A satellite is placed in equatorial orbit above Mars, which has a radius of 3397km and a mass M=6.4×10^23 kg. The mission of the satellite is to observe the Martian climate from an altitude of 488km. What is the orbital period of the satellite?

The satellite’s orbital period is 2h 2min 41.8s

In order for the satellite to stay in orbit, its vertical must be null. Therefore, its centrifugal acceleration must be the opposite of Mars’ gravitationnal acceleration.

The satellite is ##488##km above Mars’ surface and the planet’s radius is ##3397##km. Therefore, Mars’ gravitationnal acceleration is:

##g=(GcdotM)/d^2=(6.67*10^(-11)cdot6.4*10^23)/(3397000+488000)^2=(6.67cdot6.4*10^6)/(3397+488)^2~~2.83##m/s²

The satellite’s centrifugal acceleration is:

##a=v^2/r=g=2.83##

##rarr v=sqrt(2.83*3885000)=sqrt(10994550)=3315.8##m/s

If the satellite’s orbit is circular, then the orbit’s perimeter is:

##Pi=2pi*3885000~~24410174.9##m

Therefore the satellite’s orbital period is:

##P=Pi/v=24410174.9/3315.8=7361.8s##

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