If R is the radius of the earth and the acceleration due to gravity on the surface of earth is g = π² m s⁻², then the length of the seconds pendulum at a height h = 2R from the surface of earth will be:
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Question

If R is the radius of the earth and the acceleration due to gravity on the surface of earth is g = π² m s⁻², then the length of the seconds pendulum at a height h = 2R from the surface of earth will be:

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G1 Physics · Accepted Answer

At height h = 2R, distance from Earth's centre is 3R. g' = g(R/3R)² = g/9 For a seconds pendulum, T = 2 s. T = 2π√(ℓ/g') 2 = 2π√(ℓ/(g/9)) Using g = π² m s⁻²: ℓ = 1/9 m