posted by Chan on .
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.
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.