An artificial satellite revolves around the Earth in a circle. The satellite’s height above the Earth’s surface is 3200 km

| July 6, 2021 | education

An artificial satellite revolves around the Earth in a circle. The satellite’s height above the Earth’s surface is 3200 km. What is the acceleration due to gravity at this altitude? How fast is the satellite moving? the radius of the Earth is taken equal to 6400 km.

Task data: Нс (the height of the indicated satellite above the surface) = 3200 km = 3.2 * 106 m.
Constants: G (gravitational constant) ≈ 6.67 * 10 ^ -11 m2 / (s2 * kg); Mz (mass of the Earth) = 5.9726 * 10 ^ 24 kg; by condition, we take Rz (radius of the Earth) = 6400 km = 6.4 * 10 ^ 6 m.
1) Acceleration of gravity at 3200 km: gc = G * Ms / (Hs + Rs) 2 = 6.67 * 10-11 * 5.9726 * 10 ^ 24 / (3.2 * 106 + 6.4 * 10 ^ 6) ^ 2 = 4.32 m / s2.
2) The speed of the satellite: V = √ (G * Ms / (Hs + Rs)) = √ (6.67 * 10 ^ -11 * 5.9726 * 10 ^ 24 / (3.2 * 10 ^ 6 + 6 , 4 * 10 ^ 6)) = 6441.8 m / s.



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