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A 38.0 kg satellite has a circular orbit with a period of 1.30 h and a radius of 7.90 Ɨ 106 m around a planet of unknown mass. If the magnitude of the gravitational acceleration on the surface of the planet is 30.0 m/s2, what is the radius of the planet

Respuesta :

MathPhys

Answer:

5.44Ɨ10ā¶ m

Explanation:

For a satellite with period t and orbital radius r, the velocity is:

v = 2Ļ€r/t

So the centripetal acceleration is:

a = vĀ² / r

a = (2Ļ€r/t)Ā² / r

a = (2Ļ€/t)Ā² r

This is equal to the acceleration due to gravity at that elevation:

g = MG / rĀ²

(2Ļ€/t)Ā² r = MG / rĀ²

M = (2Ļ€/t)Ā² rĀ³ / G

At the surface of the planet, the acceleration due to gravity is:

g = MG / RĀ²

Substituting our expression for the mass of the planet M:

g = [(2Ļ€/t)Ā² rĀ³ / G] G / RĀ²

g = (2Ļ€/t)Ā² rĀ³ / RĀ²

RĀ² = (2Ļ€/t)Ā² rĀ³ / g

R = (2Ļ€/t) āˆš(rĀ³ / g)

Given that t = 1.30 h = 4680 s, r = 7.90Ɨ10ā¶ m, and g = 30.0 m/sĀ²:

R = (2Ļ€ / 4680 s) āˆš((7.90Ɨ10ā¶ m)Ā³ / 30.0 m/sĀ²)

R = 5.44Ɨ10ā¶ m

Notice we didn't need to know the mass of the satellite.

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