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Earth, Moon and High Tide

The mean dist...

Question

The mean distance of Jupiter from the sun is nearly $5.2$ times the corresponding distance between earth and sun. Using Kepler's Law, find the period of revolution of Jupiter in its orbit.

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Solution

According to Kepler's Law of period $T^{2} \propto r^{3}$
where $T =$ period of revolution, $r =$ mean distance from sun
$\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{r_{1}^{3}}{r_{2}^{3}}; \frac{T_{1}}{T_{2}} = {(\frac{r_{1}}{r_{2}})}^{3 / 2}$ .
Let $T_{2}$ be the period of revolution of earth around sun and $r_{2}$ be the mean distance between earth and sun.
Given: $(r_{1} / r_{2}) = 5.2$ and $T_{2} = 365 d a y s$ (one year).
Hence, $T_{1} = T_{2} {(\frac{r_{1}}{r_{2}})}^{3 / 2} = 1 (5.2)^{3 / 2} = 11.85 y e a r s$ .

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