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A simple pendulum has a period T on the surface of planet Earth. When taken to the surface of planet X, the same pendulum has a period 2T.

a) If the radius of planet X is 1/2 the radius of Earth, what is the mass of Planet X, relative to the mass of Earth?
b) Suppose a vertical mass and spring oscillator has a period T on Earth. What period would the oscillator have on planet X? Explain.

  • Physics -

    a) If the period on planet X is 2T, the acceleration of gravity (g) at the surface is 4 times less than Earth's, because
    Period = 2 pi sqrt(L/g)

    Since g = GM/R^2, if g is 1/4 as large as Earth's, and R is half the Earth's radius, M is 1/16 of Earth's.

    g(X) = g(earth)*[M(X)/M(earth)]*[Rearth/RX]^2
    = g(earth)*(1/16)*2^2 = g(earth)/4

    b) The period of a spring-mass oscillator system would not change. It does not depend upon gravity.

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