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

 Q)  A Satellite is revolving around the earth at a height of 600 km. Considering the radius of the earth as 6400 km, the mass of the earth is $6 \times 10^{24} kg $ and gravitational constant (G) as $ 6.67 \times 10^{-11} Nm^{2}/kg^{2}$ the time period of the satellite will be

 (Special thanks to Chandrasekhar, AKNU)

Solution:

 Given that radius of the earth (R) = 6400 km

Height of the satellite from the earth (h) = 600km

 Gravitational constant (G) =$ 6.67 \times 10^{-11} Nm^{2}/kg^{2}$

Mass of the earth (M)= $6 \times 10^{24} kg $

 Time period of the satellite (T) =?

 

 From Kepler’s third law

$T^{2}=(\frac{4\pi^{2}}{GM})a^{3}$-----------------------------------------------------------(1)

 

 Where a is the semi-major axis of the satellite where we will get from the center of the earth

i.e., a= R + h= 6400+ 600

                    = 7000 km

                    =$7\times 10^{3} km$

                    =$7\times 10^{6} meter$

 

$G\times M=(6.67 \times 10^{-11} Nm^{2}/kg^{2})(6 \times 10^{24} kg )$

                $=(6.67)(6)(10^{13})m^{3}/s^{2}$

                $=40.02\times10^{13}m^{3}/s^{2}$

 

 By substituting the above values in equation (1)

 

$T^{2}=(\frac{4\pi^{2}}{GM})a^{3}$

 

$T^{2}=(\frac{(4)(3.14)(3.14)}{40.02\times10^{13}m^{3}/s^{2}})(7\times 10^{6} meter)^{3}$

 

$T^{2}=\frac{(4)(3.14)(3.14)(7)(7)(7)10^{18}m^{3}}{40.02\times10^{13}m^{3}/s^{2}}$

 

$T^{2}=(\frac{13527.3712}{40.02})10^{5}s^{2}$

 

$T^{2}=338.01527\times 10^{5}s^{2}$

 

$T^{2}=33.801527\times 10^{6}s^{2}$

 

$T=\sqrt{33.801527\times 10^{6}s^{2}}$

 

$T=5.813\times 10^{3} seconds$