Question

Calculate the gravitational potential at the centre of base of a solid hemisphere of mass $M$ and radius $R$.

• A. $-\frac{GM}{2R}$
• B. $-\frac{3GM}{2R}$
• C. $-\frac{GM}{3R}$
• D. $-\frac{GM}{4R}$

Answer 1

The correct option is B $-\frac{3GM}{2R}$

Consider a hemispherical shell of radius $r$ and thickness $dr$.

The mass of the element is,

$dm=\frac{M}{\text{Volume of the hemisphere}}\times \text{Volume of the element}$

$\Rightarrow dm=\frac{M}{\frac{2}{3}\pi R^3}\left(2\pi r^{2}dr\right)$

$\Rightarrow dm=\frac{3M{r}^{2}dr}{R^3}$

Since all points of this hemispherical shell are at the same distance $r$ from centre $\text{O}$, the potential at $\text{O}$ due to it is ,

$dV=-\frac{Gdm}{r}$

$\Rightarrow dV=-\frac{3GM{r}^{2}dr}{R^{3}r}$

$\Rightarrow dV=-\frac{3GMrdr}{R^3}$

Total potential at centre due to whole mass of the sphere,

$V={\int }_0^{R}dV$

$\Rightarrow V=-{\int }_0^R\frac{3GMrdr}{R^3}$

$\Rightarrow V=-\frac{3GM}{R^3}{\int }_0^{R}rdr$

$\Rightarrow V=-\frac{3GM}{R^3}{\left[\frac{r^2}{2}\right]}_0^R$

$\therefore V=\frac{-3GM}{2R}$

Hence, option (b) is the correct answer.

Why this question: To familiarize students with the calculation of potential energy for continuous mass distribution