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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 \(v_2\) )

  1. A \(v_1 = v_2\)
  2. B \(\begin{array}{l}
    {v_1}{\mkern 1mu}  = {\mkern 1mu} {m_2}{\mkern 1mu} \sqrt {\frac{{2G}}{{d({m_1} + {m_2})}}} \\
    {v_2}{\mkern 1mu}  = {\mkern 1mu} {m_1}{\mkern 1mu} \sqrt {\frac{{2G}}{{d({m_1} + {m_2})}}} 
    \end{array}\)
  3. C \(\begin{array}{l}
    {v_1}{\mkern 1mu}  = {\mkern 1mu} {m_1}{\mkern 1mu} \sqrt {\frac{{2G}}{{d({m_1} + {m_2})}}} \\
    {v_2}{\mkern 1mu}  = {\mkern 1mu} {m_2}{\mkern 1mu} \sqrt {\frac{{2G}}{{d({m_1} + {m_2})}}} 
    \end{array}\)
  4. D \(\begin{array}{l}
    {v_1}\, = \,{m_2}\,\sqrt {\frac{{2G}}{{{m_1}}}} \\
    {v_2}\, = \,{m_2}\,\sqrt {\frac{{2G}}{{{m_2}}}} 
    \end{array}\)
Verified Solution

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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