# KEY QZ4 SPS5030 Fall 2020 Gravity and Magnetic Fields-3

```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 &amp; 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.
Μ = 0 (i.e. the
Major Assumption(s): (1) time-averaged for the declination angle is ZERO i.e. π·
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 π‘πππΌ = 2π‘πππ ); (3)
the dynamics of tectonics plates is possible. #1 is the most important assumption.
o
o
(b) [5 points] Magnetic measurements were made on a rock sample found at 47 N, 20 E. The angle of inclination of
o
the remanent magnetization on this sample is 30 , and the direction of its magnetization (angle of declination) is D
= 800. Determine the position (latitude &amp; 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&gt; Identify the parameters clearly and show ALL steps
including the needed test to select the right paleolongitude.
πππ  = 47, πππ  = +20 , π·ππ  = +80 , πΌπ€ππ  = +30 → tan(+30) = 2 tan ππ€ππ  → ππ€ππ  = +16.1
sin ππ = π πππππ  π ππππ€ππ  + cos πππ  cos ππ€ππ  cos π·ππ
sin ππ = sin (47) sin (16.1) + cos(47 ) cos(16.1) cos(+80) → ππ = +18.46 N
Test: sin ππ€ππ  ? sin ππ sin πππ  → sin(16.1) ? sin(18.46) sin(47) → 0.277 &gt; 0.228 ⇒ sin ππ€ππ  &gt; sin ππ sin πππ
cos ππ€ππ  sin π·ππ
cos(16.1) sin(80)
sin(ππ − πππ  ) =
→ sin(ππ − (+20)) =
→ ππ = +106 = 106 E
cos ππ
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 &amp; ~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 πΌ = 2 tan π → πΏ(tan πΌ) = πΏ(2 tan π) → |
1 cos π 2
1
πΏπ
3.1
π(tan πΌ)
π(tan π)
πΏπΌ
2 πΏπ
| πΏπΌ = 2 |
| πΏπ →
=
2
ππΌ
ππ
πππ  πΌ πππ  2 π
cos (16.1&deg;) 2
πΏπ = 2 ( cos πΌ ) πΏπΌ → πΏπ = 2 ( cos (30&deg;) ) (5&deg;) = 3.08&deg; → πΏπ ≅ 3.1&deg;
OR %πΏπ = π &times; 100 = 16.1 &times; 100 = %19.3
∴ πΌ = 30&deg; &plusmn; 5&deg; = 30&deg; &plusmn; %16.7 β π = 16.1&deg; &plusmn; 3.1&deg; = 30&deg; &plusmn; %19.3.
EXTRA CREDIT (2 POINTS):
Consider the total gravitational potential π(π, π) up to its π½2 -term for a rotating spheroidal planet of mass π and
the equatorial and polar radii of π and π.
(a) Show that the equatorial radius of the planet is: π = −
πΊπ
π0
(1 +
1
2
π½2 −
1
2
π), where π0 is the total gravitational
potential on the planet’s surface at the equator. The parameter π (or π) is the geodynamical constant of the
planet.
πΊπ πΊππ2
1
+
π½2 (3 πππ  2 π − 1) + π2 π 2 (1 − πππ  2 π)
3
π
2π
2
π2 π 3
π0 ≡ π(π = π, π = 90) and π ≡ πΊπ
πΊπ πΊππ2
1
π0 = π(π = π, π = 90) = −
+
π½2 (3 πππ  2 90 − 1) + π2 π2 (1 − πππ  2 90)
3
π
2π
2
πΊπ πΊπ
1 2 2
πΊπ
1
1 π2 π3
πΊπ
1
1
π0 = π(π = π, π = 90) = −
−
π½2 + π π = −
+
π½
−
(1 + π½2 − π)
(1
)=−
2
π
2π
2
π
2
2 πΊπ
π
2
2
π = π(π, π) = −
πΊπ
1
1
π = − π (1 + 2 π½2 − 2 π)
0
(b) &amp; (c) were NOT part of this quiz.
```