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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