The average distances of two planets (Neptune and Saturn) from the sun are $10^{31} m$ and $10^{32} m$ respectively. The ratio of time periods of the planets will be:

100 : 1

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1 : 10 \sqrt{10}

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\sqrt{10} : 1

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10 \sqrt{10} : 1

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Solution

The correct option is B $1 : 10 \sqrt{10}$
By Kepler's third law, $T^{2} \propto a^{3}$
Thus, $\frac{T_{n}^{2}}{T_{s}^{2}} = \frac{a_{n}^{3}}{a_{s}^{3}} = \frac{10^{93}}{10^{96}} = \frac{1}{1000}$
So, $\frac{T_{n}}{T_{s}} = \frac{1}{10 \sqrt{10}}$

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The average distances of two planets (Neptune and Saturn) from the sun are 1031m and 1032m respectively. The ratio of time periods of the planets will be:

The correct option is B 1:10√10By Kepler's third law, T2∝a3Thus, T2nT2s=a3na3s=10931096=11000So, TnTs=110√10

The average distances of two planets (Neptune and Saturn) from the sun are $10^{31} m$ and $10^{32} m$ respectively. The ratio of time periods of the planets will be:

The correct option is B $1 : 10 \sqrt{10}$
By Kepler's third law, $T^{2} \propto a^{3}$
Thus, $\frac{T_{n}^{2}}{T_{s}^{2}} = \frac{a_{n}^{3}}{a_{s}^{3}} = \frac{10^{93}}{10^{96}} = \frac{1}{1000}$
So, $\frac{T_{n}}{T_{s}} = \frac{1}{10 \sqrt{10}}$