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Taking that earth revolves round the sun in circular orbit of radius $15 \times 10^{10} m$ with a time period of 1year. The time taken by another planet which is at a distance of $540 \times 10^{10} m$ , to revolve round the sun in circular orbit once will be

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Detailed Solution

r_{1} = 15 \times 10^{10} {m,   r}_{2} = 540 \times 10^{10} m

$T_{1} = 1 {year, T}_{2} = ?$

$T^{2} \propto R^{3}$

$\frac{T_{2}}{T_{1}} = {(\frac{R_{2}}{R_{1}})}^{3 / 2}$

$= {(\frac{540}{15})}^{3 / 2} = {(36)}^{3 / 2} = 6^{3}$

$T_{2} = 216 year$

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