A huge circular arc of length 4.4 ; ly subten | vedclass

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(B) The radius $R$ of the circle is given by $R = \frac{\ell}{	heta}$,where $\ell$ is the arc length and $	heta$ is the angle in radians.Given $\ell

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$4.5 	imes 10^{10} \; s$

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(B) The radius $R$ of the circle is given by $R = \frac{\ell}{	heta}$,where $\ell$ is the arc length and $	heta$ is the angle in radians.Given $\ell = 4.4 \; ly = 4.4 	imes 9.46 	imes 10^{15} \; m$.The angle $	heta = 4'' = \frac{4}{3600} 	imes \frac{\pi}{180} \; rad$.Substituting these,$R = \frac{4.4 	imes 9.46 	imes 10^{15}}{\frac{4}{3600} 	imes \frac{\pi}{180}} \; m$.The circumference of the circle is $C = 2\pi R$.For $4$ revolutions,the total distance $D = 4 	imes 2\pi R = 8\pi R$.The speed $v = 8 \; AU/s = 8 	imes 1.5 	imes 10^{11} \; m/s$.The time taken $t = \frac{D}{v} = \frac{8\pi R}{v} = \frac{8\pi}{v} 	imes \frac{\ell}{	heta}$.Substituting the values: $t = \frac{8 	imes \pi 	imes 4.4 	imes 9.46 	imes 10^{15}}{8 	imes 1.5 	imes 10^{11} 	imes (\frac{4}{3600} 	imes \frac{\pi}{180})}$.Simplifying the expression,we get $t \approx 4.5 	imes 10^{10} \; s$.

$A$ huge circular arc of length $4.4 \; ly$ subtends an angle of $4''$ (arcseconds) at the centre of the circle. How long would it take for a body to complete $4$ revolutions if its speed is $8 \; AU/s$?
Given: $1 \; ly = 9.46 \times 10^{15} \; m$,$1 \; AU = 1.5 \times 10^{11} \; m$.

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$A$ huge circular arc of length $4.4 \; ly$ subtends an angle of $4''$ (arcseconds) at the centre of the circle. How long would it take for a body to complete $4$ revolutions if its speed is $8 \; AU/s$?Given: $1 \; ly = 9.46 \times 10^{15} \; m$,$1 \; AU = 1.5 \times 10^{11} \; m$.