Find The Distance Of A Point On The Axis Of A Uniform Ring From Its Centre Where The Gravitational Attraction Is Maximum.

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Let us consider a uniform ring of radius \( a \) with centre at \( O \). Let a point \( P \) is on the axis of the ring, at a distance \( r \) from the centre of the ring.

So the Gravitational Intensity at the point \( P \) due to the ring is,

\( \displaystyle{E=\frac{GMr}{(r^2+a^2)^{3/2}}} \), where, \( M \) is the total mass of the ring, \( G \) is the gravitational constant. READ IN DETAIL.

Fig. 1

Now we want to find the point on the axis where the gravitational force is maximum, i.e, \( \frac{dE}{\,dr}=0 \).

\( \frac{d}{dr}\left[ \frac{GMr}{(r^2+a^2)^{3/2}} \right] =0 \)

or, \( \frac{ (r^2+a^2)^{3/2} -r\frac{3}{2} (r^2+a^2)^{1/2}2r}{ (r^2+a^2)^3 }=0 \)

or, \( (r^2+a^2)-3r^2=0 \\or,\ a^2=2r^2 \)

or, \( \displaystyle{r=\pm\frac{a}{\sqrt{2}}} \)

This the position on the axis on the ring, where the gravitational attraction is maximum.

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