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(A) For a stationary satellite,its orbital period $T$ must be equal to the rotational period of the planet about its axis. The angular velocity of the

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$2/3$

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(A) For a stationary satellite,its orbital period $T$ must be equal to the rotational period of the planet about its axis. The angular velocity of the planet is $\omega = 2\pi / T$. Since the angular velocity is halved $(\omega' = \omega / 2)$,the new period $T'$ becomes $2T$ because $T = 2\pi / \omega$. According to Kepler's Third Law,the square of the orbital period is proportional to the cube of the orbital radius: $T^2 \propto r^3$. Therefore,$(T'/T)^2 = (r'/r)^3$. Substituting $T' = 2T$,we get $(2)^2 = (r'/r)^3$,which implies $4 = (r'/r)^3$. Taking the cube root on both sides,$r'/r = 4^{1/3} = (2^2)^{1/3} = 2^{2/3}$. Comparing this with $r'/r = 2^n$,we find $n = 2/3$.

If the angular velocity of a planet about its axis is halved,the distance of the stationary satellite of this planet from the centre of the planet becomes $2^{n}$ times the initial distance. Then the value of '$n$' is

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