# 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$$sin\theta$$ . πππ = 2$$\pi$$$$π^2$$ $$sin\theta$$ . ππ $$\therefore$$ The mass of the slice = 2$$\pi$$$$π^2$$$$sin\theta$$. ππ. $$\sigma$$ = 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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