A point mass m lies on the axis of a circular disk

A point mass $m$ lies on the axis of a circular disk of surface density $\mu$, at distance $a$ from the disk's center. The distance from the point to the edge of the disk is $R$. Find the attractive force between the disk and the point.

Step 1: Radius of the disk

Every point on the rim is at distance $R$ from $m$, and at (perpendicular) distance $a$ from the axial point's foot on the disk, so the disk's radius $r_0$ satisfies $r_0^2+a^2=R^2$, i.e. $r_0=\sqrt{R^2-a^2}$.

Step 2: Mass of a thin ring

A ring of radius $r$ and thickness $dr$ has area $2\pi r\,dr$ and mass $dm=2\pi\mu r\,dr$; every point on it is at distance $d=\sqrt{r^2+a^2}$ from $m$.

Step 3: Axial component of the ring's attraction

By Newton's law, $dF=G\dfrac{m\,dm}{d^2}$, directed toward the ring; by symmetry the sideways components cancel and only the axial component $dF\cos\theta$ survives, with $\cos\theta=a/d$:

\[dF_{\text{axial}}=G\frac{m\cdot2\pi\mu r\,dr}{r^2+a^2}\cdot\frac{a}{\sqrt{r^2+a^2}}=\frac{2\pi Gm\mu a\,r\,dr}{(r^2+a^2)^{3/2}}.\]

Step 4: Integrate over the disk

\[F=2\pi Gm\mu a\int_0^{r_0}\frac{r\,dr}{(r^2+a^2)^{3/2}}.\]

With $u=r^2+a^2$: at $r=0$, $u=a^2$; at $r=r_0=\sqrt{R^2-a^2}$, $u=R^2$. So

\[F=\pi Gm\mu a\int_{a^2}^{R^2}u^{-3/2}\,du=\pi Gm\mu a\Big[-2u^{-1/2}\Big]_{a^2}^{R^2}=2\pi Gm\mu a\left(\frac{1}{a}-\frac{1}{R}\right).\]

Step 5: Simplify

\[\boxed{F=2\pi Gm\mu\left(1-\frac{a}{R}\right)}\]

Step 6: Limiting checks

As $a\to0$ (point pressed against the disk's center), $F\to2\pi Gm\mu$, the attraction of an infinite plane sheet, which is sensible since the disk then fills the point's entire field of view. As $a\to R$ (disk radius $r_0\to0$, i.e. the disk shrinks to a point at distance $a$), $F\to0$, consistent with a vanishing source mass.