Asked by j
Mass of an imaginary planet is double earth's mass and radius is 1.5 times earth's radius. Using mass and radius, calculate the acceleration of gravity for the imaginary planet.
Using the pendulum formula for the period of oscillation, and your new
acceleration of gravity, determine the period of a pendulum that is 1 meter long
and it is on this planet.
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Answered by
GPT-5 mini
AI
Use g = GM/R^2. If M' = 2M and R' = 1.5R then
g' = G(2M)/(1.5R)^2 = (2 / 1.5^2) · (GM/R^2) = (2 / 2.25) g = (8/9) g.
With g = 9.81 m/s^2:
g' = (8/9)·9.81 ≈ 8.72 m/s^2.
For a small-angle simple pendulum, T = 2π sqrt(L/g'). For L = 1.00 m:
T = 2π sqrt(1 / 8.72) ≈ 2.13 s (more precisely ≈ 2.127 s).
(Assumes the small-angle approximation and a simple pendulum.)
g' = G(2M)/(1.5R)^2 = (2 / 1.5^2) · (GM/R^2) = (2 / 2.25) g = (8/9) g.
With g = 9.81 m/s^2:
g' = (8/9)·9.81 ≈ 8.72 m/s^2.
For a small-angle simple pendulum, T = 2π sqrt(L/g'). For L = 1.00 m:
T = 2π sqrt(1 / 8.72) ≈ 2.13 s (more precisely ≈ 2.127 s).
(Assumes the small-angle approximation and a simple pendulum.)
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