered}$$

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.

$\begin{array}{rcl}{\stackrel{→}{\mathrm{τ}}}_{net}& =& \frac{d\stackrel{→}{L}}{dt}\\ 0& =& \frac{d\stackrel{→}{L}}{dt}\\ & & \end{array}$

${\stackrel{→}{L}}_f={\stackrel{→}{L}}_i$

….. (1)

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

$\begin{array}{rcl}{L}_f& =& {\stackrel{→}{L}}_i\\ m{v}_2{r}_2\sin \mathrm{θ}& =& m{v}_1{r}_1\\ v_2& =& \frac{v_1{r}_1}{r_2\sin \mathrm{θ}}\\ & & \end{array}$

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