A body revolves around the sun 27 times faste | vedclass

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(B) According to Kepler's third law,the square of the orbital period $T$ is proportional to the cube of the semi-major axis $r$: $T^2 \propto r^3$.Sin

Vedclass · Accepted Answer

$1/9$

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(B) According to Kepler's third law,the square of the orbital period $T$ is proportional to the cube of the semi-major axis $r$: $T^2 \propto r^3$.Since angular velocity $\omega = 2\pi / T$,we have $T = 2\pi / \omega$,which implies $T^2 \propto 1/\omega^2$.Substituting this into Kepler's law: $1/\omega^2 \propto r^3$,which gives $\omega^2 \propto 1/r^3$ or $\omega \propto r^{-3/2}$.Rearranging for $r$,we get $r \propto \omega^{-2/3}$.Given that the body revolves $27$ times faster than the earth,$\omega_{	ext{body}} = 27 \omega_{	ext{earth}}$.The ratio of the radii is $\frac{r_{	ext{body}}}{r_{	ext{earth}}} = \left( \frac{\omega_{	ext{earth}}}{\omega_{	ext{body}}} ight)^{2/3}$.Substituting the values: $\frac{r_{	ext{body}}}{r_{	ext{earth}}} = \left( \frac{1}{27} ight)^{2/3} = \left( (3^3)^{-1} ight)^{2/3} = (3^{-3})^{2/3} = 3^{-2} = \frac{1}{9}$.

$A$ body revolves around the sun $27$ times faster than the earth. What is the ratio of their orbital radii?

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