The period of revolution of the satellite around the Earth is 4 hours, the height H above the Earth’s surface

| August 12, 2021 | education

The period of revolution of the satellite around the Earth is 4 hours, the height H above the Earth’s surface is equal to the Earth’s radius of 6400 km. The speed of the satellite in orbit is approximately

1) Determine the radius of the satellite’s orbit: R = 2Rz, where, according to the condition, Rz is the value of the Earth’s radius (Rz = 6400 km = 6400 * 10 ^ 3 m).

R = 2Rz = 2 * 6400 * 10 ^ 3 = 12800 * 10 ^ 3 m.

2) Calculate the length of the satellite’s orbit: L = 2ΠR = 2 * 3.14 * 12800 * 10 ^ 3 = 80384 * 10 ^ 3 m.

3) The speed of the satellite in its orbit: V = L / t, where, according to the condition, t is the time of one revolution of the satellite (t = 4 h = 4 * 3600 s = 14400 s).

V = L / t = 80384 * 10 ^ 3/14400 = 5582.2 m / s ≈ 5.6 km / s.

Answer: The satellite speed is 5.6 km / s.



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