Two particles of mass $m_{1}$ and $m_{2}$ are initially at rest at infinite distance. Find their velocity of approach due to gravitational attraction, when separation is d :

\sqrt{\frac{2 G (m_{1} + m_{2})}{d}}

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\sqrt{\frac{G (2 m_{1} + m_{2})}{3 d}}

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\sqrt{\frac{3 G (2 m_{1} + m_{2})}{d}}

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\sqrt{\frac{G (m_{1} + m_{2})}{d}}

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Solution

The correct option is A $\sqrt{\frac{2 G (m_{1} + m_{2})}{d}}$
from the law of conservation of energy
decrease in potential energy = increase in kinetic energy

$\frac{G m_{1} m_{2}}{d^{2}} = \frac{1}{2} [\frac{m_{1} m_{2}}{m_{1} + m_{2}}] {v_{r}}^{2}$

$v_{r} = \sqrt{\frac{2 G m_{1} m_{2}}{(\frac{m_{1} m_{2}}{m_{1} + m_{2}}) d}}$

$v_{r} = \sqrt{\frac{2 G (m_{1} + m_{2})}{d}}$ where $v_{r}$ is the relative velocity of approach

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Two particles of mass m1 and m2 are initially at rest at infinite distance. Find their velocity of approach due to gravitational attraction, when separation is d :

The correct option is A √2G(m1+m2)dfrom the law of conservation of energy decrease in potential energy = increase in kinetic energyGm1m2d2=12[m1m2m1+m2]vr2vr=  ⎷2Gm1m2(m1m2m1+m2)dvr=√2G(m1+m2)d where vr is the relative velocity of approach

Two particles of mass $m_{1}$ and $m_{2}$ are initially at rest at infinite distance. Find their velocity of approach due to gravitational attraction, when separation is d :