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[Post New]24 Mar 2008 16:00:51 IST
    Subject: Re:Gravitation
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elastiboysai (2327)

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Olaaa!! Perrrfect answer. 421  [532 rates]
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\text{Let v be velocity of approach}\\ \text{Increase in kinetic energy=decrease in potential energy}\\ \frac{1}{2}\mu\cdot v_r^2=\frac{Gm_1m_2}{r}\\ v_r=\sqrt\frac{2G\cdot(m_1+m_2)}{r}\\ \\ Here \mu=\frac{m_1\cdot m_2}{m_1+m_2} \\ \text{represents the reduced mass of the system}\\ \text{this equation can be obtained in a genl case by working}\\ \text{in the center of mass frame}
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