A star of mass $M$ (equal to the solar mass) with a planet (much smaller than the star) revolves around the star in a circular orbit. The velocity of the star with respect to the center of mass of the star-planet system is shown below:
The radius of the planet's orbit is closest to $(1 A . U . = E a r t h - S u n d i s t a n c e)$ .

0.004 A . U .

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0.008 A . U .

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0.04 A . U .

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0.12 A . U .

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Solution

The correct option is D $0.04 A . U .$
From KEPLER'S III- Law we have suquare of thime period is directly propositional to cube of radius of orbit
Let $T_{1} =$ time period of plant
from graph
$T_{1} = 3$ days
$T_{2} =$ Time period of earth
$T_{3} = 365$
$r_{1} =$ radius of planent orbit
$r_{2} =$ radius of earth orbit
$r_{2} = 1 A . U$ (given)
By kepler's law we have
$\frac{(T_{1})^{2}}{(T_{2})^{2}} = \frac{(r_{1})^{3}}{(r_{2})^{3}}$
${(\frac{3}{365})}^{2} = \frac{(r_{1})^{3}}{(1 A . U)^{3}}$
${(\frac{3}{365})}^{\frac{2}{3}} = \frac{r_{1}}{1 A . U}$
$0.04 = \frac{r_{1}}{1 A . U}$
$r_{1} = 0.04 A . U$

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