The time taken by the satellite to orbit the Earth is 1.51 seconds.
Given:
The satellite with mass 1,500 kg and travels at [tex]6.8\times 10^3 m/s[/tex] , 8,500 km above the center of the Earth.
The mass of the Earth is [tex]5.97\times 10^{24} kg[/tex].
To find:
The time taken by a satellite to orbit around the Earth.
Solution:
(Under the action of zero gravity, the effect of the mass of the satellite on the orbital time becomes negligible)
[tex]1 km = 1000 m\\\\r=8,500 km= 8,500\times 1000m=8.5\times 10^6 m[/tex]
The orbital period of a satellite is given by:
[tex]\frac{T^2}{r^3}=\frac{4 \pi}{GM}\\\\\frac{T^2}{( 8.5\times 10^6 m)^2}=\frac{4\times 3.14}{6.67\times 10^{-11} m^3 kg^{-1}s^{-2}\times 5.97\times 10^{24} kg}\\\\T^2=\frac{4\times 3.14\times ( 8.5\times 10^6 m)^2}{6.67\times 10^{-11} m^3 kg^{-1}s^{-2}\times 5.97\times 10^{24} kg}\\\\T^2=2.28 s^2\\\\T=1.51 s[/tex]
The time taken by the satellite to orbit the Earth is 1.51 seconds.
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