JEE PYQ: Gravitation Question 44
Question 44 - 2019 (10 Jan Shift 2)
Two stars of masses $3 \times 10^{31}$ kg each, and at distance $2 \times 10^{11}$ m rotate in a plane about their common centre of mass O. A meteorite passes through O moving perpendicular to the star’s rotation plane. In order to escape from the gravitational field of this double star, the minimum speed that the meteorite should have at O is: (Take Gravitational constant $G = 6.67 \times 10^{-11}$ Nm$^2$ kg$^{-2}$)
(1) $2.4 \times 10^4$ m/s
(2) $1.4 \times 10^5$ m/s
(3) $3.8 \times 10^4$ m/s
(4) $2.8 \times 10^5$ m/s
Show Answer
Answer: (4)
Solution
At O, distance from each star $= r = 10^{11}$ m. $v = \sqrt{\frac{4GM}{r}} = \sqrt{\frac{4 \times 6.67 \times 10^{-11} \times 3 \times 10^{31}}{10^{11}}} = \sqrt{8 \times 10^{10}} \approx 2.8 \times 10^5$ m/s.
BETA
