(a) A uniform rod of mass `M` and length `L` is placed at distance `L` from a point mass `m` as shown. Find force on `m`

(b) A semiconductor wire has a length `L` and mass `M`. A particle of mass `m` is placed at the centre of the circle. find the gravitational attraction on the particle due to the wire.

Explanation

Let us consider an elementary mass dm of length dr at a distance r from the particle of mass m. Here, `dm = M//Ldr` . The gravitational force dF on m due to this elementary mass dm is

`dF=(Gm)/(r^(2))dm=(GmM)/(Lr^(2))dr`
`:.` Force on the particle `F=(GmM)/(L) underset(d)overset(L+d)(int) (dr)/(r^(2))=(GmM)/(L) [-(1)/(r)]_(d)^(L+d)`
`=-(GmM)/(L)[(1)/(L+d)-(1)/(d)]=(GmM)/(d(L+d))`.