Chapter 18, Problem 9PEB

A survey of a mid-oceanic ridge determined the distance from the ridge axis to the reversal event as shown in the following table. What is the rate of seafloor spreading at this ridge?

Magnetic Polarity	Reversal Event Age (106 Years)	Distance from Ridge Axis to Reversal Episode (km)
Normal	0.715	87.5
Reverse	0.850	109.7
Normal	1.015	119.8
Reverse	1805	211.2
Normal	2.055	246.6

To determine

The rate of seafloor spreading at the given ridge data.

Answer to Problem 9PEB

Solution:

$12.11\text{ cm/yr}$.

Explanation of Solution

Given data:

The following data table has been provided:

Magnetic Polarity	Reversal Event Age (106 Years)	Distance from Ridge Axis to Reversal Episode (km)
Normal	0.715	87.5
Reverse	0.850	109.7
Normal	1.015	119.8
Reverse	1.805	211.2
Normal	2.055	246.6

Formulae used:

The speed of the moving plate can be determined by the relation between distance velocity and time:

$v=\frac{d}{t}$ …… (1)

Here, $d$ is the distance covered and $t$ is the time taken.

Writing an expression for the average velocity if the time duration for each velocity is same:

$v_{avg}=\frac{v_1+v_2+v_3+\dots +v_n}{n}$

Explanation:

The average speed can be calculated by taking the sum of all the velocities and diving by the number of observations.

$\overline{v}=\frac{v_{0.715}+v_{0.85}+v_{1.015}+v_{1.805}+v_{2.055}}{5}$

Here, $v_{0.715}$, $v_{0.85}$, $v_{1.015}$, $v_{1.805}$ and $v_{2.055}$ are the velocities at different time frames.

Case 1

Convert 87.5 km to cm.

$\begin{array}{c}d=87\text{km}\left(\frac{{10}^5\text{ cm}}{1\text{km}}\right)\\ =8.7\times {10}^6\text{ cm}\end{array}$

Hence 87 km in cm is $8.7\times {10}^6\text{ cm}$.

Substituting $8.7\times {10}^6\text{ cm}$ for $d$ and $0.715\times {10}^6\text{ years}$ for $t$ in equation (1).

$\begin{array}{c}{v}_{0.715}=\frac{8.7\times {10}^6\text{ cm}}{0.715\times {10}^6\text{ yr}}\\ =12.16\text{ cm/yr}\end{array}$

Hence the velocity is $12.16\text{ cm/yr}$.

Case 2

Convert 109.7 km to cm.

$\begin{array}{c}d=109.7\text{km}\left(\frac{{10}^5\text{ cm}}{1\text{km}}\right)\\ =1.097\times {10}^7\text{ cm}\end{array}$

Hence 109.7 km in cm is $1.097\times {10}^7\text{ cm}$.

Substituting $1.097\times {10}^7\text{ cm}$ for $d$ and $0.85\times {10}^6\text{ years}$ for $t$ in equation (1).

$\begin{array}{c}{v}_{0.85}=\frac{1.097\times {10}^7\text{ cm}}{0.85\times {10}^6\text{ yr}}\\ =12.9\text{ cm/yr}\end{array}$

Hence the velocity is $12.9\text{ cm/yr}$.

Case 3

Convert 119.8 km to cm.

$\begin{array}{c}d=119.8\text{km}\left(\frac{{10}^5\text{ cm}}{1\text{km}}\right)\\ =1.198\times {10}^7\text{ cm}\end{array}$

Hence 119.8 km in cm is $1.198\times {10}^7\text{ cm}$.

Substitute $1.198\times {10}^7\text{ cm}$ for $d$ and $1.015\times {10}^6\text{ years}$ for $t$ in equation (1).

$\begin{array}{c}{v}_{1.015}=\frac{1.198\times {10}^7\text{ cm}}{1.015\times {10}^6\text{ yr}}\\ =11.8\text{ cm/yr}\end{array}$

Hence the velocity is $11.8\text{ cm/yr}$.

Case 4

Convert 211.2 km to cm.

$\begin{array}{c}d=211.2\text{km}\left(\frac{{10}^5\text{ cm}}{1\text{km}}\right)\\ =2.112\times {10}^7\text{ cm}\end{array}$

Hence 211.2 km in cm is $2.112\times {10}^7\text{ cm}$.

Substituting $2.112\times {10}^7\text{ cm}$ for $d$ and $1.805\times {10}^6\text{ yr}$ for $t$ in equation (1).

$\begin{array}{c}{v}_{1.805}=\frac{2.112\times {10}^7\text{ cm}}{1.805\times {10}^6\text{ yr}}\\ =11.7\text{ cm/yr}\end{array}$

Hence the velocity is $11.7\text{ cm/yr}$.

Case 5

Convert 246.6 km to cm.

$\begin{array}{c}d=246.6\text{km}\left(\frac{{10}^5\text{ cm}}{1\text{km}}\right)\\ =2.466\times {10}^7\text{ cm}\end{array}$

Hence 211.2 km in cm is $2.466\times {10}^7\text{ cm}$.

Substituting $2.466\times {10}^7\text{ cm}$ for $d$ and $2.055\times {10}^6\text{ yr}$ for $t$ in equation (1).

$\begin{array}{c}{v}_{2.055}=\frac{2.466\times {10}^7\text{ cm}}{2.055\times {10}^6\text{ yr}}\\ =12\text{ cm/yr}\end{array}$

Hence the velocity is $12\text{ cm/yr}$.

Calculate the average velocity.

Substituting $12.16\text{ cm/yr}$ for $v_{0.715}$, $12.9\text{ cm/yr}$ for $v_{0.85}$, $11.8\text{ cm/yr}$ for $v_{1.015}$, $11.7\text{ cm/yr}$ for $v_{1.805}$ and $12\text{ cm/yr}$ for $v_{2.055}$ in equation (2).

$\begin{array}{c}\overline{v}=\frac{12.16\text{ cm/yr}+12.9\text{ cm/yr}+11.8\text{ cm/yr}+11.7\text{ cm/yr}+12\text{ cm/yr}}{5}\\ =12.11\text{ cm/yr}\end{array}$

Conclusion:

Hence, the velocity of the plate moving is $12.11\text{ cm/yr}$.