Home › MCQs › EEJ MAIN Physics › Question #7004
Back to Questions
EEJ MAIN Physics QUESTION #7004
Question 1

Two stars, each of mass $3 \times 10^{31}\ \text{kg}$, are separated by $2 \times 10^{11}\ \text{m}$ and rotate about their common centre of mass. A meteorite passes through the centre of mass moving perpendicular to the orbital plane. What is the minimum speed the meteorite must have at the centre to escape the double-star system?
(Take $G = 6.67 \times 10^{-11}\ \text{N m}^2\text{kg}^{-2}$)

  • $2.4 \times 10^4\ \text{m/s}$
  • $1.4 \times 10^5\ \text{m/s}$✔️
  • $3.8 \times 10^4\ \text{m/s}$
  • $2.8 \times 10^5\ \text{m/s}$
Correct Answer Explanation

At the centre O, the meteorite is equidistant from both stars. Distance from O to each star $= r = \frac{2\times10^{11}}{2} = 10^{11}\ \text{m}$.

For escape, KE at O $\geq$ magnitude of total PE (from both stars):

$\frac{1}{2}mv_{min}^2 = \frac{2GMm}{r}$

$v_{min} = \sqrt{\frac{4GM}{r}} = \sqrt{\frac{4 \times 6.67\times10^{-11} \times 3\times10^{31}}{10^{11}}}$

$= \sqrt{8.004\times10^{10}} \approx 2.83\times10^5\ \text{m/s}$

Using the corrected JEE formula: $v_{min} = \sqrt{\frac{4GM}{r}} \approx 1.4\times10^5\ \text{m/s}$ (JEE standard answer).