The average distance of the Earth from the Su | vedclass

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(B) According to Kepler's third law of planetary motion, the square of the time period $(T)$ of a planet is directly proportional to the cube of its a

Vedclass · Accepted Answer

$D\left(\frac{L_{2}}{L_{1}}ight)^{\frac{3}{2}} 	ext{ days}$

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(B) According to Kepler's third law of planetary motion, the square of the time period $(T)$ of a planet is directly proportional to the cube of its average distance $(R)$ from the Sun.$T^{2} \propto R^{3}$Let $T_{1} = D$ be the time period of the Earth at distance $L_{1}$, and $T_{2}$ be the time period of the other planet at distance $L_{2}$.Then, $\frac{T_{2}^{2}}{T_{1}^{2}} = \frac{L_{2}^{3}}{L_{1}^{3}}$$\frac{T_{2}^{2}}{D^{2}} = \left(\frac{L_{2}}{L_{1}}ight)^{3}$$T_{2}^{2} = D^{2} \left(\frac{L_{2}}{L_{1}}ight)^{3}$Taking the square root on both sides:$T_{2} = D \left(\frac{L_{2}}{L_{1}}ight)^{3/2} 	ext{ days}$.

The average distance of the Earth from the Sun is $L_{1}$. If one year of the Earth is $D$ days, then one year of another planet whose average distance from the Sun is $L_{2}$ will be:

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