Physics - Gravitation Question with Solution | TestHub

$\mathit{\text{W}}=\Delta \mathit{\text{U}}={\mathit{\text{U}}}_{\mathit{\text{f}}}-{\mathit{\text{U}}}_{\mathit{\text{i}}}={\mathit{\text{U}}}_{\infty }-{\mathit{\text{U}}}_{\mathit{\text{P}}}$

$=-{\mathit{\text{U}}}_{\mathit{\text{P}}}=-\mathit{\text{m}}{\mathit{\text{V}}}_{\mathit{\text{P}}}=-{\mathit{\text{V}}}_{\mathit{\text{P}}}\left(\text{as}\mathit{\text{m}}=1\right)$

Upon integration, potential at point P will be obtained as given below

Let dM be the mass of small ring as shown

$\mathit{\text{dM}}=\frac{\mathit{\text{M}}}{\pi {\left(4\mathit{\text{R}}\right)}^2-\pi {\left(3\mathit{\text{R}}\right)}^2}\left(2\pi \mathit{\text{r}}\right)\mathit{\text{dr}}=\frac{2\mathit{\text{Mr}}\mathit{\text{dr}}}{7{\mathit{\text{R}}}^2}$

${\mathit{\text{dV}}}_{\text{P}}=-\frac{\mathit{\text{G}}\mathit{\text{d}}\mathit{\text{M}}}{\sqrt{16{\mathit{\text{R}}}^2+{\text{r}}^2}}$



$=-\frac{2\mathit{\text{GM}}}{7{\mathit{\text{R}}}^2}\underset{3\mathit{\text{R}}}{\overset{4\mathit{\text{R}}}{\int }}\frac{\mathit{\text{r}}}{\sqrt{16{\mathit{\text{R}}}^2+{\mathit{\text{r}}}^2}}\mathit{\text{dr}}$

$=-\genfrac{}{}{0.1ex}{}{2GM}{7R}(4\sqrt{2}-5)$

$\therefore \mathit{\text{W}}=+\frac{2\mathit{\text{GM}}}{7\mathit{\text{R}}}\left(4\sqrt{2}-5\right)$$$