GATE-2018 (24) :Wyllie – Time average equation

Problem on Remote Sensing

Ex:- The space shuttle is a example of the low Earth orbit satellite. sometimes, it's orbit around 250 km form the earth's surface. where there is still a finite number of molecules from the atmosphere.The mean earth radius approximately 6378.14 km. Using this figure, to calculate the period of the shuttle orbit when the altitude of 250 km and the orbit is circular. find the linear velocity of the shuttle along its orbit.

Solution:

Period of the Orbit is

$T^{2} = \frac{4*\pi^{2}*a^{3}}{\mu}$

$T^{2} = \frac{4*\pi^{2}*(6628.14)^{3}}{3.986*10^{5}}$

$T^{2}= 2.88*10^{7} s^{2}$

Therefore, T= 5370.305 s

= 89 min 30.3 s

This orbit period is about as small as possible. At a lower altitude, friction with the earth’s atmosphere will quickly slow the shuttle down and it will return to earth. Thus, all spacecraft in stable earth orbit have orbital periods exceeding 89 min, 30 s.

The circumference of the orbit is =$ 2\times π\times a$

                                              = $2\times 3.14\times 6628.14$

                                             = 41,624.7192 km

Velocity of the shuttle in orbit is $V=\frac{Distance}{time }$

$V=\frac{41,624.719}{5370}$

      = 7.75 km/ s

The velocity of 7.8 km/s is the typical velocity for the low earth orbit satellite. as the altitude of the satellite increases the velocity of the satellite decreases.