Two particles of masses `m_(1)` and `m_(2)` are initially at rest infinite distance apart. If they approach each other under inverse square law of force `(F=k//r^(2))` find their speed of approach at the instant when they are distance d apart.

Explanation

Let the particles be at a distance r apart at any instant. Then as `(F=k//r^(2))`, the accelrations of the particles
`a_(1)=(F)/(m_(1))=(K)/(m_(1)r^(2))`
and `a_(2)=(F)/(m_(2))=(K)/(m_(2)r^(2))`
As they are approaching each other, the acc. of one relative to toehr
`a=a_(1)+a_(2)=(k)/(r^(2))[(1)/(m_(1)+(1)/(m_(2)))]` ..........(i)
or `(dv)/(dt)=(k)/(r^(2))[(1)/(m_(1))+(1)/(m_(2))] ["as" alpha=(dv)/(dt)]`
or `(dv)/(dr).(dr)/(dt)=(K)/(r^(2))[(1)/(m_(1))+(1)/(m_(2))]` [by chain ru le]
Now as the particles are approaching each other, with time r is ecreasing so
`(dr//dt)=-v`
So, `-v(dv)/(dr)=(K)/(r^(2))[(1)/(m_(1))+(1)/(m_(2))]`
Integrating the above equation
`(v^(2))/(2)=(K)/(r)[(1)/(m_(1))+(1)/(m_(2))]+C`
But as initially v=0 at r= `porp,C=0`.
`therefore v=sqrt((2k)/(r)[(1)/(m_(1))+(1)/(m_(2))])`
So, for r=d,v=`sqrt((2k)/(d)[(1)/(m_(1))+(1)/(m_(2))])`