Answer:
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
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;
[tex]T = 2 \times \pi \times \sqrt{\dfrac{(R + h)^3}{g \times R^2} }[/tex]
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;
[tex]T = 2 \times \pi \times \sqrt{\dfrac{((6.371 + 0.3)\times 10^6)^3}{9.81 \times (6.371 \times 10^6)^2} } \approx 5,425.305[/tex]
T ≈ 5,425.305 seconds ≈ 1.507 hours
b) The orbital velocity of the satellite is given as follows;
[tex]v_0 = \sqrt{\dfrac{G \times M_{central}}{R + h} } = \sqrt{\dfrac{g \times R^2}{R + h} }[/tex]
Where;
v₀ = The orbital velocity of the satellite
Which by substitution gives;
[tex]v_0 = \sqrt{\dfrac{9.81 \times (6.371 \times 10^6)^2}{(6.371 + 0.3) \times 10^6} } = 7,725.8565[/tex]
The orbital velocity, v₀ = 7,725.8565 m/s.
