Q. The maximum and minimum distances of a comet from the sun are $1.4 \times 10^{12}$m and $6 \times 10^{10}$ m respectively. If its velocity nearest to the sun is $7 \times 10^{4}ms^{- 1}$, what is the velocity in the farthest position? Assume the paths of comet in both instantaneous positions are circular
A) $3 \times 10^{3}ms^{- 1}$
B) $5 \times 10^{3}ms^{- 1}$
C) $7 \times 10^{3}ms^{- 1}$
D) $7 \times 10^{3}ms^{- 1}$
Explanation
(a) As a real velocity of comet is constant, therefore,
$r_{1}v_{1} = r_{2}v_{2}\ \ or\ \ v_{2} = \frac{r_{1}v_{1}}{r_{2}}$
$= \frac{6 \times 10^{10} \times 7 \times 10^{4}}{1.4 \times 10^{12}} = {3 \times 10}^{3}ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.