Question

Two bodies of masses `m_(1)` and `m_(2)` are placed at a distance `r` apart. Show that the position where the gravitational field due to them is zero, the potential is given by
`-G (m_(1) + m_(2) + 2 sqrt(m_(1) + m_(2)))//r`

Answer 1

Consider the two bodies `A` and `B` of masses `m_(1)` and `m_(2)` produce zero resultant gravitational field at `C` where `AC = x`.
`:. BC = (r - x)`.
Hence, `(Gm_(1))/(x^(2)) = (Gm_(2))/((r - x)^(2))` or `(sqrt(m_(1)))/(x) = (sqrt(m_(2)))/((r - x))`
or `(r - x) sqrt(m_(1)) = x sqrt(m_(2))`
`:. x = (r sqrt(m_(1)))/(sqrt(m_(1)) + sqrt(m_(2)))`
and `(r - x) = r - ((r sqrt(m_(1)))/(sqrt(m_(1)) + sqrt(m_(2))))`
`= ((r sqrt(m_(2)))/(sqrt(m_(1)) + sqrt(m_(2))))`
`:. (1)/(x) = (sqrt(m_(1)) + sqrt(m_(2)))/(r sqrt(m_(1)))`
and `(1)/((1 - x)) = (sqrt(m_(1)) + sqrt(m_(2)))/(r sqrt(m_(2)))` ltbrltgt Gravitational potential at `P`
`= - (Gm_(1))/(x) + (-(G m_(2))/(r - x)) = - G[(m_(1))/(x) + (m_(2))/(r - x)]`
`= - G[m_(1)((sqrt(m_(1)) + sqrt(m_(2))))/(r sqrt(m_(1))) +(m_(2) (sqrt(m_(1)) + sqrt(m_(2))))/(rsqrt(m_(2)))]`
`= - (G)/(r ) [sqrt(m_(1)) (sqrt(m_(2)) + sqrt(m_(2))) + sqrt(m_(2)) (sqrt(m_(1)) + sqrt(m_(2)))]`
`= - (G)/(r ) [m_(1) + m_(2) + 2 sqrt(m_(1) m_(2))]`