# Gravitational Potential due to Spherical Shell of Matter

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\(\pi\)π\(sin\theta\)
\(\therefore\) The surface area of the slice = 2\(\pi\)π\(sin\theta\) . πππ
= 2\(\pi\)\(π^2\) \(sin\theta\) . ππ
\(\therefore\) The mass of the slice = 2\(\pi\)\(π^2\)\(sin\theta\). ππ. \(\sigma\) = 2\(\pi\)\(\sigma\)\(π^2\)\(sin\theta\) . \(d\theta\)
(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

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