Question #49be6

1 Answer
Olivier B.
Oct 29, 2015

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

Explanation:

In order for the satellite to stay in orbit, its vertical acceleration 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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