# Question #f608f

Nov 5, 2017

OK, some algebra required but not too hard. Incidentally, I think the ratio should read ${r}^{3} / {T}^{2}$ not ${T}^{3}$ but on that basis, we can proceed.

#### Explanation:

We know that ${r}^{3} / {T}^{2}$ is constant for the system.

So for any body orbiting in this system ${r}^{3} / {T}^{2} = 3.5 \times {10}^{-} 18$ and for planet y: ${r}^{3} / {T}^{2} = \frac{3 \times {10}^{10}}{T} ^ 2$ putting these two equal to each other and rearranging for T squared gives:

${T}^{2} = \frac{3 \times {10}^{10}}{3.5 \times {10}^{-} 18}$

So $T = 9.25 \times {10}^{13}$ s

Voilà! (Wasn’t Kepler wonderful?)