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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
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 gravitational potential of the disc of mass M and radius R on the axis at a distance x from the center of the disc.
$$\mathrm{Gravitational}\:\mathrm{potential}\:\mathrm{of}\:\mathrm{a}\:\mathrm{ring}\:\mathrm{of} \\ $$$$\mathrm{radius}\:{R}\:\mathrm{and}\:\mathrm{mass}\:{M}\:\mathrm{on}\:\mathrm{the}\:\mathrm{axis}\:\mathrm{at}\:\mathrm{a} \\ $$$$\mathrm{distance}\:{x}\:\mathrm{from}\:\mathrm{the}\:\mathrm{center}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by} \\ $$$${v}\left({x}\right)\:=\:−\:\frac{{GM}}{\:\sqrt{{R}^{\mathrm{2}} \:+\:{x}^{\mathrm{2}} }}\:\mathrm{Nm}/\mathrm{kg} \\ $$$$\mathrm{Using}\:\mathrm{the}\:\mathrm{above}\:\mathrm{expression}\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{gravitational}\:\mathrm{potential}\:\mathrm{of}\:\mathrm{the}\:\mathrm{disc}\:\mathrm{of} \\ $$$$\mathrm{mass}\:{M}\:\mathrm{and}\:\mathrm{radius}\:{R}\:\mathrm{on}\:\mathrm{the}\:\mathrm{axis}\:\mathrm{at}\:\mathrm{a} \\ $$$$\mathrm{distance}\:{x}\:\mathrm{from}\:\mathrm{the}\:\mathrm{center}\:\mathrm{of}\:\mathrm{the}\:\mathrm{disc}. \\ $$
Commented by Tinkutara last updated on 07/Jun/17
Commented by Tinkutara last updated on 07/Jun/17
But answer is ((−2M)/R^2 )[(√(R^2 + x^2 )) − x]
$$\mathrm{But}\:\mathrm{answer}\:\mathrm{is}\:\frac{−\mathrm{2}{M}}{{R}^{\mathrm{2}} }\left[\sqrt{{R}^{\mathrm{2}} \:+\:{x}^{\mathrm{2}} }\:−\:{x}\right] \\ $$
Commented by mrW1 last updated on 07/Jun/17
((−2GM)/R^2 )[(√(R^2 + x^2 )) − x] =((−2GM)/R^2 )[(√(R^2 + x^2 )) − x]×(([(√(R^2 +x^2 ))+x])/( (√(R^2 +x^2 ))+x)) =((−2GM)/R^2 )[R^2 + x^2 − x^2 ]×(1/( (√(R^2 +x^2 ))+x)) =−((2GM)/( (√(R^2 +x^2 ))+x))
$$\frac{−\mathrm{2G}{M}}{{R}^{\mathrm{2}} }\left[\sqrt{{R}^{\mathrm{2}} \:+\:{x}^{\mathrm{2}} }\:−\:{x}\right] \\ $$$$\:=\frac{−\mathrm{2G}{M}}{{R}^{\mathrm{2}} }\left[\sqrt{{R}^{\mathrm{2}} \:+\:{x}^{\mathrm{2}} }\:−\:{x}\right]×\frac{\left[\sqrt{\mathrm{R}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }+\mathrm{x}\right]}{\:\sqrt{\mathrm{R}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }+\mathrm{x}} \\ $$$$\:=\frac{−\mathrm{2G}{M}}{{R}^{\mathrm{2}} }\left[{R}^{\mathrm{2}} \:+\:{x}^{\mathrm{2}} \:−\:{x}^{\mathrm{2}} \right]×\frac{\mathrm{1}}{\:\sqrt{\mathrm{R}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }+\mathrm{x}} \\ $$$$\:=−\frac{\mathrm{2GM}}{\:\sqrt{\mathrm{R}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }+\mathrm{x}} \\ $$
Commented by Tinkutara last updated on 07/Jun/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$
Answered by ajfour last updated on 07/Jun/17
dv=−((GdM)/( (√(r^2 +x^2 ))))=−((G(σ)(2πrdr))/( (√(r^2 +x^2 )))) v =−πσG∫_0 ^( R) (((2rdr)/( (√(r^2 +x^2 ))))) =−πσG[2(√(r^2 +x^2 )) ]_(r=0) ^(r=R) =−2πσG((√(R^2 +x^2 ))−R) =−2πG((M/(πR^2 )))((√(R^2 +x^2 ))−R) v(x)_(disc) =−((2GM)/R^2 )((√(R^2 −x^2 ))−R) .
