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(A) According to Kepler's third law of planetary motion, the square of the time period $T$ is proportional to the cube of the orbital radius $r$:$T^2

Vedclass · Accepted Answer

$2^{-2 / 3} R$

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(A) According to Kepler's third law of planetary motion, the square of the time period $T$ is proportional to the cube of the orbital radius $r$:$T^2 \propto r^3$Since angular speed $\omega = \frac{2\pi}{T}$, we have $T = \frac{2\pi}{\omega}$. Substituting this into the law:$(\frac{2\pi}{\omega})^2 \propto r^3 \Rightarrow \frac{1}{\omega^2} \propto r^3 \Rightarrow r^3 \omega^2 = 	ext{constant}$.For the earth $(E)$ and the planet $(P)$:$r_E^3 \omega_E^2 = r_P^3 \omega_P^2$Given $r_E = R$ and $\omega_P = 2\omega_E$:$R^3 \omega_E^2 = r_P^3 (2\omega_E)^2$$R^3 \omega_E^2 = r_P^3 (4\omega_E^2)$$R^3 = 4 r_P^3$$r_P^3 = \frac{R^3}{4} = \frac{R^3}{2^2}$Taking the cube root on both sides:$r_P = \frac{R}{2^{2/3}} = 2^{-2/3} R$.

Assume that the earth moves around the sun in a circular orbit of radius $R$ and there exists a planet which also moves around the sun in a circular orbit with an angular speed twice as large as that of the earth. The radius of the orbit of the planet is

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