JEE Mains · Physics · STD 11 - 7. gravitation
Two hypothetical planets of masses \(m_1\) and \(m_2\) are at rest when they are infinite distance apart. Because of the gravitational force they move towards each other along the line joining their centres . What is their speed when their separation is \('d'\) ? (Speed of \(m_1\) is \(v_1\) and that of \(m_2\) is \(v_2\) )

- A \(v_1 = v_2\)
- B \(\begin{array}{l}
{v_1}{\mkern 1mu} = {\mkern 1mu} {m_2}{\mkern 1mu} \sqrt {\frac{{2G}}{{d({m_1} + {m_2})}}} \\
{v_2}{\mkern 1mu} = {\mkern 1mu} {m_1}{\mkern 1mu} \sqrt {\frac{{2G}}{{d({m_1} + {m_2})}}}
\end{array}\) - C \(\begin{array}{l}
{v_1}{\mkern 1mu} = {\mkern 1mu} {m_1}{\mkern 1mu} \sqrt {\frac{{2G}}{{d({m_1} + {m_2})}}} \\
{v_2}{\mkern 1mu} = {\mkern 1mu} {m_2}{\mkern 1mu} \sqrt {\frac{{2G}}{{d({m_1} + {m_2})}}}
\end{array}\) - D \(\begin{array}{l}
{v_1}\, = \,{m_2}\,\sqrt {\frac{{2G}}{{{m_1}}}} \\
{v_2}\, = \,{m_2}\,\sqrt {\frac{{2G}}{{{m_2}}}}
\end{array}\)
Answer & Solution
Correct Answer
(B) \(\begin{array}{l}
{v_1}{\mkern 1mu} = {\mkern 1mu} {m_2}{\mkern 1mu} \sqrt {\frac{{2G}}{{d({m_1} + {m_2})}}} \\
{v_2}{\mkern 1mu} = {\mkern 1mu} {m_1}{\mkern 1mu} \sqrt {\frac{{2G}}{{d({m_1} + {m_2})}}}
\end{array}\)
Step-by-step Solution
Detailed explanation
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, \(Initial\, energy=Final\, energy\)…
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