Answer:
Orbital period of satellite is 5.83 x 10³ s
Explanation:
The orbital period of satellite revolving around Earth is given by the equation :
[tex]T=\sqrt{\frac{4\pi ^{2} (R+h)^{3} }{GM} }[/tex] .....(1)
Here R is radius of Earth, h is height of satellite from the Earth's surface, M is mass of Earth and G is gravitational constant.
In this problem,
Height of satellite, h = 500 km = 500 x 10³ m
Substitute 6378.1 x 10³ m for R, 500 x 10³ m for h, 5.972 x 10²⁴ kg for M and 6.67 x 10⁻¹¹ m³ kg⁻¹ s⁻² for G in equation (1).
[tex]T=\sqrt{\frac{4\pi ^{2} [(6378.1+500)\times10^{3} ]^{3} }{6.67\times10^{-11} \times5.972\times10^{24} } }[/tex]
T = 5.83 x 10³ s
