Two hypothetical planets of masses m_1 and m_ | vedclass

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(B) The initial total energy of the system at an infinite distance is $0$.By the law of conservation of energy,the total energy at separation $d$ must

Vedclass · Accepted Answer

$v_1 = m_2 \sqrt{\frac{2G}{d(m_1 + m_2)}}, v_2 = m_1 \sqrt{\frac{2G}{d(m_1 + m_2)}}$

Vedclass · Answer

(B) The initial total energy of the system at an infinite distance is $0$.By the law of conservation of energy,the total energy at separation $d$ must also be $0$:$\frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 - \frac{G m_1 m_2}{d} = 0$$\frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 = \frac{G m_1 m_2}{d} \quad ... (i)$By the law of conservation of linear momentum (as no external force acts on the system):$m_1 v_1 - m_2 v_2 = 0 \implies v_2 = \frac{m_1}{m_2} v_1$Substituting $v_2$ into equation $(i)$:$\frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 \left( \frac{m_1}{m_2} v_1 \right)^2 = \frac{G m_1 m_2}{d}$$\frac{1}{2}m_1 v_1^2 \left( 1 + \frac{m_1}{m_2} \right) = \frac{G m_1 m_2}{d}$$\frac{1}{2} v_1^2 \left( \frac{m_1 + m_2}{m_2} \right) = \frac{G m_2}{d}$$v_1^2 = \frac{2 G m_2^2}{d(m_1 + m_2)} \implies v_1 = m_2 \sqrt{\frac{2G}{d(m_1 + m_2)}}$Similarly,$v_2 = m_1 \sqrt{\frac{2G}{d(m_1 + m_2)}}$.

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