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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. Earth-Sun distance)
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The radius of the planet's orbit is closest to (1 A. U. Earth-Sun distance)
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The correct answer is:
$0.004$ A.U.
$\mathrm{T}^{2}=\frac{4 \pi^{2}}{\mathrm{GM}} \mathrm{a}^{2}$
$\frac{4 \pi^{2}}{\mathrm{GM}}=1$
$\mathrm{~T}=\mathrm{in}$ year
$\mathrm{a}=$ radius in A.U.
$\therefore \mathrm{T}=3$ days $=\frac{3}{365}$ year
$\therefore \mathrm{a}=\left(\frac{3}{365}\right)^{2 / 3}$
$\mathrm{a}=0.04 \mathrm{A.U}$
$\frac{4 \pi^{2}}{\mathrm{GM}}=1$
$\mathrm{~T}=\mathrm{in}$ year
$\mathrm{a}=$ radius in A.U.
$\therefore \mathrm{T}=3$ days $=\frac{3}{365}$ year
$\therefore \mathrm{a}=\left(\frac{3}{365}\right)^{2 / 3}$
$\mathrm{a}=0.04 \mathrm{A.U}$
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