Answer: 0.107
Explanation:
We can solve this problem with Kepler's Third Law of Planetary motion:
[tex]T^{2}=4 \pi^{2} \frac{r^{3}}{G M_{MARS}}[/tex] (1)
Where:
[tex]T=7 h 39 min[/tex] is the orbital period of Phobos around Mars
[tex]G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex] is the Gravitational Constant
[tex]M_{MARS}[/tex] is the mass of Mars
[tex]r=9380 km \frac{1000 m}{1 km}=9,380,000 m[/tex] is the semimajor axis of the orbit Phobos describes around Mars (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)
Well, firstly we have to convert the orbital period to seconds:
[tex]T=7 h 39 min=(7 h \frac{3600 s}{1 h}) + (39 min \frac{60 s}{1 min})=25200 s + 2340 s=27540 s[/tex]
Now, we have to find [tex]M_{MARS}[/tex] from (1):
[tex]M_{MARS}=\frac{4 \pi^{2} r^{3}}{G T^{2}}[/tex] (2)
[tex]M_{MARS}=\frac{4 \pi^{2} (9,380,000 m)^{3}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}) (27540 s)^{2}}[/tex] (3)
[tex]M_{MARS}=6.436(10)^{23} kg[/tex] (4) This is the mass of Mars
On the other hand, it is known the mass of the Earth is:
[tex]M_{EARTH}=5.972(10)^{24} kg[/tex] (5)
Then, if we want to know the ratio of Mars’s mass to the mass of the earth, we have to divide [tex]M_{MARS}[/tex] by [tex]M_{EARTH}[/tex]:
[tex]\frac{M_{MARS}}{M_{EARTH}}=\frac{6.436(10)^{23} kg}{5.972(10)^{24} kg}[/tex]
Finally:
[tex]\frac{M_{MARS}}{M_{EARTH}}=0.107[/tex]
