Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The measured declination and inclination of the paleomagnetic field in Lower Cretaceous rocks at \(45.5^{\circ} \mathrm{N}\) and \(73^{\circ} \mathrm{W}\) are \(D=154^{\circ}\) and \(I=-58^{\circ}\). Determine the paleomagnetic pole position.

Short Answer

Expert verified
The paleomagnetic pole position is approximately \(33.6^{\circ} \mathrm{S}\) and \(47^{\circ} \mathrm{E}\).

Step by step solution

01

Calculate the Paleolatitude

Use the inclination value to find the paleolatitude using the formula \( \tan(I) = 2\tan(\text{paleolatitude}) \). Solve for paleolatitude: \( \text{paleolatitude} = \tan^{-1}\left(\frac{\tan(-58^{\circ})}{2}\right) \). Compute this to get the paleolatitude as approximately \(-33.6^{\circ}\).
02

Determine the Paleolongitude

The declination, \( D = 154^{\circ} \), needs to be adjusted by adding or subtracting \(180^{\circ}\) to align with the current location longitude. For \( D = 154^{\circ} \), the corrected value becomes \(154^{\circ} + 180^{\circ} \approx 334^{\circ}\). Adjust within the \([0^{\circ}, 360^{\circ}]\) range if needed. Since the site longitude is \(73^{\circ} \mathrm{W}\), the paleolongitude is \(73^{\circ} + (334^{\circ} - 180^{\circ}) = 227^{\circ} \approx 47^{\circ} \mathrm{E} \).
03

Locate the Paleomagnetic Pole Position

Combine the paleolatitude and paleolongitude to express the paleomagnetic pole position. Paleolatitude indicates a southern latitude, so the paleomagnetic pole position calculated is approximately \(33.6^{\circ} \mathrm{S}\) and \(47^{\circ} \mathrm{E}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Paleolatitude calculation
Paleolatitude calculation is an essential process used to determine the ancient latitude of a location based on the Earth's past magnetic field. It employs the idea that the inclination of the magnetic field can inform us about the latitude. The inclination, denoted by \(I\), is the angle made by the magnetic field with the horizontal.
The formula used here is \( \tan(I) = 2 \tan(\text{paleolatitude}) \). This formula expresses the relationship between the Earth's magnetic field inclination and the latitude at which rocks formed. To find the paleolatitude, the solution involves rearranging the formula to solve for paleolatitude:
\[ \text{paleolatitude} = \tan^{-1}\left(\frac{\tan(I)}{2}\right) \].
For the problem given, with an inclination of \(-58^{\circ}\), we compute:
\[ \text{paleolatitude} = \tan^{-1}\left(\frac{\tan(-58^{\circ})}{2}\right) \],
resulting in a paleolatitude of approximately \(-33.6^{\circ}\). This negative value suggests the location was south of the equator at that time.
Paleolongitude determination
Determining the paleolongitude involves adjusting the measured declination to reflect the ancient geographic position. The declination, \(D\), tells us how much the magnetic north has shifted from the geographic north at a specific location and can be used to derive the paleolongitude.
First, adjust the declination by accounting for the earth's current position. This is done by adding or subtracting \(180^{\circ}\) to line up with the present-day reference. For a measured declination of \(154^{\circ}\), the corrected declination is:
\(154^{\circ} + 180^{\circ} = 334^{\circ}\).

Since this value should fit within a \([0^{\circ}, 360^{\circ}]\) range, the corrected declination remains \(334^{\circ}\).
Next, use the site’s current longitude, in this case, \(73^{\circ} \text{W}\). Since west is negative in the conventional setting, the paleolongitude calculation becomes:
\[ 73^{\circ} + (334^{\circ} - 180^{\circ}) = 227^{\circ} \].
Within the west-east spectrum, this results in \(47^{\circ} \text{E}\), indicating that the ancient longitude was shifted eastward.
Paleomagnetic pole position
Paleomagnetic pole position is a method to locate where the magnetic north or south pole was on Earth in the geologic past. This helps scientists understand continental drift and plate tectonics by providing insights into how landmasses have shifted over millions of years.
In the context of the exercise, combining the previous calculations gives us the paleomagnetic pole's position. The paleolatitude calculation pointed to \(-33.6^{\circ}\), confirming a southern location. Similarly, the determined paleolongitude \(47^{\circ} \text{E}\) identified the eastward position on the globe.

