Answer :
Answer:
[tex]0.2528[/tex]
Explanation:
To calculate the period we need the formula:
[tex]T=\frac{2\pi r^{3/2}}{\sqrt{GM}}[/tex]
Where [tex]r[/tex] is the radius of the moon, [tex]G[/tex] is the universal constant of gravitation and [tex]M[/tex] is the mass of mars.
The period of Phobos:
[tex]T_{p}=\frac{2\pi r_{p}^{3/2}}{\sqrt{GM}}[/tex]
The period of Deimos:
[tex]T_{D}=\frac{2\pi r_{D}^{3/2}}{\sqrt{GM}}[/tex]
The ratio of the period of Phobos and Deimos:
[tex]\frac{T_{p}}{T_{D}}=\frac{\frac{2\pi r_{p}^{3/2}}{\sqrt{GM}}}{\frac{2\pi r_{D}^{3/2}}{\sqrt{GM}}}[/tex]
[tex]\frac{T_{p}}{T_{D}}=\frac{\sqrt{GM}2\pi r_{p}^{3/2}}{\sqrt{GM}2\pi r_{D}^{3/2}}[/tex]
Most terms get canceled and we have:
[tex]\frac{T_{p}}{T_{D}}=\frac{r_{p}^{3/2}}{r_{D}^{3/2}}[/tex]
According to the problem
[tex]r_{p}=9,378km\\r_{D}=23,459km[/tex]
so the ratio will be:
[tex]\frac{T_{p}}{T_{D}}=\frac{(9,378)^{3/2}}{(23,459)^{3/2}}=\frac{908166.22}{3593058.125}=0.25275[/tex] ≈ [tex]0.2528[/tex]
the ratio of the period of revolution of Phobos to that of Deimos is 0.2528