Gabe asked in Science & MathematicsPhysics · 2 months ago

What is the moon's orbital period in earth days?

A 1.4 x 10^21 kg moon orbits a distant planet in a circular orbit of radius 1.5 x 10^8 m. It experiences a 1.1 x 10^19 N gravitational pull from the planet. What is the moon's orbital period in earth days?

1 Answer

  • 2 months ago

    F = G m₁m₂/r² = (6.674e-11)(1.4e21)M/(1.5e8)² = 1.1e19N

    where M is planet mass

    M = (1.1e19)(1.5e8)² / (6.674e-11)(1.4e21)

    T = 2π√[R³/GM]

    T = 2π√[(1.5e8)³/(6.674e-11)M]

    T = 2π√[(1.5e8)³ / (6.674e-11)((1.1e19)(1.5e8)² / (6.674e-11)(1.4e21))]

    T = 2π√[(1.5e8)³ (1.4e21) / (1.1e19)(1.5e8)²]

    T = 2π√[(1.5e8)³(1.4e21) / (1.1e19)]

    you can do the rest. Note that this is in seconds.

    Gravitational attraction in newtonsF = G m₁m₂/r²G = 6.674e-11 m³/kgs²m₁ and m₂ are the masses of the two objects in kgr is the distance in meters between their centers

    Satellite motion, circularV = √(GM/R)V = √(gR)T = 2π√[R³/GM]g = GM/R²  T is period of satellite in sec  V = velocity in m/s  g = acceleration of gravity in m/s²    (9.8 m/s² at ground level)  G = 6.673e-11 Nm²/kg²  M is mass of central body in kg  R is radius of orbit in m

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