Question

At an orbital height of 500 km, find the orbital period of the satellite.

Answer 1

h = 500 × 10³ m, R = 6371 × 10³ m, 

v = 7616 × 10³ kms^-1

T = \(\frac{(R+h)}{V}\) = 2 x \(\frac{22}{7}\) x \(\frac{6371 + 500}{7616}\)

= 5.6677 x 10³s =5667 s.

This is T ≈ 95 min