Suppose P is a particle having unit mass, outside the spherical shell of radius r. Let, A = distance between the particle P and the centre of the spherical shell = OP. \(\sigma\) = surface density of the material of the spherical shell. Let us cut the spherical shell at the planes LM and KL perpendicular to OP & close to each other to get the slice KLMN. Let LP = x, \(\angle\)LOP = Q & \(\angle\)KOL = dQ. \(\therefore\) The width of the slice = KL = r.dQ In the \(\triangle\) LQO, LQ = r\(sin\phi\) \(\therefore\) The length of the slice = 2\(\pi\)𝑟\(sin\theta\) \(\therefore\) The surface area of the slice = 2\(\pi\)𝑟\(sin\theta\) . 𝑟𝑑𝑄 = 2\(\pi\)\(𝑟^2\) \(sin\theta\) . 𝑑𝑄 \(\therefore\) The mass of the slice = 2\(\pi\)\(𝑟^2\)\(sin\theta\). 𝑑𝑄. \(\sigma\) = 2\(\pi\)\(\sigma\)\(𝑟^2\)\(sin\theta\) . \(d\theta\) (Here, the thickness of the slice or spherical shell has been neglected). As the particle P is at the same distance from every mass element of the slice, the gravitational potential