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(A) The surface mass density $\sigma$ of the annular disc is $\sigma = \frac{M}{\pi(4R)^2 - \pi(3R)^2} = \frac{M}{7\pi R^2}$.Consider a thin ring elem

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$\frac{2 GM}{7 R}(4 \sqrt{2}-5)$

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(A) The surface mass density $\sigma$ of the annular disc is $\sigma = \frac{M}{\pi(4R)^2 - \pi(3R)^2} = \frac{M}{7\pi R^2}$.Consider a thin ring element of radius $r$ and width $dr$. The mass of this element is $dm = \sigma (2\pi r dr)$.The gravitational potential $V_P$ at point $P$ (at distance $h=4R$ from the center) due to this ring is $dV_P = -\frac{G dm}{\sqrt{r^2 + h^2}} = -\frac{G \sigma 2\pi r dr}{\sqrt{r^2 + (4R)^2}}$.Integrating from $r=3R$ to $r=4R$:$V_P = -\int_{3R}^{4R} \frac{G \sigma 2\pi r dr}{\sqrt{r^2 + 16R^2}} = -2\pi G \sigma [\sqrt{r^2 + 16R^2}]_{3R}^{4R}$.$V_P = -2\pi G \left(\frac{M}{7\pi R^2}\right) [\sqrt{16R^2 + 16R^2} - \sqrt{9R^2 + 16R^2}] = -\frac{2GM}{7R^2} [4R\sqrt{2} - 5R] = -\frac{2GM}{7R}(4\sqrt{2}-5)$.The work required to move a unit mass from $P$ to infinity is $W = U_{\infty} - U_P = 1 \cdot V_{\infty} - 1 \cdot V_P = 0 - V_P = -V_P$.Therefore,$W = \frac{2GM}{7R}(4\sqrt{2}-5)$.

$A$ thin uniform annular disc (see figure) of mass $M$ has outer radius $4 R$ and inner radius $3 R$. The work required to take a unit mass from point $P$ on its axis to infinity is

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