The distance of Neptune and Saturn from Sun are nearly 1013 and 1012 meters respectively.

Assuming that they move in circular orbit, the ratio of their time period is ____.

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Detailed Solution

We know that square of time period is directly proportional to cubic square of radius of orbit in which the satellite is revolving i.e.

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

w h ere, T_{n}^{2} = square of time period taken by Naptune to revolve around sun,

T_{s}^{2} = square of time period taken by saturn to revolve around sun,

r_{n}^{3} = square cube of radius of o rbit ofnaptune,

r_{s}^{3} = square cube of radius of orbit of saturn

Given, T_{n} = 10^{13}, T_{s} {= 10}^{12}

(\frac{T_{n}^{2}}{T_{s}^{2}}) = (\frac{r_{n}^{3}}{r_{s}^{3}})

= (\frac{10^{13}}{10^{12}}) = 10 \sqrt{10}

Therefore, the correct answer is

10 \sqrt{10}

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