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

Mam how mew is equal to m1m2/m1+m2

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.

Ask Question

Tag:MathPhyChemBioOther

Take Test

...