A point mass body of mass `2 kg` is placed at a distance `20 cm` from one end of a uniform rod of length `2 m` and mass `10 kg`. Calculate
(i) gravitational intensity at the location of point mass body due to the rod.
(ii) gravitational force on the body due to the rod. Use `G = 6.67 xx 10^(-11) Nm^(2) kg^(-2)`.

Explanation

Let `M, L` be the mass and length of the rod and `m` be the mass of body. Refer to Fig.
Mass per unit length of rod `= (M)/(L)`
Take a small element of the rod of mass `dm`
and length `dr` at a distance `r` from the body at `O`.
Mass of the element, `dm = (M)/(L) dr`
(i) Gravitational field intensity at `O` due to element is
`dI = (G dm)/(r^(2)) = (GM)/(L) (dr)/(r^(2))`
Total gravitational field intensity at `O` due to whole rod is
`I = int_(r=d)^(r=d+L) (GM)/(L) (dr)/(r^(2)) = (GM)/(L) [ - (1)/(r)]_(d)^(d + L)`
`= - (GM)/(L) [(1)/(d + L) - (1)/(d)] = (GM)/(d(d + L))`
Here, `M = 10` kg, `d = 20 cm = 0.20 m , L = 2 m`
`I = (6.67 xx 10^(-11) xx 10)/(0.20 (0.20 + 2)) = (6.67 xx 106(-10))/(0.2 xx 2.2)`
`= 1.5 xx 10^(-9) N kg^(-1)`
(ii) Force on the body at `O`,
`F = m xx I = 2 xx 1.5 xx 10^(-9)`
`= 3.0 xx 10^(-9) N`