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

0.107

Step-by-step explanation:

The expert on top solved it correctly.

Answer 2

Answer: 0.107

Step-by-step explanation:

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

`T^(2)=4 \pi^(2) (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)(m^(3))/(kgs^(2))` is the Gravitational Constant

`M_(MARS)` is the mass of Mars

`r=9380 km (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 (3600 s)/(1 h)) + (39 min (60 s)/(1 min))=25200 s + 2340 s=27540 s`

Now, we have to find
`M_(MARS)` from (1):

`M_(MARS)=(4 \pi^(2) r^(3))/(G T^(2))` (2)

`M_(MARS)=(4 \pi^(2) (9,380,000 m)^(3))/((6.674(10)^(-11)(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)`:

`(M_(MARS))/(M_(EARTH))=(6.436(10)^(23) kg)/(5.972(10)^(24) kg)`

Finally:

`(M_(MARS))/(M_(EARTH))=0.107`