A 5,000 kg satellite is orbiting the Earth in a circular path.The height of the satellite above the surface of the Earth is 800 km.The angular velocity of the satellite as it orbits the Earth is (Me = 5.98 $\times$ 10²⁴ kg, Re = 6.37 $\times$ 10⁶ m, G = 6.67 $\times$ 10^-11 N.m²/kg²) A) 9.50 $\times$ 10^-4 rad/s. B) 1.04 $\times$ 10^-3 rad/s. C) 1.44 $\times$ 10^-3 rad/s. D) 1.90 $\times$ 10^-3 rad/s. E) 2.20 $\times$ 10^-3 rad/s.

Question

Quizplus · Accepted Answer

The angular velocity ω can be calculated using the formula ω = sqrt(G * Me / r^3), where r is the distance from the center of the Earth to the satellite. Here, r = Re + height = 6.37 * 10^6 m + 800 * 10^3 m = 7.17 * 10^6 m. Plugging in the values, ω = sqrt(6.67 * 10^-11 * 5.98 * 10^24 / (7.17 * 10^6)^3) ≈ 1.04 * 10^-3 rad/s.

A 5,000 Kg Satellite Is Orbiting the Earth in a Circular