- #1
Vitani11
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Homework Statement
A pendulum clock is adjusted so that it keeps excellent time on the ground. The clock is brought to a mine of depth h below the ground and then raised a height h above the ground to see the differences. In which case is the error larger?
Homework Equations
T = 2π √(L/g)
L = length of the pendulum
g = gravity on Earth's surface
T = period
g = GMm/Re2
G = gravitational constant
Re = Earth's radius
M = mass of earth
m = mass of pendulum (negligible)
The Attempt at a Solution
g1 = GMm/(Re - h)
g2 = GMm/(Re+h)
g1 = GMm/(Re - h) is for the pendulum underground. Plugging into the period equation gives T = 2π (Re - h) √(L/GMm)
g2 = GMm/(Re+h) is for the pendulum above ground. Plugging into the period equation gives T = 2π (Re + h) √(L/GMm)
It's obvious that the pendulum will tick slower below ground and faster above - but what about the error? I think that the errors would be the same. The question seems to imply that they are not though. I think there is a factor that I am not taking into account?