Problem 17 - Classical Dynamics of Particles and Systems

Question

Approximate Lake Superior by a circle of radius \(162 \mathrm{km}\) at a latitude of \(47^{\circ}\) Assume the water is at rest with respect to Earth and find the depth that the center is depressed with respect to the shore due to the centrifugal force.

Step-by-Step Solution

Answer

The depth of depression at the center of Lake Superior due to centrifugal force can be found using the formula \( d = \frac{F_{c}}{F_{ge}} \times r \). By calculating the Earth's rotational speed at a latitude of 47°, converting it to radians per second, and finding the centrifugal and effective gravitational forces, we can determine the depth of depression at the lake's center.

Step 1: Calculate the Earth's rotational speed at the given latitude

At the given latitude of 47°, we need to find the Earth's rotational speed (angular velocity). The Earth's rotational speed at the equator is approximately 1670 km/h. The rotational speed tends to decrease as we move towards the poles due to the Earth's spherical shape. We can calculate the rotational speed at the given latitude using the following formula: \(v = v_{\mathrm{equator}} \times \cos(latitude)\), where \(v_{\mathrm{equator}}\) is the Earth's rotational speed at the equator, and latitude is given in degrees.

Step 2: Convert the Earth's rotational speed to radians per second

To perform calculations more comfortably, we should convert the Earth's rotational speed obtained in Step 1 from km/h to radians per second. The conversion factor for km/h to m/s is \( \frac{1000}{3600} \) and for radians per second, it is \( \frac{2 \pi}{24 \times 3600} \). Apply these conversion factors to the Earth's rotational speed to obtain the speed in radians per second.

Step 3: Calculate the centrifugal force

Now that we have the Earth's rotational speed at the given latitude, we can calculate the centrifugal force acting on the water in Lake Superior. The formula for centrifugal force is: \(F_{c} = m \times r \times \omega^2\), where \(F_{c}\) is the centrifugal force, \(m\) is the mass of the water, \(r\) is the radius of Lake Superior (162 km), and \(\omega\) is the Earth's rotational speed in radians per second.

Step 4: Calculate the effective gravitational force

The effective gravitational force is the difference between the actual gravitational force acting on the water in Lake Superior and the centrifugal force calculated in Step 3. The formula for gravitational force is: \(F_{g} = G \frac{M_{Earth} m}{(R_{Earth} + h)^2}\), where \(F_{g}\) is the gravitational force, \(G\) is the gravitational constant (approximately \(6.674 \times 10^{-11} Nm^2/kg^2\)), \(M_{Earth}\) is the mass of Earth (\(5.972 \times 10^{24} kg\)), and \(R_{Earth}\) is the Earth’s radius. The effective gravitational force is then: \(F_{ge} = F_{g} - F_{c}\), where \(F_{ge}\) is the effective gravitational force.

Step 5: Find the depth of depression at the center of the lake

Finally, to find the depth of depression at the center of Lake Superior, we can use the following formula: \(d = \frac{F_{c}}{F_{ge}} \times r\), where \(d\) is the depth of depression, \(F_{c}\) is the centrifugal force, \(F_{ge}\) is the effective gravitational force, and \(r\) is the radius of Lake Superior. By plugging in our calculated values from Steps 1 through 4, we can compute the depth of depression at the center of Lake Superior due to centrifugal force.

Key Concepts

Gravitational Force

Gravity is a force that pulls objects toward each other. On Earth, it gives us our weight by pulling us towards the planet’s center. The force of gravity can be calculated using a formula that involves the gravitational constant, the mass of the Earth, and the distance from the center of the Earth. This is given by:

• \(F_{g} = G \frac{M_{Earth} m}{(R_{Earth} + h)^2}\)

Where:

• \(G\) is the gravitational constant (\(6.674 \times 10^{-11} \text{Nm}^2/\text{kg}^2\)).
• \(M_{Earth}\) is the mass of Earth (\(5.972 \times 10^{24} \text{kg}\)).
• \(m\) is the mass of the object being considered, which in this case is the water in Lake Superior.
• \(R_{Earth}\) is the average radius of Earth.
• \(h\) is the height above Earth's surface, though it's often negligible.

Gravity acts towards Earth's center, counteracting the outward centrifugal force caused by Earth's rotation.

Rotational Speed

The rotational speed of Earth decreases as one moves from the equator to the poles. This is why the speed is not uniform everywhere on the planet. At the equator, Earth rotates at nearly 1670 km/h. This speed can be adjusted for other latitudes using:

• \(v = v_{\mathrm{equator}} \times \cos(latitude)\)

Here, \(v_{\mathrm{equator}}\) is the speed at the equator, and \(\cos(latitude)\) adjusts this speed for any given latitude. At Lake Superior’s latitude of \(47^{\circ}\), this formula helps find the local rotational speed. It's then converted into radians per second for calculations involving centrifugal force. The steps include:

• Converting km/h to meters per second using \(\frac{1000}{3600}\).
• Converting to radians per second using \(\frac{2 \pi}{24 \times 3600}\).

This allows one to work comfortably within the system used in physics formulas.

Depth Depression Calculation

When calculating how much the center of a lake is "depressed" or pushed down due to Earth's rotation, we primarily deal with centrifugal force. Centrifugal force acts outward from Earth's axis of rotation. It can make the lake's center appear lower than the shore. The formula used is:

• \(d = \frac{F_{c}}{F_{ge}} \times r\)

In this formula:

• \(d\) is the depth of depression.
• \(F_{c}\) is the centrifugal force (calculated from rotational speed, mass, and lake radius).
• \(F_{ge}\) is the effective gravitational force (the net force considering both gravity and the outward centrifuge effect).
• \(r\) is the lake's radius.

By putting calculated values of forces into this formula, you can discover the depth by which the lake's center dips compared to the shoreline. Understanding how different forces affect physical geography gives insight into natural phenomena.