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Expected GATE-2022 Geophysics Solution (13)

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 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

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