Four particles each of mass .m. are located at the four corners of a square of side a. If the system is revolving in a circle due to their mutual force of attraction then the angular velocity and time period of each particle is

Four particles each of mass .m. are located at the four corners of a s

1.	Four particles each of mass .m. are located at the four corners of a square of side a. If the system is revolving in a circle due to their mutual force of attraction then the angular velocity and time period of each particle is
Answer» Solution : The gravitational force between the masses provides the necessary centripetal force `(Gm_1m_2)/(d^2)=m_(1)r_(1)omega^2 RARR(1)` The distance of CENTRE of mass from `m_1` is `r_(1)=(m_2d)/(m_1+m_2)rarr(2)` From (1) and (2) `(Gm_1m_2)/d^2 =(m_1m_2d)/(m_1+m_2).omega^2` (or) `omega^(2)=(G(m_1+m_2))/d^3 ` (or) `omega=sqrt((G(m_1+m_2))/d^3)` The resultant gravitational force on one of the particles GM^2)/a^2+(Gm^2)/(2a^2)` This gravitational force provides the necessary centripetal force `mromega^2=(Gm)/(a^2)(SQRT2+1/2)implies` `a/sqrt2omega^2=(Gm)/a^2((2sqrt2+1)/2)impliesomega=sqrt((Gm)/a^3(2+1/sqrt2)) and T =(2pi)/omega=2pisqrt((a^3)/(Gm)((sqrt2)/(2sqrt2+1))`