Consider a ring of mass m and radius r. The m | vedclass

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(C) The gravitational intensity $E$ at a distance $x$ from the center of a ring of mass $m$ and radius $r$ along its axis is given by:$E = \frac{Gmx}{

Vedclass · Accepted Answer

$\frac{2Gm}{3\sqrt{3}r^2}$

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(C) The gravitational intensity $E$ at a distance $x$ from the center of a ring of mass $m$ and radius $r$ along its axis is given by:$E = \frac{Gmx}{(r^2 + x^2)^{3/2}}$To find the maximum intensity,we differentiate $E$ with respect to $x$ and set it to zero:$\frac{dE}{dx} = Gm \left[ \frac{(r^2 + x^2)^{3/2} - x \cdot \frac{3}{2}(r^2 + x^2)^{1/2} \cdot 2x}{(r^2 + x^2)^3} ight] = 0$$(r^2 + x^2)^{3/2} - 3x^2(r^2 + x^2)^{1/2} = 0$$r^2 + x^2 = 3x^2 \implies 2x^2 = r^2 \implies x = \frac{r}{\sqrt{2}}$Substituting $x = \frac{r}{\sqrt{2}}$ back into the expression for $E$:$E_{	ext{max}} = \frac{Gm(r/\sqrt{2})}{(r^2 + r^2/2)^{3/2}} = \frac{Gmr/\sqrt{2}}{(3r^2/2)^{3/2}} = \frac{Gmr/\sqrt{2}}{(3\sqrt{3}r^3)/(2\sqrt{2})} = \frac{Gmr}{\sqrt{2}} \cdot \frac{2\sqrt{2}}{3\sqrt{3}r^3} = \frac{2Gm}{3\sqrt{3}r^2}$

Consider a ring of mass $m$ and radius $r$. The maximum gravitational intensity on the axis of the ring has a value of:

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