Question

Mars’s moon Phobos orbits the planet at a distance of 9380 km from its center, and it takes 7 hours and 39 minutes to complete one orbit. What is the ratio of Mars’s mass to the mass of the earth?

Answer 1

Answer: 0.107

Explanation:

We can solve this problem with Kepler's Third Law of Planetary motion:

$T^{2}=4 \pi^{2} \frac{r^{3}}{G M_{MARS}}$ (1)

Where:

$T=7 h 39 min$ is the orbital period of Phobos around Mars

$G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the Gravitational Constant  

$M_{MARS}$ is the mass of Mars

$r=9380 km \frac{1000 m}{1 km}=9,380,000 m$ 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:

$T=7 h 39 min=(7 h \frac{3600 s}{1 h}) + (39 min \frac{60 s}{1 min})=25200 s + 2340 s=27540 s$

Now, we have to find $M_{MARS}$ from (1):

$M_{MARS}=\frac{4 \pi^{2} r^{3}}{G T^{2}}$ (2)

$M_{MARS}=\frac{4 \pi^{2} (9,380,000 m)^{3}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}) (27540 s)^{2}}$ (3)

$M_{MARS}=6.436(10)^{23} kg$ (4) This is the mass of Mars

On the other hand, it is known the mass of the Earth is:

$M_{EARTH}=5.972(10)^{24} kg$ (5)

Then, if we want to know the ratio of Mars’s mass to the mass of the earth, we have to divide $M_{MARS}$ by $M_{EARTH}$:

$\frac{M_{MARS}}{M_{EARTH}}=\frac{6.436(10)^{23} kg}{5.972(10)^{24} kg}$

Finally:

$\frac{M_{MARS}}{M_{EARTH}}=0.107$