Question

A spy satellite of mass of 1000 kg is orbiting the earth at an altitude of 300 km/s

a) what is its period
b) what is its orbital speed​

Answer 1

a) The period of the satellite is approximately 5,425.305 seconds (1.507 hours)

b) The orbital velocity of the satellite is approximately 7,725.8565 m/s

Step-by-step explanation:

The given parameters are;

The mass of the satellite, m = 1,000 kg

The altitude at which the satellite is orbiting, h = 300 km

a) The period of the satellite is given as follows;

`T = 2 * \pi * \sqrt{((R + h)^3)/(g * R^2) }`

Where;

R = The radius of the Earth = 6.371 × 10⁶ m

g = The acceleration due to gravity = 9.81 m/s²

h = The altitude of the orbit of the satellite = 300 km = 300,000 m

By substitution, we have;

`T = 2 * \pi * \sqrt{(((6.371 + 0.3)* 10^6)^3)/(9.81 * (6.371 * 10^6)^2) } \approx 5,425.305`

T ≈ 5,425.305 seconds ≈ 1.507 hours

b) The orbital velocity of the satellite is given as follows;

`v_0 = \sqrt{(G * M_(central))/(R + h) } = \sqrt{(g * R^2)/(R + h) }`

Where;

v₀ = The orbital velocity of the satellite

Which by substitution gives;

`v_0 = \sqrt{(9.81 * (6.371 * 10^6)^2)/((6.371 + 0.3) * 10^6) } = 7,725.8565`

The orbital velocity, v₀ = 7,725.8565 m/s.