Answer:
Orbital period of satellite is 5.83 x 10³ s
Step-by-step explanation:
The orbital period of satellite revolving around Earth is given by the equation :
.....(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).
![T=\sqrt{(4\pi ^(2) [(6378.1+500)*10^(3) ]^(3) )/(6.67*10^(-11) *5.972*10^(24) ) }](https://img.qammunity.org/2021/formulas/physics/high-school/kob69vlajv45b2dpgwln86l4xkynmit800.png)
T = 5.83 x 10³ s