Gravitational potential and inensity at a point on the axis of a circular disk or plate:
Let us consider a uniform circular disk of radius \( a \). Let there is a point \( P \) on the axis of the circular disc at a distance r from the centre \( O \) of the disc.
Now we want to calculate gravitational potential and gravitational intensity at the point \( P \) due to the disc or plate.
In order to do that, let’s draw an elementary strip or ring of radius \( x \) and thickness \( dx \) with the centre at \( O \) as shown in the adjoining Fig. 1.
The mass of that elementary strip or ring is \( 2\pi{x}\,dx\cdot\sigma \) where \( \sigma \) is the surface mass density of the disk.
Now the gravitational potential at \( P \) due to this elementary strip or ring is,
\( dv=-G\frac{ 2\pi{x}\,dx\cdot\sigma }{PQ} \\=-G\frac{ 2\pi{x}\,dx\cdot\sigma }{\sqrt{r^2+x^2}} \), where \( G \) is the gravitational constant.
Now, the gravitational potential at the point \( P \) due to the whole disc is,
\( V=-G2\pi\sigma\displaystyle\int_0^a\frac{x\,dx}{\sqrt{r^2+x^2}} \\=-G2\pi\sigma\left(\sqrt{r^2+x^2}\right)|_0^a \\=-2\pi{G}\sigma(\sqrt{r^2+x^2}-r) \)Solution of \( \displaystyle\int\frac{x\,dx}{\sqrt{r^2+x^2}} \) :
Put \( x^2+r^2=z^2, \\ or,\ 2x\,dx=2z\,dz\\or,\ x\,dx=z\,dz \)
\( \displaystyle\int\frac{x\,dx}{\sqrt{r^2+x^2}} \\=\displaystyle\int\frac{z\,dz}{z} \\=\displaystyle\int{dz} \\=z \\=\sqrt{r^2+x^2} \)
So the potential \( V =2\pi{G}\sigma(r-\sqrt{r^2+a^2}) \\=\pi{a^2}G\frac{2G}{a^2}(r-\sqrt{r^2+a^2}) \\or,\ V=\frac{2MG}{a^2}(r-\sqrt{r^2+a^2}) \)
where, \( M=\pi{a^2}\sigma \) is the total mass of the disc.
Gravitational Inensity at \( P \) is:
\( E=-\frac{dv}{dr} \\=-\frac{2MG}{a^2}\frac{d}{\,dr}(r-\sqrt{r^2+a^2}) \\= -\frac{2MG}{a^2}\left(1-\frac{2r}{2\sqrt{r^2+a^2}}\right) \)Or, \( E= -\frac{2MG}{a^2}\left(1-\frac{r}{\sqrt{r^2+a^2}}\right) \)
Hence the intensity along PO is \( E= \frac{2MG}{a^2}\left(1-\frac{r}{\sqrt{r^2+a^2}}\right) \)
The maximum intensity will be when \( r=0 \),
\( E_{max}=\frac{2MG}{a^2} \)