# Mars rotates on its axis once every 24.8 hours. What is the altitude of a geosynchronous...

## Question:

Mars rotates on its axis once every 24.8 hours. What is the altitude of a geosynchronous satellite orbiting Mars?

## Geosynchronous Satellite:

A geosynchronous satellite of a planet is a satellite which is orbiting in a geosynchronous orbit of that planet.

The radius of the satellite from the centre of its planet is given by;

{eq}r^3 = \dfrac{GM}{4 \pi^2} T^2 {/eq}

The orbital speed of the satellite is given by;

{eq}v = \dfrac{2\pi r}{T} {/eq}

where {eq}M {/eq} is the mass of the planet, {eq}T {/eq} is the period of rotation of the planet and {eq}G = 6.67 \times 10^{-11} \ Nm^2/kg^2 {/eq} is the gravitational constant.

Given:

• A planet Mars.
• The period of rotation of the Mars is {eq}T = 24.8 \ Hours \ = 24.8 \times 3600 \ s. {/eq}
• The mass of the Mars is {eq}M = 6.39 \times 10^{23} \ kg {/eq}
• The radius of the Mars is {eq}R = 3.3895 \times 10^{6} \ m {/eq}

The radius of the satellite from the centre of the Mars is given by;

{eq}\begin{align} r^3 &= \dfrac{GM}{4 \pi^2} T^2\\ &= \dfrac{6.67 \times 10^{-11} \times 6.39 \times 10^{23}}{4 \pi^2} (24.8 \times 3600)^2\\ \implies r &= 2.05 \times 10^7 \ m \\ \end{align} {/eq}

The altitude of a geosynchronous satellite orbiting Mars is

{eq}\begin{align} h &= r-R\\ &= 2.05 \times 10^7 \ m - 3.3895 \times 10^{6} \ m \\ &= 17.1 \times 10^{6} \ m \\ \end{align} {/eq} 