Thus, the paleomagnetic pole was located approximately at \(33.6^{\circ} \text{S}\) and \(47^{\circ} \text{E}\). This indicates that the measured location was once positioned more southward and eastward relative to today, reflecting ancient continental positions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the TTT triple junction illustrated in Figure \(1-38 .\) This triple junction is acceptable because the relative velocity between plates \(\mathrm{C}\) and \(\mathrm{A}, \mathbf{u}_{\mathrm{CA}},\) is parallel to the trench in which plate \(\mathrm{B}\) is being subducted beneath plate \(\mathrm{C}\). The trench between plates \(\mathrm{C}\) and \(\mathrm{B}\) has an azimuth of \(180^{\circ}\) so that \(\mathbf{u}_{\mathrm{CA}}\) has an azimuth of \(0^{\circ}\); assume that \(u_{\mathrm{CA}}=\) \(50 \mathrm{~mm} \mathrm{yr}^{-1}\). Also assume that the azimuth and magnitude of \(\mathbf{u}_{B A}\) are \(315^{\circ}\) and \(60 \mathrm{~mm} \mathrm{yr}^{-1} .\) Determine the azimuth and magnitude of \(\mathbf{u}_{B C}\).

If we assume that the current rate of subduction, \(0.09 \mathrm{~m}^{2} \mathrm{~s}^{-1}\), has been applicable in the past, what thickness of sediments would have to have been subducted in the last 3 Gyr if the mass of subducted sediments is equal to one-half the present mass of the continents? Assume the density of the continents \(\rho_{c}\) is \(2700 \mathrm{~kg} \mathrm{~m}^{-3},\) the density of the sediments \(\rho_{s}\) is \(2400 \mathrm{~kg} \mathrm{~m}^{-3}\), the continental area \(A_{c}\) is \(1.9 \times 10^{8} \mathrm{~km}^{2},\) and the mean continental thickness \(h_{c}\) is \(35 \mathrm{~km}\)

At what depth will ascending mantle rock with a temperature of \(1600 \mathrm{~K}\) melt if the equation for the solidus temperature \(T\) is $$ T(K)=1500+0.12 p(\mathrm{MPa}) $$ Assume \(\rho=3300 \mathrm{~kg} \mathrm{~m}^{-3}, g=10 \mathrm{~m} \mathrm{~s}^{-2}\), and the man- tle rock ascends at constant temperature.

Consider an RRR triple junction of plates \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\). The ridge between plates \(\mathrm{A}\) and \(\mathrm{B}\) lies in a north-south direction (an azimuth of \(0^{\circ}\) with respect to the triple junction) and has a relative velocity of \(60 \mathrm{~mm} \mathrm{yr}^{-1}\). The ridge between plates \(\mathrm{B}\) and C has an azimuth of \(120^{\circ}\) with respect to the triple junction, and the ridge between plates \(\mathrm{A}\) and \(\mathrm{C}\) has an azimuth of \(270^{\circ}\) with respect to the triple junction. Determine the azimuths and magnitudes of the relative velocities between plates \(\mathrm{B}\) and \(\mathrm{C}\) and \(\mathrm{C}\) and \(\mathrm{A}\).

Determine the declination and inclination of the Earth's magnetic field at Boston \(\left(\phi=42.5^{\circ}, \psi=-71^{\circ}\right)\). Use the dipole approximation to the field, but do not assume that the geographic and magnetic poles coincide.

See all solutions

Recommended explanations on Geography Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free