Question #1666

It is the year 2305 and the tallest structure in the world has an insane height of 3.19×106 m above the surface of the Earth. A pendulum clock that keeps perfect time on the surface of the Earth is placed at the top of the tower. How long does the clock take to register one elapsed hour? The radius of the Earth is 6.38×106 m and its mass is 5.97×1024 kg. Answer in minutes.

Expert's answer

The periods on the ground and on the top of building are related by the expression:

T_{1/}T_{2} = 2 π(√ (L/g_{1})) / 2π (√(L/g_{2}) = √(g_{2}/g_{1})

As g = GM/R^{2}, g_{1} = GM/R^{2}, g_{2} = GM/(R+h)^{2}, where h is the height of building, R the Earth's radius (6.38x10^{6} m).

T_{1}/T_{2} = √(g_{2}/g_{1}) = √((GM/(R+h)^{2}) / (GM/R^{2})) = √(R^{2}/(R+h)^{2}) = R^{2}/(R+h)^{2}.

T_{1} = 60 min,

60/T_{2} = (6.38/(6.38+3.19))^{2}

**Answer: T**_{2} = 135 min.

T

As g = GM/R

T

T

60/T

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