Mass Mis distributed uniformly over a disk of radius a. Find the

gravitational force (magnitude and direction) between this disk-shaped mass

and a particle with mass mlocated a distance xabove the centre of the disk

(Fig. P13.81). Does your result reduce to the correct expression as xbecomes

very large? (Hint:Divide the disk into infinitesimally thin concentric rings, use

the expression derived in Exercise 13.35 for the gravitational force due to each

ring, and integrate to find the total force.)

• The mass of the disk is,M.
• The radius of the disk is, a.
• The mass of the particle is,m.
• The distance of particle above the disk is,x.

Whenever two bodies come into contact, the attraction between them is equal

to product of respective masses and inversely proportional to square of their

distance from one another.

The gravitational force can be given by,

$F=\frac{GMm}{r^2}$… (i)

The radius of the disk is,r.

The thickness of the disk is,dr.

The area of each will be,$dA=2\mathrm{\pi }$.

The mass of the disk will be,

$$\begin{gathered} dM=\frac{M}{{\mathrm{\pi a}}^2}dA \\ dM=\frac{2M}{a^2}rdr \end{gathered}$$

Thus, the amount of the force exerted on mass by this little ring is,

$F=\left(GmdM\right)\left(\frac{x}{{\left(r^2+x^2\right)}^{{\displaystyle \frac{3}{2}}}}\right)$

In terms of force,is a significant factor,

$dF=\frac{2GMmx}{a^2}\frac{rdr}{{\left(x^2+r^2\right)}^{{\displaystyle \frac{3}{2}}}}$

The total force is evaluated over the range r ,

$F=\int dF=\frac{2GMmx}{a}\frac{rdr}{{\left(x^2+r^2\right)}^{{\displaystyle \frac{3}{2}}}}dr$

Substitute $u=r^2+a^2$in above equation,

${\int }_0^a\frac{r}{{\left(x^2+r^2\right)}^{{\displaystyle \frac{3}{2}}}}dr=\left[\frac{1}{x}-\frac{1}{\sqrt{a^2+x^2}}\right]=\frac{1}{x}\left[1-\frac{x}{\sqrt{a^2+x^2}}\right]$

Substitute $F=\frac{2GMm}{a^2}\left[1+\frac{x}{\sqrt{a^2+x^2}}\right]$. Here, the force on is towards the centre of

the ring. Therefore,

$$\begin{gathered} F=\frac{1}{\sqrt{1+{\left({\displaystyle \frac{a}{x}}\right)}^2}} \\ F={\left(1+{\left(\frac{a}{x}\right)}^2\right)}^{-\frac{1}{2}} \\ F\approx 1-\frac{1}{2}{\left(\frac{a}{x}\right)}^2 \end{gathered}$$

If $x>>a$,

$F\approx \frac{GMm}{x^2}$

Thus, the gravitational force between the disk-shaped mass and particle is,$\frac{GMm}{x^2}$.