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In free space, two particles of mass m each are initially both at rest at a distance a from each other. They start moving towards each other due to their mutual gravitational attraction. The time after which the distance between them has reduced to (a)/(2) is: |
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Answer» `((pi+2)/(4sqrt(2)))((a^(3))/(Gm))^(1//2)` Let the instantaneous velocity of each particle be v Let the time after which the distance between the particles has reduced to `(a)/(2)` be T Then, for the particle that was initially at `x=(a)/(2)`, `(Gm^(2))/((2R)^(2))=-m("DVD")/(dr)""implies""(Gm)/(r^(2))=-(4vdv)/(dr)` `implies""-4int_(0)^(v)dvd=Gmint_(a//2)^(r)(dr)/(r^(2))impliesv^(2)=(Gm)/(2)((1)/(r)-(2)/(a))impliesv=-[(Gm)/(2)((1)/(r)-(2)/(a))]^(1//2)` [v is negative because the velocity is towards the -X direction] `implies""(dr)/(dt)=-[(Gm)/(2)((1)/(r)-(2)/(a))]^(1//2)implies-int_(a//2)^(a//4)sqrt((r)/(a-2r))dr=sqrt((Gm)/(2A))int_(0)^(T)dt` `implies""int_(a//2)^(a//4)sqrt((r)/(a-2r))dr=-sqrt((Gm)/(2a))T""...(i)` Let us now evaluate the integral `I=int sqrt((r)/(a-2r))dr` Let `r=(a sin^(2)theta)/(2)""implies""dr=a sin theta cos theta d theta` Therefore, `I=(a)/(sqrt(2))int sin^(2)theta d theta=(a)/(2sqrt(2))int(1-cos2theta)d theta=(a)/(2sqrt(2))[theta-(1)/(2)sin 2theta]` Since `r=(a sin^(2)theta)/(2),theta=sin^(-1)sqrt((2r)/(a))` and `sin2theta=2sin theta cos ttheta=2(sqrt((2r)/(a)))(sqrt(1-(2r)/(a)))=(sqrt(8r(a-2r)))/(a)` So, `I=(a)/(2sqrt(2))(sin^(-1)sqrt((2r)/(a))-(sqrt(2r(a-2r)))/(a))=(a)/(2sqrt(2))sin^(-1)sqrt((2r)/(a))-(sqrt(r(a-2r)))/(2)` Therefore, from equation (i), sin `T=-sqrt((2a)/(Gm))((a)/(2sqrt(2))sin^(-1)sqrt((2r)/(a))-(sqrt(r(a-2r)))/(2))_(a//2)^(a//4)=-sqrt((2a)/(Gm))(-(a)/(2sqrt(2))((pi)/(4))-(1)/(2)((a)/(2sqrt(2))))` Hence, `""T=sqrt((a)/(Gm))((api)/(8)+(a)/(4))=((pi+2)/(8))sqrt((a^(3))/(Gm))` |
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