What is the time period of a simple pendulum of infinite length oscillating near the Earth's surface? (Mₑ = mass of Earth, R = radius of Earth.)

2*pi*sqrt(R²/(G*Mₑ))
2*pi*sqrt(R³/(G*Mₑ))
2*pi*sqrt(R/(G*Mₑ))
2*pi*sqrt(R⁴/(G*Mₑ))

Correct answer: 2*pi*sqrt(R³/(G*Mₑ))

Solution

For a pendulum of length L near a spherical Earth, the period is T = 2*pi*sqrt(1/(g*(1/L + 1/R))). As L -> infinity, 1/L -> 0, giving T = 2*pi*sqrt(R/g). With g = G*Mₑ/R², this becomes T = 2*pi*sqrt(R/(G*Mₑ/R²)) = 2*pi*sqrt(R³/(G*Mₑ)).

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