JEE PYQ: Gravitation Question 1
Question 1 - 2021 (16 Mar Shift 1)
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 $6 \times 10^{4}$ ms$^{-1}$, the speed at the farthest point is:
(1) $1.5 \times 10^{3}$ m/s
(2) $6.0 \times 10^{3}$ m/s
(3) $3.0 \times 10^{3}$ m/s
(4) $4.5 \times 10^{3}$ m/s
Show Answer
Answer: (3)
Solution
By angular momentum conservation: $mv_1 r_1 = mv_2 r_2$. $v_2 = \frac{v_1 r_1}{r_2} = \frac{6 \times 10^4 \times 8 \times 10^{10}}{1.6 \times 10^{12}} = 3 \times 10^3$ m/s.
BETA
