Two masses m_1 and m_2 are at rest at infinit | vedclass

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Question

(B) According to the law of conservation of energy,the initial total energy at infinity is $0$. At a separation $d$,the total energy is the sum of kin

Vedclass · Accepted Answer

$\sqrt{\frac{2G(m_1 + m_2)}{d}}$

Vedclass · Answer

(B) According to the law of conservation of energy,the initial total energy at infinity is $0$. At a separation $d$,the total energy is the sum of kinetic energy and potential energy: $K.E. + P.E. = 0$. Using the reduced mass $\mu = \frac{m_1 m_2}{m_1 + m_2}$,the kinetic energy is $\frac{1}{2} \mu v_{rel}^2$. The potential energy is $-\frac{G m_1 m_2}{d}$. Thus,$\frac{1}{2} \left( \frac{m_1 m_2}{m_1 + m_2} \right) v_{rel}^2 - \frac{G m_1 m_2}{d} = 0$. Solving for $v_{rel}$: $\frac{1}{2} \left( \frac{m_1 m_2}{m_1 + m_2} \right) v_{rel}^2 = \frac{G m_1 m_2}{d}$. $v_{rel}^2 = \frac{2G(m_1 + m_2)}{d}$. $v_{rel} = \sqrt{\frac{2G(m_1 + m_2)}{d}}$.

Two masses $m_1$ and $m_2$ are at rest at infinity. Find their relative velocity of approach due to gravitational attraction when their separation is $d$.

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