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Two bodies of masses $m_1$ and $m_2$ are initially at rest placed infinite distance apart. They are then allowed to move towards each other under their mutual gravitational attraction. Their relative velocity when they are 'r' distance apart is

\[(a)\;\sqrt {\frac{2G(m_1+m_2)}{r}} \quad (b)\; \sqrt {\frac{2Gm_1m_2}{(m_1+m_2)r}}\quad (c)\;\sqrt {\frac{G(m_1+m_2)}{r}} \quad (d)\;\sqrt {\frac{G(m_1m_2)}{(m_1+m_+2)r}}\]

Mam how mew is equal to m1m2/m1+m2

1 Answer

From conservation of mechanical energy
Decrease in potential energy is equal to increase in kinetic energy
$\large\frac{Gm_1m_2}{r}=\frac{1}{2}$$ \mu {v_r}^2$
$\mu -reduced\; mass=\large\frac{m_1m_2}{m_1+m_2}$
$v_r$ - relative velocity of approach
$v_r=\sqrt {\large\frac{2Gm_1m_2}{\mu r}}$
$\qquad=\sqrt {\large\frac{2Gm_1m_2}{\bigg(\Large\frac{m_1m_2}{m_1+m_2}\bigg)r}}$
$\qquad=\sqrt {\large\frac{2G(m_1+m_2)}{r}}$
Hence a is the correct answer.


answered Aug 20, 2013 by meena.p
edited Feb 17, 2014 by meena.p

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