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A planet revolves around the sun whose mean distance is $1.588$ times the mean distance between the earth and the sun. Calculate the time period of the revolution of the planet.

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answer is B.

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Detailed Solution

The time period of revolution of the given planet is

2 years .

From Kepler’s law of orbits we have

T^{2} \propto r^{3}

Where

T

represents time period and

r

represents orbital radius.
Where

T_{1}

is the time period of revolution of the given planet and

r_{1}

is the mean distance between the planet and the sun.

T_{E}

is the time period of revolution for the earth and

r_{E}

is the radius of the earth.

{(\frac{T_{1}}{T_{E}})}^{2} = {(\frac{r_{1}}{r_{E}})}^{3}

Given,

T_{E} = 1 year, r_{1} = 1.588

r_{E}

Then,

\Rightarrow {T_{1}}^{2} = {1.588}^{3};

{\Rightarrow T}_{1} = {1.588}^{\frac{3}{2}}

\Rightarrow T_{1} = 2 years

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