Average distance of the earth from the sun is $L_{1}$ . If one year of the earth = D days, one year of another planet whose average distance from the sun is $L_{2}$ will be

D {(\frac{L_{2}}{L_{1}})}^{1 / 2}

days

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D {(\frac{L_{2}}{L_{1}})}^{3 / 2}

days

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D {(\frac{L_{2}}{L_{1}})}^{2 / 3}

days

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D (\frac{L_{2}}{L_{1}})

days

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Solution

The correct option is B $D {(\frac{L_{2}}{L_{1}})}^{3 / 2}$ days
According to Kepler's third law,
$T^{2} \propto r^{3} o r T \propto r^{3 / 2}$

$∴ \frac{D^{'}}{D} = {(\frac{L_{2}}{L_{1}})}^{3 / 2}$ or
$D^{'} = D {(\frac{L_{2}}{L_{1}})}^{3 / 2}$ days

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Average distance of the earth from the sun is L1. If one year of the earth = D days, one year of another planet whose average distance from the sun is L2 will be

The correct option is B D(L2L1)3/2daysAccording to Kepler's third law,T2∝r3orT∝r3/2∴D′D=(L2L1)3/2orD′=D(L2L1)3/2 days

Average distance of the earth from the sun is $L_{1}$ . If one year of the earth = D days, one year of another planet whose average distance from the sun is $L_{2}$ will be

The correct option is B $D {(\frac{L_{2}}{L_{1}})}^{3 / 2}$ days
According to Kepler's third law,
$T^{2} \propto r^{3} o r T \propto r^{3 / 2}$

$∴ \frac{D^{'}}{D} = {(\frac{L_{2}}{L_{1}})}^{3 / 2}$ or
$D^{'} = D {(\frac{L_{2}}{L_{1}})}^{3 / 2}$ days