Two planets are at mean distances d_1 and d_2 | vedclass

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(B) According to Kepler's Third Law of Planetary Motion,the square of the time period $T$ is proportional to the cube of the mean distance $d$ from th

Vedclass · Accepted Answer

$n_1^2 d_1^3 = n_2^2 d_2^3$

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(B) According to Kepler's Third Law of Planetary Motion,the square of the time period $T$ is proportional to the cube of the mean distance $d$ from the sun: $T^2 \propto d^3$.Since frequency $n$ is the reciprocal of the time period $(n = 1/T)$,we have $T = 1/n$.Substituting this into the law: $(1/n)^2 \propto d^3$,which implies $1/n^2 \propto d^3$.Rearranging this gives $n^2 d^3 = 	ext{constant}$.Therefore,for two planets,we have $n_1^2 d_1^3 = n_2^2 d_2^3$.

Two planets are at mean distances $d_1$ and $d_2$ from the sun,and their orbital frequencies are $n_1$ and $n_2$ respectively. Then:

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