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Chapter 5: Problem 62

You stand on a spring scale at the north pole and again at the equator. Which scale reading will be lower, and by what percentage will it be lower than the higher reading? Assume \(g\) has the same value at pole and equator.

Short Answer

Expert verified
The scale reading at the equator will be lower. The approximate percentage by which it will be lower depends on the values of \(g, \omega,\) and \(R\). With a typical value of \(g = 9.81 m/s^2, \omega = 7.27 \times 10^{-5} rad/s,\) and \(R = 6.38 \times 10^6 m\), the percentage difference is approximately 0.3%.

Step by step solution

01

- Determine the forces

At the equator, apart from the gravitational force (\(mg\)), there's also a centrifugal force (\(m\omega^2R\)), where \(\omega\) is the angular velocity of the earth and \(R\) its radius. The net force acting on the body is \(mg - m\omega^2R = m(g -\omega^2R)\). This is the force that the scale measures.
02

- Calculate the scale readings

Let \(\omega\) be the angular velocity of the earth, \(R_e\) the radius of the earth, and \(m\) the mass of the body. At the pole, the scale reading is \(mg\), and at the equator, the scale reading is \(m(g -\omega^2R)\).
03

- Calculate the percentage difference

Next, find the percentage difference between the two weight readings. You do this by subtracting the weights, dividing the result by the weight at the pole and multiplying by 100 to get the percentage. Using the known values of \(g, \omega,\) and \(R\), you can estimate this difference.

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