JEE Main 201910 JanuaryEvening Shift
- Exam
- JEE Main
- Year
- 2019
- Date
- 10 January
- Shift
- Evening Shift
- Subject
- Physics
- Chapter
- Gravitation
JEE Main 2019 · 10 January · Evening Shift · Physics · Gravitation
Two stars of masses 3 × 10 ^ 31 kg each, and at distance 2 × 10 ^...
Two stars of masses each, and at distance rotate in a plane about their common centre of mass A meteorite passes through moving perpendicular to the stars,s rotation plane. In order to escape from the gravitational field of this double star, the minimum speed that meteorite should have at is ( Take Gravitational constant )
Options
- A
- B
- C
- D
Correct answer
C.
Step-by-step solution
To escape, the total energy of small particle must be zero.
\times 2⇒ V = \sqrt{\dfrac{8GM}{d}}= \sqrt{\dfrac{8 \times 6.67 \times {10}^{- 11} \times 3 \times {10}^{31}}{2 \times {10}^{11}}}= 2.8 \times {10}^{5}m{s}^{- 1}
\times 2⇒ V = \sqrt{\dfrac{8GM}{d}}= \sqrt{\dfrac{8 \times 6.67 \times {10}^{- 11} \times 3 \times {10}^{31}}{2 \times {10}^{11}}}= 2.8 \times {10}^{5}m{s}^{- 1}