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A satellite of mass 230 kg is placed in Earth orbit at a height of 500 km above the surface. (a) Assuming a circular orbit, how long does the satellite take to complete one orbit

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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 :


T=\sqrt{(4\pi ^(2) (R+h)^(3) )/(GM) } .....(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) ) }

T = 5.83 x 10³ s

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