Question

The maximum and minimum distances of a comet from the Sun are $1.6\times {10}^{12}m$ and $8.0\times {10}^{10}m$respectively. If the speed of the comet at the nearest point is ,$6\times {10}^{4}m{s}^{-1}$ the speed at the farthest point is :

• A. $1.5\times {10}^{3}m/s$
• B. $4.5\times {10}^{3}m/s$
• C. $3.0\times {10}^{3}m/s$
• D. $6.0\times {10}^{3}m/s$

Answer 1

The correct option is C

$3.0\times {10}^{3}m/s$

Step 1: Given Data:

The maximum and minimum distances of a comet from the Sun $=1.6\times {10}^{12}m$ and $8.0\times {10}^{10}m$

$r_1=8.0\times {10}^{10}m$

$r_2=1.6\times {10}^{12}m$

The speed of the comet at the nearest point $v_1=6\times {10}^{4}m{s}^{-1}$

Step 2: Formula Used:

$L=mvr$

where $L=$Angular momentum, $m=$Mass, $v=$Velocity, and $r=$Radius.

Step 3: Calculation of speed at the farthest point:

Let point 1 is nearest point,

and point 2 is the farthest point

By angular momentum conservation

$$\begin{gathered} L_1=L_2 \\ m{r}_1{v}_1=m{r}_2{v}_2 \\ \begin{array}{l}{v}_2=\frac{r_1{v}_1}{r_2}\end{array} \\ \begin{array}{l}{v}_2=\frac{8\times {10}^{10}\times 6\times {10}^4}{1.6\times {10}^{12}}\end{array} \\ \therefore v_2=3.0\times {10}^{3}m/s \end{gathered}$$

Hence option C is the correct answer.