91Ó°ÊÓ

Mars rotates on its axis once every$24.8hours.$

When a satellite revolves around a planet, the necessary centripetal force is provided by the gravitational force of attraction between the planet and the satellite.

Geosynchronous satellites are those satellites which appear stationary over a point above the surface of a planet. For this to happen the time period of the satellite should be equal to the time period of rotation of planet about its own axis.

Now, time period of satellite is $24.8\mathrm{hrs}=89280\mathrm{s}.$

Radius of mars $R=3.39×{10}^6\mathrm{m}.$

Mass of Mars = $6.39×{10}^{23}\mathrm{kg}.$

Substituting values and solving we get,

$$\begin{gathered} r=20.4×{10}^6\mathrm{m} \\ R+h=20.4×{10}^6\mathrm{m} \\ h=20.4×{10}^6\mathrm{m}-3.39×{10}^6\mathrm{m} \\ h=17.1×{10}^6\mathrm{m} \end{gathered}$$

Now calculating velocity from the equation : $v=\sqrt{\frac{GM}{r}}$

Substituting values of G, M and r we get : $v=1.44×{10}^3\mathrm{m}/\mathrm{s}.$

The speed of the satellite is$1.4×{10}^3\mathrm{m}/\mathrm{s}$and the altitude is$17.1×{10}^6\mathrm{m}.$