Question

Objects that are at rest relative to the Earth's surface are in circular motion due to Earth's rotation. (a) What is the radial acceleration of an object at the equator? (b) Is the object's apparent weight greater or less than its weight? Explain. (c) By what percentage does the apparent weight differ from the weight at the equator? (d) Is there any place on Earth where a bathroom scale reading is equal to your true weight? Explain.

Short answer

Answer: The radial acceleration at the equator is approximately 0.0339 m/s². Due to Earth's rotation, the centrifugal force is greatest at the equator, causing the object's apparent weight to be less than its true weight. The difference in weight is approximately 0.346%.

Step-by-step solution

(a) Radial acceleration at the equator

To calculate the radial acceleration at the equator, we can use the formula: \( a_r = \frac{v^2}{r} \) where \(a_r\) is the radial acceleration, \(v\) is the velocity of the object due to Earth's rotation, and \(r\) is the radius of the Earth. We know that the Earth completes one rotation every 24 hours, and its circumference at the equator is given by: \( C = 2\pi r \) From these two pieces of information, we can determine the velocity, \(v\): \( v = \frac{C}{T} = \frac{2\pi r}{24 \times 3600 \, \text{seconds}} \) where \(T\) is the period of rotation (24 hours). Now, we can plug this expression for \(v\) back into the formula for the radial acceleration: \( a_r = \frac{(2\pi r)^2}{24^2 \times 3600^2 \, r} \) Simplifying this and using the Earth's radius (approximately 6371 km), we find: \( a_r ≈ 0.0339 \, \text{m/s^2} \) So, the radial acceleration of an object at the equator is approximately 0.0339 m/s².

(b) Apparent weight vs. true weight

The apparent weight of an object is the force it feels due to gravity, with any centrifugal force caused by the Earth's rotation subtracted from its true weight. At the equator, this centrifugal force is greatest, so objects have the lowest apparent weight. Therefore, the object's apparent weight is less than its true weight.

(c) Percentage difference in weight at the equator

We can calculate the percentage difference in weight at the equator using the formula: \( \% \, \text{difference} = \frac{\text{centrifugal force}}{\text{true weight}} \times 100 \) Centrifugal force on an object of mass \(m\) is given by: \( F_c = m a_r \) We know that true weight is the force due to gravity, which can be calculated as: \( W = mg \) where \(g = 9.81 \, \text{m/s^2}\) is the acceleration due to gravity. Using these expressions, we find the percentage difference: \( \% \, \text{difference} = \frac{m a_r}{mg} \times 100 = \frac{0.0339}{9.81} \times 100 ≈ 0.346 \% \) So, the apparent weight at the equator differs from the true weight by approximately 0.346%.

(d) Bathroom scale reading equal to true weight

There is a place on Earth where the bathroom scale reading would be equal to the true weight, and that is at the poles. This is because the centrifugal force caused by rotation is zero at the poles, so the object's apparent weight is equal to its true weight. Thus, the bathroom scale reading would be equal to the object's true weight at the poles.