91Ó°ÊÓ

The given data can be listed below,

• The height of Aura satellite above the earth is h = 705 km.
• The radius of the earth is $\mathrm{R}=6.37×{10}^6\mathrm{m}$.

Satellite motion is the circular motion around the earth, tangential to the earth's surface, and the velocity and acceleration are directed towards the center of the earth.

The time to complete one orbit in time period T is given by,

$T=\frac{2{\mathrm{Ï€°ù}}^{3/2}}{\sqrt{G{M}_E}}$

Here, r is the distance of the satellite from the center of the earth, which is ( R+h) Mis the mass of the planet whose value is localid="1657179502689" $5.97×{10}^{24}\mathrm{kg}$G is the constant of gravitation whose value is $\mathrm{G}=6.673×{10}^{-11}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2$.

Substitute values in the above,

$$\begin{gathered} \mathrm{T}=\frac{2\mathrm{π}{(6.37×{10}^6\mathrm{m}+7.05×{10}^5\mathrm{m})}^{3/2}}{\sqrt{(6.673×{10}^{-11}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2)(5.97×{10}^{24}\mathrm{kg})}}\left(\frac{1\mathrm{N}}{1\mathrm{kg}-\mathrm{m}/{\mathrm{s}}^2}\right) \\ =5.93×{10}^3\mathrm{s}\left(\frac{1\mathrm{hr}}{3600\mathrm{s}}\right) \\ =1.64\mathrm{hr} \end{gathered}$$

Thus, the time taken by the satellite to complete one revolution in T time is 1.64 hr.

The velocity of satellite can be given by,

$v=\frac{2\mathrm{Ï€°ù}}{T}$

Here, r ther is the distance of the satellite from the center of the earth, and T is the time of satellite to complete one revolution.

Put all the values in the above,

localid="1657179627957" $$\begin{gathered} \mathrm{v}=\frac{2\mathrm{π}(6.37×{10}^6\mathrm{m}+7.05×{10}^5\mathrm{m})}{5.97×{10}^3\mathrm{s}} \\ =7.50×{10}^3\mathrm{m}/\mathrm{s}\left(\frac{1\mathrm{km}}{1000\mathrm{m}}\right) \\ =7.50\mathrm{km} \end{gathered}$$

Thus, the velocity at which the Aura satellite moving is 7.50 km.