$$ \\ $$$${dv}=−\frac{{GdM}}{\:\sqrt{{r}^{\mathrm{2}} +{x}^{\mathrm{2}} }}=−\frac{{G}\left(\sigma\right)\left(\mathrm{2}\pi{rdr}\right)}{\:\sqrt{{r}^{\mathrm{2}} +{x}^{\mathrm{2}} }} \\ $$$$\:\:\:{v}\:=−\pi\sigma{G}\int_{\mathrm{0}} ^{\:\:{R}} \left(\frac{\mathrm{2}{rdr}}{\:\sqrt{{r}^{\mathrm{2}} +{x}^{\mathrm{2}} }}\right) \\ $$$$\:\:\:\:\:\:=−\pi\sigma{G}\left[\mathrm{2}\sqrt{{r}^{\mathrm{2}} +{x}^{\mathrm{2}} }\:\right]_{{r}=\mathrm{0}} ^{{r}={R}} \\ $$$$\:\:\:\:\:\:=−\mathrm{2}\pi\sigma{G}\left(\sqrt{{R}^{\mathrm{2}} +{x}^{\mathrm{2}} }−{R}\right)\: \\ $$$$\:\:\:\:\:\:=−\mathrm{2}\pi{G}\left(\frac{{M}}{\pi{R}^{\mathrm{2}} }\right)\left(\sqrt{{R}^{\mathrm{2}} +{x}^{\mathrm{2}} }−{R}\right) \\ $$$$\:\:\:{v}\left({x}\right)_{{disc}} \:=−\frac{\mathrm{2}{GM}}{{R}^{\mathrm{2}} }\left(\sqrt{{R}^{\mathrm{2}} −{x}^{\mathrm{2}} }−{R}\right)\:. \\ $$
Commented by Tinkutara last updated on 07/Jun/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$
Answered by mrW1 last updated on 07/Jun/17
m=(M/(πR^2 )) [kg/m^2 ] v_d (x)=−G∫_0 ^R ((m2πrdr)/( (√(r^2 +x^2 ))))=−((GM)/R^2 )∫_0 ^R ((2rdr)/( (√(r^2 +x^2 )))) =−((GM)/R^2 )∫_0 ^R (dr^2 /( (√(r^2 +x^2 )))) =−((2GM)/R^2 )[(√(r^2 +x^2 ))]_0 ^R =−((2GM)/R^2 )[(√(R^2 +x^2 ))−x] =−((2GM)/R^2 )[(√(R^2 +x^2 ))−x]×(([(√(R^2 +x^2 ))+x])/([(√(R^2 +x^2 ))+x])) =−((2GM)/(x+(√(R^2 +x^2 ))))
$$\mathrm{m}=\frac{\mathrm{M}}{\pi\mathrm{R}^{\mathrm{2}} }\:\left[\mathrm{kg}/\mathrm{m}^{\mathrm{2}} \right] \\ $$$$\mathrm{v}_{\mathrm{d}} \left(\mathrm{x}\right)=−\mathrm{G}\int_{\mathrm{0}} ^{\mathrm{R}} \:\frac{\mathrm{m2}\pi\mathrm{rdr}}{\:\sqrt{\mathrm{r}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }}=−\frac{\mathrm{GM}}{\mathrm{R}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\mathrm{R}} \frac{\mathrm{2rdr}}{\:\sqrt{\mathrm{r}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }} \\ $$$$=−\frac{\mathrm{GM}}{\mathrm{R}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\mathrm{R}} \frac{\mathrm{dr}^{\mathrm{2}} }{\:\sqrt{\mathrm{r}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }} \\ $$$$=−\frac{\mathrm{2GM}}{\mathrm{R}^{\mathrm{2}} }\left[\sqrt{\mathrm{r}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\mathrm{R}} \\ $$$$=−\frac{\mathrm{2GM}}{\mathrm{R}^{\mathrm{2}} }\left[\sqrt{\mathrm{R}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }−\mathrm{x}\right] \\ $$$$=−\frac{\mathrm{2GM}}{\mathrm{R}^{\mathrm{2}} }\left[\sqrt{\mathrm{R}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }−\mathrm{x}\right]×\frac{\left[\sqrt{\mathrm{R}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }+\mathrm{x}\right]}{\left[\sqrt{\mathrm{R}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }+\mathrm{x}\right]} \\ $$$$=−\frac{\mathrm{2GM}}{\mathrm{x}+\sqrt{\mathrm{R}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }} \\ $$
Commented by mrW1 last updated on 07/Jun/17
I write the formula in this form to make its physical meaning more clear: the distance to the center of disc is x, the distance to the edge of disc is (√(R^2 +x^2 )) the average distance to disc is d_m =((x+(√(R^2 +x^2 )))/2) v_d (x)=−((GM)/d_m )=−((2GM)/(x+(√(R^2 +x^2 ))))
$$\mathrm{I}\:\mathrm{write}\:\mathrm{the}\:\mathrm{formula}\:\mathrm{in}\:\mathrm{this}\:\mathrm{form}\:\mathrm{to} \\ $$$$\mathrm{make}\:\mathrm{its}\:\mathrm{physical}\:\mathrm{meaning}\:\mathrm{more}\:\mathrm{clear}: \\ $$$$\mathrm{the}\:\mathrm{distance}\:\mathrm{to}\:\mathrm{the}\:\mathrm{center}\:\mathrm{of} \\ $$$$\mathrm{disc}\:\mathrm{is}\:\mathrm{x},\:\mathrm{the}\:\mathrm{distance}\:\mathrm{to}\:\mathrm{the}\:\mathrm{edge}\:\mathrm{of} \\ $$$$\mathrm{disc}\:\mathrm{is}\:\sqrt{\mathrm{R}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{the}\:\mathrm{average}\:\mathrm{distance}\:\mathrm{to}\:\mathrm{disc}\:\mathrm{is}\: \\ $$$$\mathrm{d}_{\mathrm{m}} =\frac{\mathrm{x}+\sqrt{\mathrm{R}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\mathrm{v}_{\mathrm{d}} \left(\mathrm{x}\right)=−\frac{\mathrm{GM}}{\mathrm{d}_{\mathrm{m}} }=−\frac{\mathrm{2GM}}{\mathrm{x}+\sqrt{\mathrm{R}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }} \\ $$

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