Question

Calculate the gravitational field intensity and potential at the centre of the base of a solid hemisphere of mass m, radius R.

Answer 1

We consider the shaded elemental disc of radius R$\sin \theta$ and thickness $Rd\theta$
It mass,
$dM=\frac{M}{\frac{2}{3}\pi R^3}\pi \left(R\sin \theta {)}^2\right(Rd\theta \sin \theta )$
or $dM=\frac{3M}{2}{\sin }^3\theta d\theta$
Field due to this plate at O,
$dE=\frac{2GdM(1-\cos \theta )}{(R\sin \theta {)}^2}$ (see field due to a uniform disc)
or $dE=\frac{3GM\sin \theta (1-\cos \theta )d\theta }{R^2}$
Therefore, $E={\int }_0^{\frac{\pi }{2}}dE={\int }_0^{\frac{\pi }{2}}\frac{3GM\sin \theta (1-\cos \theta )}{R^2}d\theta$
$=\frac{3GM}{R^2}{[-\cos \theta +\frac{{\cos }^2\theta }{2}]}_0^{\frac{\pi }{2}}$
or $E=\frac{3GM}{2R^2}$
Now potential due to the element under consideration at the centre of the base of the hemisphere,
$dV=-2GdM/r(cosec\theta -\cot \theta )$ (see potential due to a circular plate)
or $dV=\frac{3GM{\sin }^3\theta (cosec\theta -\cot \theta )d\theta }{\left(R\sin \theta \right)}$
Therefore, $V=-\frac{3GM}{R}{\int }_0^{\frac{\pi }{2}}(\sin \theta -\cos \theta \sin \theta )d\theta$
$=-\frac{3GM}{R}{[-\cos \theta +\frac{{\cos }^2\theta }{2}]}_0^{\frac{\pi }{2}}$
or $V=\frac{-3GM}{2R}$.