Question:
The maximum and minimum distances of a comet from the Sun are $1.6 \times 10^{12} \mathrm{~m}$ and $8.0 \times 10^{10} \mathrm{~m}$ respectively. If the speed of the comet afthe nearest point is $6 \times 10^{4} \mathrm{~ms}^{-1}$, the speed at the farthest point is:
Correct Option: , 3
Solution:
By angular momentum conservation :
$\mathrm{mv}_{1} \mathrm{r}_{1}=\mathrm{mv}_{2} \mathrm{r}_{2}$
$v_{1}=\frac{48 \times 10^{14}}{1.6 \times 10^{12}}=3000 \mathrm{~m} / \mathrm{sec}$
$=3 \times 10^{3} \mathrm{~m} / \mathrm{sec}$