Gravitational-potential-of-a-ring-of-radius-R-and-mass-M-on-the-axis-at-a-distance-x-from-the-center-is-given-by-v-x-GM-R-2-x-2-Nm-kg-Using-the-above-expression-find-the-gravitationa

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Question Number 15052 by Tinkutara last updated on 07/Jun/17

$$\mathrm{Gravitational}\mathrm{potential}\mathrm{of}\mathrm{a}\mathrm{ring}\mathrm{of}$$$$\mathrm{radius}R\mathrm{and}\mathrm{mass}M\mathrm{on}\mathrm{the}\mathrm{axis}\mathrm{at}\mathrm{a}$$$$\mathrm{distance}x\mathrm{from}\mathrm{the}\mathrm{center}\mathrm{is}\mathrm{given}\mathrm{by}$$$$v\left(x\right)=-\frac{GM}{\sqrt{R^2+x^2}}\mathrm{Nm}/\mathrm{kg}$$$$\mathrm{Using}\mathrm{the}\mathrm{above}\mathrm{expression}\mathrm{find}\mathrm{the}$$$$\mathrm{gravitational}\mathrm{potential}\mathrm{of}\mathrm{the}\mathrm{disc}\mathrm{of}$$$$\mathrm{mass}M\mathrm{and}\mathrm{radius}R\mathrm{on}\mathrm{the}\mathrm{axis}\mathrm{at}\mathrm{a}$$$$\mathrm{distance}x\mathrm{from}\mathrm{the}\mathrm{center}\mathrm{of}\mathrm{the}\mathrm{disc}.$$

Commented by Tinkutara last updated on 07/Jun/17

Commented by Tinkutara last updated on 07/Jun/17

$$\mathrm{But}\mathrm{answer}\mathrm{is}\frac{-2M}{R^2}[\sqrt{R^2+x^2}-x]$$

Commented by mrW1 last updated on 07/Jun/17

$$\frac{-2\mathrm{G}M}{R^2}[\sqrt{R^2+x^2}-x]$$$$=\frac{-2\mathrm{G}M}{R^2}[\sqrt{R^2+x^2}-x]\times \frac{[\sqrt{{\mathrm{R}}^2+{\mathrm{x}}^2}+\mathrm{x}]}{\sqrt{{\mathrm{R}}^2+{\mathrm{x}}^2}+\mathrm{x}}$$$$=\frac{-2\mathrm{G}M}{R^2}[R^2+x^2-x^2]\times \frac{1}{\sqrt{{\mathrm{R}}^2+{\mathrm{x}}^2}+\mathrm{x}}$$$$=-\frac{2\mathrm{GM}}{\sqrt{{\mathrm{R}}^2+{\mathrm{x}}^2}+\mathrm{x}}$$

Commented by Tinkutara last updated on 07/Jun/17

$$\mathrm{Thanks}\mathrm{Sir}!$$

Answered by ajfour last updated on 07/Jun/17

$$$$$$dv=-\frac{GdM}{\sqrt{r^2+x^2}}=-\frac{G\left(\sigma \right)\left(2\pi rdr\right)}{\sqrt{r^2+x^2}}$$$$v=-\pi \sigma G{\int }_0^R\left(\frac{2rdr}{\sqrt{r^2+x^2}}\right)$$$$=-\pi \sigma G{\left[2\sqrt{r^2+x^2}\right]}_{r=0}^{r=R}$$$$=-2\pi \sigma G(\sqrt{R^2+x^2}-R)$$$$=-2\pi G\left(\frac{M}{\pi R^2}\right)(\sqrt{R^2+x^2}-R)$$$$v{\left(x\right)}_{disc}=-\frac{2GM}{R^2}(\sqrt{R^2-x^2}-R).$$

Commented by Tinkutara last updated on 07/Jun/17

$$\mathrm{Thanks}\mathrm{Sir}!$$

Answered by mrW1 last updated on 07/Jun/17

$$\mathrm{m}=\frac{\mathrm{M}}{\pi {\mathrm{R}}^2}\left[\mathrm{kg}/{\mathrm{m}}^2\right]$$$${\mathrm{v}}_{\mathrm{d}}\left(\mathrm{x}\right)=-\mathrm{G}{\int }_0^{\mathrm{R}}\frac{\mathrm{m}2\pi \mathrm{rdr}}{\sqrt{{\mathrm{r}}^2+{\mathrm{x}}^2}}=-\frac{\mathrm{GM}}{{\mathrm{R}}^2}{\int }_0^{\mathrm{R}}\frac{2\mathrm{rdr}}{\sqrt{{\mathrm{r}}^2+{\mathrm{x}}^2}}$$$$=-\frac{\mathrm{GM}}{{\mathrm{R}}^2}{\int }_0^{\mathrm{R}}\frac{{\mathrm{dr}}^2}{\sqrt{{\mathrm{r}}^2+{\mathrm{x}}^2}}$$$$=-\frac{2\mathrm{GM}}{{\mathrm{R}}^2}{\left[\sqrt{{\mathrm{r}}^2+{\mathrm{x}}^2}\right]}_0^{\mathrm{R}}$$$$=-\frac{2\mathrm{GM}}{{\mathrm{R}}^2}[\sqrt{{\mathrm{R}}^2+{\mathrm{x}}^2}-\mathrm{x}]$$$$=-\frac{2\mathrm{GM}}{{\mathrm{R}}^2}[\sqrt{{\mathrm{R}}^2+{\mathrm{x}}^2}-\mathrm{x}]\times \frac{[\sqrt{{\mathrm{R}}^2+{\mathrm{x}}^2}+\mathrm{x}]}{[\sqrt{{\mathrm{R}}^2+{\mathrm{x}}^2}+\mathrm{x}]}$$$$=-\frac{2\mathrm{GM}}{\mathrm{x}+\sqrt{{\mathrm{R}}^2+{\mathrm{x}}^2}}$$

Commented by mrW1 last updated on 07/Jun/17

$$\mathrm{I}\mathrm{write}\mathrm{the}\mathrm{formula}\mathrm{in}\mathrm{this}\mathrm{form}\mathrm{to}$$$$\mathrm{make}\mathrm{its}\mathrm{physical}\mathrm{meaning}\mathrm{more}\mathrm{clear}:$$$$\mathrm{the}\mathrm{distance}\mathrm{to}\mathrm{the}\mathrm{center}\mathrm{of}$$$$\mathrm{disc}\mathrm{is}\mathrm{x},\mathrm{the}\mathrm{distance}\mathrm{to}\mathrm{the}\mathrm{edge}\mathrm{of}$$$$\mathrm{disc}\mathrm{is}\sqrt{{\mathrm{R}}^2+{\mathrm{x}}^2}$$$$\mathrm{the}\mathrm{average}\mathrm{distance}\mathrm{to}\mathrm{disc}\mathrm{is}$$$${\mathrm{d}}_{\mathrm{m}}=\frac{\mathrm{x}+\sqrt{{\mathrm{R}}^2+{\mathrm{x}}^2}}{2}$$$${\mathrm{v}}_{\mathrm{d}}\left(\mathrm{x}\right)=-\frac{\mathrm{GM}}{{\mathrm{d}}_{\mathrm{m}}}=-\frac{2\mathrm{GM}}{\mathrm{x}+\sqrt{{\mathrm{R}}^2+{\mathrm{x}}^2}}$$

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