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, the speed at the farthest point is:

(A) $3.0 \times 10^3 $ m/s
(B) $1.5 \times 10^3 $ m/s
(C) $4.5 \times 10^3 $ m/s
(D) $6.0 \times 10^3 $ m/s

Solution

We have, $v_1 r_1 = v_2 r_2 $

$\therefore v_1 \times 1.6 \times 10^{12} = 6 \times 10^4 \times 8.0 \times 10^ {10} $

$\Rightarrow v_1 = 3.0 \times 10^3 $ m/s

Answer: (A)