Ask Question
26 November, 00:43

A 38.0 kg satellite has a circular orbit with a period of 1.30 h and a radius of 7.90 * 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

+2
Answers (1)
  1. 26 November, 01:09
    0
    5.44*10⁶ m

    Explanation:

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

    v = 2πr/t

    So the centripetal acceleration is:

    a = v² / r

    a = (2πr/t) ² / r

    a = (2π/t) ² r

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

    g = MG / r²

    (2π/t) ² r = MG / r²

    M = (2π/t) ² r³ / G

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

    g = MG / R²

    Substituting our expression for the mass of the planet M:

    g = [ (2π/t) ² r³ / G] G / R²

    g = (2π/t) ² r³ / R²

    R² = (2π/t) ² r³ / g

    R = (2π/t) √ (r³ / g)

    Given that t = 1.30 h = 4680 s, r = 7.90*10⁶ m, and g = 30.0 m/s²:

    R = (2π / 4680 s) √ ((7.90*10⁶ m) ³ / 30.0 m/s²)

    R = 5.44*10⁶ m

    Notice we didn't need to know the mass of the satellite.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “A 38.0 kg satellite has a circular orbit with a period of 1.30 h and a radius of 7.90 * 106 m around a planet of unknown mass. If the ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers