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Two stars of masses 3×10^31kg each, and at distance 2×10^11m 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 meteorite should have at O is: (Take Gravitational constant G=6.67×10^−11Nm^2kg^−2)

Textbook solution:

$$KE_i + PE_i = KE_f + PE_f$$ $$\frac{1}{2}mv^2 + (-\frac{GMm}{r} + -\frac{GMm}{r}) = 0+0$$ $$\frac{1}{2}mv^2 = 2\frac{GMm}{r}$$ you can solve for $v$ from there. My question is if we assume all three objects(star-1, star-2, and meteriod) as one system, why didn't we include kinetic energy of both stars in our conservation of mechanical energy equation? I'm unable to see the whole picture of this question. Both the stars are also rotating around each other and they have velocity too

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  • $\begingroup$ You are looking at the conservation of energy of the meteorite, not the stars. Therefore only its kinetic and potential energy appear in the first equation. The kinetic energy of the stars don't as well as their potential energy which would be of the form: \begin{equation} -\frac{GM^2}{2r} \end{equation} for one of them due to the other. (Because of the textbook solution, I assumed that with distance of the stars they actually meant radius of their orbit. Otherwise we needed to replace $r$ with $r/2$ everywhere.) $\endgroup$
    – Samuel Adrian Antz
    Commented Apr 7, 2022 at 7:57
  • $\begingroup$ If we're looking at potential energy of meteriod only then it'll be zero as meteriod is only our system as gravitational potential energy is not defined for one object but rather a system. But the fact that we didn't means we've considered both stars as our system, but we didn't calculate their kinetic energies despite them belonging to our considered system. $\endgroup$
    – Rambal heart remo
    Commented Apr 7, 2022 at 8:54
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    $\begingroup$ You assume that the effect of the meteorite on the stars is neglectable. If you have a constant distance given, the stars move in a perfect circle, so their velocity and therefore kinetic energy does not change, which makes it constant. You can either shorten it on both sides of the equation when considering $E(t_1)=E(t_2)$ or ignore it when considering $E(t)=\text{const}$. $\endgroup$
    – Samuel Adrian Antz
    Commented Apr 7, 2022 at 9:11
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    $\begingroup$ @SamuelAdrianAntz: That looks like an answer to me. $\endgroup$
    – Michael Seifert
    Commented Apr 18, 2022 at 19:02
  • $\begingroup$ You can use a frame of reference which is rotating with the stars. In this frame of reference the stars are stationary and the meteorite moves along the axis of rotation so there is no centrifugal force on it. $\endgroup$
    – sammy gerbil
    Commented Apr 20, 2022 at 22:12

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