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द्रव्यमान $m _{1}$ एवं $m _{2}$ के दो परिकल्पित उपग्रह विश्राम अवस्था में हैं जब वे एक दूसरे से अनन्त दूरी पर हैं। गुरुत्वाकर्षण बल के कारण उनके केन्द्रों को मिलाने वाली रेखा पर एक दूसरे की ओर गति करना प्रारम्भ करते हैं। जब उनके बीच दूरी $'d '$ है, तब उनकी चाल क्या है ?
$\left( m _{1}\right.$ की चाल $v_{1}$ एवं $m _{2}$ की चाल $v_{2}$ है )
$v_1 = v_2$
$\begin{array}{l}
{v_1}{\mkern 1mu} = {\mkern 1mu} {m_2}{\mkern 1mu} \sqrt {\frac{{2G}}{{d({m_1} + {m_2})}}} \\
{v_2}{\mkern 1mu} = {\mkern 1mu} {m_1}{\mkern 1mu} \sqrt {\frac{{2G}}{{d({m_1} + {m_2})}}}
\end{array}$
$\begin{array}{l}
{v_1}{\mkern 1mu} = {\mkern 1mu} {m_1}{\mkern 1mu} \sqrt {\frac{{2G}}{{d({m_1} + {m_2})}}} \\
{v_2}{\mkern 1mu} = {\mkern 1mu} {m_2}{\mkern 1mu} \sqrt {\frac{{2G}}{{d({m_1} + {m_2})}}}
\end{array}$
$\begin{array}{l}
{v_1}\, = \,{m_2}\,\sqrt {\frac{{2G}}{{{m_1}}}} \\
{v_2}\, = \,{m_2}\,\sqrt {\frac{{2G}}{{{m_2}}}}
\end{array}$
Solution
We choose reference point, infinity, where total energy of the system is zero.
So initial energy of the system $=0$
Final energy
$ = \frac{1}{2}{m_1}v_1^2 + \frac{1}{2}{m_2}v_2^2=\frac{{G{m_1}{m_2}}}{d}$
From conservation of energy,
$Initial\, energy=Final\, energy$
$\therefore 0 = \frac{1}{2}{m_1}v_1^2 + \frac{1}{2}{m_2}v_2^2 = \frac{{G{m_1}{m_2}}}{d}$
$or\,\frac{1}{2}{m_1}v_1^2 + \frac{1}{2}{m_1}v_2^2 = \frac{{G{m_1}{m_2}}}{d}\,\,…\left( i \right)$
By conservation of linear momentum
${m_1}{v_1} + {m_2}{v_2} = 0$
$or\,\frac{{{v_1}}}{{{v_2}}} = \frac{{{m_2}}}{{{m_1}}} \Rightarrow {v_2} = – \frac{{{m_1}}}{{{m_2}}}{v_1}$
Putting value of $v_2$ in equation $(1)$, we get
${m_1}v_1^2 + {m_2}{\left( { – \frac{{{m_1}{v_1}}}{{{m_2}}}} \right)^2} = \frac{{2G{m_1}{m_2}}}{d}$
$\frac{{{m_1}{m_2}v_1^2 + m_1^2v_1^2}}{{{m_2}}} = \frac{{2G{m_1}{m_2}}}{d}$
${v_1} = \sqrt {\frac{{2Gm_2^2}}{{d\left( {{m_1} + {m_2}} \right)}}} = {m_2}\sqrt {\frac{{2G}}{{d\left( {{m_1} + {m_2}} \right)}}} $
$Similarly\,{v_2} = {m_1}\sqrt {\frac{{2G}}{{d\left( {{m_1} + {m_2}} \right)}}} $
