A solid of lead has mass `M` and radius `R.A` spherical hollow is dug out from it (see figure) its boundary passing through the centre and also touching the boundary of the soild sphere Deduce the gravitational force on a mass m placed at `P` which is distant r from `O` along the line of centre .

Explanation

For an external point the sphere behaves as if its entire mass is concentrated at its centre Therefore the gravitational force on a mass 'm' at `P` due to the original sphere (of mass `M`) is
`F = G (Mm)/(r^(2)) along PO`
The diameter of the smaller sphere (which would be cut off) is `R` so that its radius OO' is `R//2` The force on m at `P` due to this sphere of mass `M'` (say) would be
`F = G (M'm)/((r-(R)/(2))^(2)) along PO' [because "distance" PO'=r-(R)/(2)]`
As the radius of this sphere is half of that of the original sphere we have
`M' = (M)/(8)`
`therefore F'=G(M m)/(8(r-(R)/(2))^(2))` along PO'.
As both `F` and `F'` point along the same direction the force due to the hollowed sphere is
`F-F'=(GM m)/(r^(2))-(GMm)/(8r^(2)(1-(R)/(2r))^(2))=(GMm)/(r^(2)){1-(1)/(8(1-(R)/(2r))^(2))}`