KEY_QZ4_SPS3010_Fall 2020-Gravity and/or Magnetic Fields
(a) [3 points] What is the “apparent” Polar Wander Path? How one can determine this path for a continent? What major
assumption(s) must be made to construct the path?
Apparent PWP: The apparent path of Earth’s magnetic poles (or magnetic axis) over a long
period of time inferred from the magnetization of rocks of different ages. The magnetic field did
not WANDER, the tectonic plates housing rocks moved over eons.
Determining PWP: Determine paleopole’s Lat & Long for rocks of various ages on the same
continent. Place these points on a map and connect them. This curve/path is the PWP for that
continent.
̅ = 0 (i.e. the
Major Assumption(s): (1) time-averaged for the declination angle is ZERO i.e. 𝐷
average location of the earth’s magnetic axis is along the Earth’s spin axis); (2) Major component
of magnetic field has always been a dipole field (so you are justified to use 𝑡𝑎𝑛𝐼 = 2𝑡𝑎𝑛𝜆 ); (3)
the dynamics of tectonics plates is possible. #1 is the most important assumption.
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(b) [5 points] Magnetic measurements were made on a rock sample found at 47 N, 20 E. The angle of inclination of
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the remanent magnetization on this sample is 30 , and the direction of its magnetization (angle of declination) is D
= 800. Determine the position (latitude & longitude) of its palaeomagnetic pole. What does this indicate about the
mother-continent on which this rock was formed? At the end of your calculations, please make sure to draw a
schematic of the rock’s past motion over the globe. NOTE> Identify the parameters clearly and show ALL steps
including the needed test to select the right paleolongitude.
𝜆𝑖𝑠 = 47, 𝜙𝑖𝑠 = +20 , 𝐷𝑖𝑠 = +80 , 𝐼𝑤𝑎𝑠 = +30 → tan(+30) = 2 tan 𝜆𝑤𝑎𝑠 → 𝜆𝑤𝑎𝑠 = +16.1
sin 𝜆𝑃 = 𝑠𝑖𝑛𝜆𝑖𝑠 𝑠𝑖𝑛𝜆𝑤𝑎𝑠 + cos 𝜆𝑖𝑠 cos 𝜆𝑤𝑎𝑠 cos 𝐷𝑖𝑠
sin 𝜆𝑃 = sin (47) sin (16.1) + cos(47 ) cos(16.1) cos(+80) → 𝜆𝑃 = +18.46 N
Test: sin 𝜆𝑤𝑎𝑠 ? sin 𝜆𝑃 sin 𝜆𝑖𝑠 → sin(16.1) ? sin(18.46) sin(47) → 0.277 > 0.228 ⇒ sin 𝜆𝑤𝑎𝑠 > sin 𝜆𝑃 sin 𝜆𝑖𝑠
cos 𝜆𝑤𝑎𝑠 sin 𝐷𝑖𝑠
cos(16.1) sin(80)
sin(𝜙𝑃 − 𝜙𝑖𝑠 ) =
→ sin(𝜙𝑃 − (+20)) =
→ 𝜙𝑃 = +106 = 106 E
cos 𝜆𝑃
cos(18.46)
The rock was solidified at ~16 N, then it moved +47 N. Its path was a small circle (latitude of
rotation) about the apparent paleopole (location of the paleo-pole is 18.5 N & ~106 E.
PM
(c) [2 points] If the measurement of the angle of inclination of the rock sample of Part “b” is in error by 5 degrees, what
is the subsequent error in the calculated palaeolatitude?
tan 𝐼 = 2 tan 𝜆 → 𝛿(tan 𝐼) = 𝛿(2 tan 𝜆) → |
1 cos 𝜆 2
1
𝛿𝜆
3.1
𝑑(tan 𝐼)
𝑑(tan 𝜆)
𝛿𝐼
2 𝛿𝜆
| 𝛿𝐼 = 2 |
| 𝛿𝜆 →
=
2
𝑑𝐼
𝑑𝜆
𝑐𝑜𝑠 𝐼 𝑐𝑜𝑠 2 𝜆
cos (16.1°) 2
𝛿𝜆 = 2 ( cos 𝐼 ) 𝛿𝐼 → 𝛿𝜆 = 2 ( cos (30°) ) (5°) = 3.08° → 𝛿𝜆 ≅ 3.1°
OR %𝛿𝜆 = 𝜆 × 100 = 16.1 × 100 = %19.3
∴ 𝐼 = 30° ± 5° = 30° ± %16.7 ➔ 𝜆 = 16.1° ± 3.1° = 30° ± %19.3.
EXTRA CREDIT (2 POINTS):
Consider the total gravitational potential 𝑈(𝑟, 𝜃) up to its 𝐽2 -term for a rotating spheroidal planet of mass 𝑀 and
the equatorial and polar radii of 𝑎 and 𝑐.
(a) Show that the equatorial radius of the planet is: 𝑎 = −
𝐺𝑀
𝑈0
(1 +
1
2
𝐽2 −
1
2
𝑞), where 𝑈0 is the total gravitational
potential on the planet’s surface at the equator. The parameter 𝑞 (or 𝑚) is the geodynamical constant of the
planet.
𝐺𝑀 𝐺𝑀𝑎2
1
+
𝐽2 (3 𝑐𝑜𝑠 2 𝜃 − 1) + 𝜔2 𝑟 2 (1 − 𝑐𝑜𝑠 2 𝜃)
3
𝑟
2𝑟
2
𝜔2 𝑎 3
𝑈0 ≡ 𝑈(𝑟 = 𝑎, 𝜃 = 90) and 𝑞 ≡ 𝐺𝑀
𝐺𝑀 𝐺𝑀𝑎2
1
𝑈0 = 𝑈(𝑟 = 𝑎, 𝜃 = 90) = −
+
𝐽2 (3 𝑐𝑜𝑠 2 90 − 1) + 𝜔2 𝑎2 (1 − 𝑐𝑜𝑠 2 90)
3
𝑎
2𝑎
2
𝐺𝑀 𝐺𝑀
1 2 2
𝐺𝑀
1
1 𝜔2 𝑎3
𝐺𝑀
1
1
𝑈0 = 𝑈(𝑟 = 𝑎, 𝜃 = 90) = −
−
𝐽2 + 𝜔 𝑎 = −
+
𝐽
−
(1 + 𝐽2 − 𝑞)
(1
)=−
2
𝑎
2𝑎
2
𝑎
2
2 𝐺𝑀
𝑎
2
2
𝑈 = 𝑈(𝑟, 𝜃) = −
𝐺𝑀
1
1
𝑎 = − 𝑈 (1 + 2 𝐽2 − 2 𝑞)
0
(b) & (c) were NOT part of this quiz.