Question

Question #49be6

Answer 1

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`