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.

Answer 1

According to Kepler's Law of period $T^2\propto r^3$
where $T=$ period of revolution, $r=$ mean distance from sun
$\frac{T_1^2}{T_2^2}=\frac{r_1^3}{r_2^3};\frac{T_1}{T_2}={\left(\frac{r_1}{r_2}\right)}^{3/2}$.
Let $T_2$ be the period of revolution of earth around sun and $r_2$ be the mean distance between earth and sun.
Given: $\left(r_1/r_2\right)=5.2$ and $T_2=365days$ (one year).
Hence, $T_1=T_2{\left(\frac{r_1}{r_2}\right)}^{3/2}=1(5.2{)}^{3/2}=11.85years$.