Answer:
The ratio is [tex]\frac{T_1}{T_2} = 3.965 [/tex]
Explanation:
From the question we are told that
The radius of Phobos orbit is R_2 = 9380 km
The radius of Deimos orbit is [tex]R_1 = 23500 \ km[/tex]
Generally from Kepler's third law
[tex]T^2 = \frac{ 4 * \pi^2 * R^3}{G * M }[/tex]
Here M is the mass of Mars which is constant
G is the gravitational constant
So we see that [tex]\frac{ 4 * \pi^2 }{G * M } = constant[/tex]
[tex]T^2 = R^3 * constant [/tex]
=> [tex][\frac{T_1}{T_2} ]^2 = [\frac{R_1}{R_2} ]^3[/tex]
Here [tex]T_1[/tex] is the period of Deimos
and [tex]T_1[/tex] is the period of Phobos
So
[tex][\frac{T_1}{T_2} ] = [\frac{R_1}{R_2} ]^{\frac{3}{2}}[/tex]
=> [tex]\frac{T_1}{T_2} = [\frac{23500 }{9380} ]^{\frac{3}{2}}][/tex]
=> [tex]\frac{T_1}{T_2} = 3.965 [/tex]
The given values are:
We know the formula,
→ [tex]v = \frac{2 \pi r}{T}[/tex]
then,
→ [tex]\sqrt{\frac{GM_{mars}}{r} } =\frac{2 \pi r}{T}[/tex]
or,
→ [tex]T = 2 \pi \sqrt{\frac{r^3}{GM_{mars}} }[/tex]
Now,
The orbital period of Deimos will be:
[tex]= 2 \pi \sqrt{\frac{(23500)^3}{GM_{mars}} }[/tex]
The orbital period of Phobos will be:
[tex]= 2 \pi \sqrt{\frac{(9380)^3}{GM_{mars}} }[/tex]
hence,
The ratio will be:
→ [tex]\frac{T_{deimos}}{T_{Phobos}}[/tex] = [tex]\frac{2 \pi \sqrt{\frac{(23500)^3}{GM_{mars}} } }{2 \pi \sqrt{\frac{(9380)^3}{GM_{mars}} }}[/tex]
= [tex]\sqrt{(\frac{23500}{9380})^3 }[/tex]
= [tex]3.96[/tex]
Thus the above answer is correct.
Learn more about orbital period here:
https://brainly.com/question/14286363
