A point of mass is at a distance x from the centre of a ring of mass M and radius R on it s axis as shown
in figure. Find the gravitational force between the two. What will this force be if x `gt gtR and "x"lt lt R?` For what value of x is the maximum ?

Explanation

Considering a small element of the ring (as shown in figure) and treating it as a point mass, by Newton.s
law of gravitation, force on the particle of mass m due to this element,
`"dF"=("GmdM")/(r^(2))=("Gm"(M//2 piR)dl)/(r^(2))" "["as dM"=(M)/(2pi R)dl]`
Now due to symmetry of the problem the component of forces perpendicular to the axis due to all elements
will cancel each other, i.e.,
`F_(y)= oint " d F"_(y)=0`
while components along x-axis will add up so that
`F= oint " d F"cos theta` along PO
`i.e.,F=("GmM")/(2 pi R)oint" "(dl)/(r^(2))("x"/(r))" "["as" cos theta=("x")/(r)]`
Now as for a given point and ring x and r `[=(R^(2)-:"x"^(2))^(1//2)]` will be the same for all elements
`F=("GmMx")/(2 piR(R^(2)+"x"^(2))^(3//2))oint dl=("GmMx")/((R^(2)+"x"^(2))^(3//2))" "["as"oint dl=2 pi R]`

Now if `"x"gt gt R,F=("GmM"//x^(2)),` then for a distant point ring behaves as a point mass (which is expected)
and for `"x"lt lt R,F=("GmM"//R^(3))"x",i.e.,`
force varies linearly with distance x. Furthermore F will be maximum when
`("dF"/("dx"))=0`
i.e., `(d)/("dx")[("GmMx")/((R^(2)+"x"^(2))^(3//2))]=0` " or"`(d)/("dx")["x"(R^(2)+"x"^(2))^(-3//2)]=0`" "`["as GmM"ne 0]`
i.e., `[(R^(2)+"x"^(2))^(-3//2)-3"x"^(2)(R^(2)+"x"^(2))^(-5//2)]=0`
or `"x"=pm(R)/sqrt(2)" "["as"(R^(2)+"x"^(2))^(-5//2)\ne0]`
Substituting this value of x in the expression for force F,
`F_(max)=((2"GMm")/(3 sqrt 3 R^(2)))`