Two stars of masses 3times10^{31} kg each,an | vedclass

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(D) Let the mass of the meteorite be $m$. The distance of each star from the center of mass $O$ is $r = d/2 = (2	imes10^{11} \ m)/2 = 1	imes10^{11}

Vedclass · Accepted Answer

$2.8	imes10^5 \ m/s$

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(D) Let the mass of the meteorite be $m$. The distance of each star from the center of mass $O$ is $r = d/2 = (2	imes10^{11} \ m)/2 = 1	imes10^{11} \ m$.The gravitational potential energy of the meteorite at point $O$ due to the two stars is $U = -\frac{GMm}{r} - \frac{GMm}{r} = -\frac{2GMm}{r}$.To escape the gravitational field,the total mechanical energy of the meteorite at $O$ must be at least zero. By the law of conservation of energy:$\frac{1}{2}mv^2 + U = 0$$\frac{1}{2}mv^2 - \frac{2GMm}{r} = 0$Solving for $v$:$v^2 = \frac{4GM}{r}$$v = \sqrt{\frac{4 	imes 6.67	imes10^{-11} 	imes 3	imes10^{31}}{1	imes10^{11}}}$$v = \sqrt{4 	imes 6.67 	imes 3 	imes 10^9}$$v = \sqrt{80.04 	imes 10^9} = \sqrt{8.004 	imes 10^{10}}$$v \approx 2.83 	imes 10^5 \ m/s$.

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