A star of mass M (equal to the solar mass) wi | vedclass

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(C) From the given graph,the time period of rotation of the planet around the star is $T = 3 	ext{ days}$.According to Kepler's third law,the relatio

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$0.04$

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(C) From the given graph,the time period of rotation of the planet around the star is $T = 3 	ext{ days}$.According to Kepler's third law,the relationship between the orbital period $T$ and the orbital radius $r$ is given by:$T^{2} = \frac{4 \pi^{2}}{G M} \cdot r^{3} \quad \dots(i)$If we express the time period in years and the radius in $AU$,then for the Earth ($T = 1 	ext{ yr}$,$r = 1 	ext{ AU}$),the constant $\frac{4 \pi^{2}}{G M} = 1$.Given $T = 3 	ext{ days} = \frac{3}{365} 	ext{ yr}$,we substitute this into equation $(i)$:$r^{3} = T^{2}$$r = T^{2/3}$$r = \left(\frac{3}{365}ight)^{2/3} \approx \left(\frac{1}{121.67}ight)^{2/3} \approx \left(0.0082ight)^{2/3} \approx 0.04 	ext{ AU}$.Thus,the radius of the planet's orbit is approximately $0.04 	ext{ AU}$.

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