Problem 3 - Classical Mechanics

Question

The water in a circular lake of radius \(1 \mathrm{~km}\) in latitude \(60^{\circ}\) is at rest relative to the Earth. Find the depth by which the centre is depressed relative to the shore by the centrifugal force. For comparison, find the height by which the centre is raised by the curvature of the Earth's surface. (Earth radius \(=6400 \mathrm{~km}\).)

Step-by-Step Solution

Answer

Answer: The center of the circular lake is raised by approximately 0.078 m due to Earth's curvature and is depressed by approximately 0.00347 m due to the centrifugal force at latitude 60°.

Step 1: Calculate the centrifugal acceleration

At latitude \(60^{\circ}\), the centrifugal acceleration, a_c, can be determined using the following formula: \[ a_c = R_e \omega^2 \cos{\phi} \] Where \(R_e = 6400\,\text{km}\) is the Earth's radius, \(\omega\) is the Earth's angular velocity (\(7.292\,x\,10^{-5}\,\text{rad/s}\)), and \(\phi\) is the latitude. Let us find \(a_c\). \[ a_c = 6400\,000\,\text{m} \times (7.292\,x\,10^{-5}\,\text{rad/s})^2 \cos{(60^{\circ})} \approx 0.034\,\text{m/s}^{2} \]

Step 2: Find the depth by which the center is depressed

Using hydrostatic equilibrium and centrifugal acceleration, we can write the pressure difference between the center and the shore of the lake as: \[ \Delta P = \rho g h_\text{centrifugal} = \rho a_c R \] Where \(\rho\) is the water density, \(g\) is the gravitational acceleration, \(h_\text{centrifugal}\) is the depth by which the center is depressed, and \(R = 1000\,\text{m}\) is the lake's radius. We can solve for \(h_\text{centrifugal}\): \[ h_\text{centrifugal} = \frac{\rho a_c R}{\rho g} \] Using density of water \(\rho = 1000\,\text{kg/m}^3\), \(g = 9.81\,\text{m/s}^2\), \(a_c = 0.034\,\text{m/s}^{2}\), and \(R = 1000\,\text{m}\), we get: \[ h_\text{centrifugal} \approx 0.00347\,\text{m} \]

Step 3: Find the height by which the center is raised by the Earth's curvature

To find the height raised by the Earth's curvature, we can consider the difference in radius between the Earth's surface and the curved surface of the lake. Let us call this height \(h_{curvature}\). Using the Pythagorean theorem, we can write: \[ (R_e + h_{curvature})^2 = R_e^2 + R^2 \] Solve for \(h_{curvature}\): \[ h_{curvature} = \sqrt{R_e^2 + R^2} - R_e \] Using \(R_e = 6400\,\text{km}\) and \(R = 1\,\text{km}\), we get: \[ h_{curvature} \approx 0.078\,\text{m} \] So, the depth by which the center is depressed relative to the shore by the centrifugal force is approximately \(0.00347\,\text{m}\) whereas the height by which the center is raised by the curvature of the Earth's surface is approximately \(0.078\,\text{m}\).

Key Concepts

Hydrostatic Equilibrium

Hydrostatic equilibrium is a state in which a volume of liquid is at rest or at a constant velocity due to the balance between gravitational forces and the pressure gradient forces. This scientific principle helps us understand why the depth at the center of a lake might differ from the depth at its edges due to the presence of centrifugal force resulting from Earth's rotation.

Let's consider our circular lake at latitude 60°. Its water molecules experience Earth's gravity pulling them downward, while at the same time, the Earth's rotation creates an outward centrifugal force. These forces combined dictate the shape of the water's surface. At the equator, where the centrifugal force is maximal, this effect is most pronounced. In fact, without the presence of these balancing forces, water wouldn't stay neatly distributed; it would be free to move until equilibrium is achieved. The step-by-step solution calculates the depression of the water's surface at the center, showing it to be much lesser than the effects due to Earth's curvature.

Earth's Angular Velocity

Earth's angular velocity is the rate of rotation of our planet around its axis. For any point on Earth's surface, this is a constant value of roughly 7.292 x 10^-5 radians per second. This doesn't change regardless of your position on Earth; however, the real effect of this angular velocity on a person or object changes with latitude.

At the poles, you're not moving in a circle relative to Earth's center, but on the equator, you're in a large circular motion due to Earth's rotation. Angular velocity is crucial when calculating the centrifugal acceleration, as seen in the exercise where this constant is part of the formula used to determine the centrifugal acceleration at a latitude of 60°. This understanding helps us ascertain how centrifugal force contributes to the apparent 'depression' of water in our fictional circular lake.

Centrifugal Acceleration Calculation

Centrifugal acceleration arises due to the rotation of Earth and it's the perceived force pushing a mass outward from the center of rotation. To calculate centrifugal acceleration, one can use the formula \( a_c = R_e \omega^2 \cos{\phi} \), where \( R_e \) is Earth's radius, \( \omega \) is Earth's angular velocity, and \( \phi \) is the latitude.

In the exercise, this calculation is essential because it determines how much the water's surface at the center of the lake is depressed. Although this depression is slight, it's indeed caused by the outward 'push' from Earth's rotation. Understanding how to calculate this acceleration provides us with a clear understanding of the forces at play on Earth's surface due to our planet's rotation.

Earth's Curvature Effects

The curvature of Earth's surface is another factor that affects the water levels in our circular lake example. Because Earth is not a perfect sphere, but an oblate spheroid, its curvature varies with latitude, and the centrifugal force due to Earth's rotation causes an equatorial bulge.

From our exercise example, we learned that Earth's curvature causes the center of the lake to appear raised relative to the edges, which we call the height due to Earth's curvature, \( h_{curvature} \). It's fascinating to see that despite the size of Earth, these effects are noticeable even in a small-scale scenario such as a circular lake with a radius of 1 km. This height difference demonstrates Earth's sphericity and, independently from the centrifugal force, contributes to the shape of water bodies.