What is the linear speed, due to the Earth’s rotation, of a point (a) on the equator, (b) on the Arctic Circle (latitude 66.5° N), and (c) at a latitude of 42.0° N?

The term "angular velocity" in physics may be defined as the rate at which a particle shifts its angular position from one position to another with respect to time.

(a)

Every point on the Earth rotates with the same angular speed as that of the Earth. As the Earth makes one revolution around its axis in\(T = 1\;{\rm{day}}\), it turns through an angle of\(\Delta \theta = 2\pi \;{\rm{rad}}\).

Draw a diagram showing the rotation of the Earth about its axis.

From the above diagram, we can see that the radius of the circular path varies with the location. And

\(r = R\cos \theta \).

Here,\(R\)is the radius of the Earth; its value is\(6.38 \times {10^6}\;{\rm{m}}\);\(r\)is the radius at the latitude\(\theta \).

The linear speed of the point on the equator can be calculated as follows:

\(\begin{aligned}{v_{\rm{e}}} &= r\omega \\{v_{\rm{e}}} &= R\left( {\frac{{2\pi \;{\rm{rad}}}}{T}} \right)\\{v_{\rm{e}}} &= \left( {6.38 \times {{10}^6}\;{\rm{m}}} \right)\left({\frac{{2\pi\;{\rm{rad}}}}{{1\;{\rm{day}}}}}\right)\left({\frac{{1\;{\rm{day}}}}{{86400\;{\rm{s}}}}}\right)\\{v_{\rm{e}}} &= 464\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)

Thus, the linear speed of the point on the equator is\(464\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}}{\rm{s}}}} \right.}{\rm{s}}}\).

(b)

The linear speed of the point on theArctic Circle at latitude 66.5° N can be calculated as follows:

\(\begin{aligned}{l}{\left( {{v_{\rm{a}}}} \right)_{\theta = 66.5^\circ }} &= r\omega \\{\left( {{v_{\rm{a}}}} \right)_{\theta = 66.5^\circ }} &= R\cos \theta \left( {\frac{{2\pi \;{\rm{rad}}}}{T}} \right)\\{\left( {{v_{\rm{a}}}} \right)_{\theta = 66.5^\circ }} &= \left( {6.38 \times {{10}^6}\;{\rm{m}}} \right)\cos \left( {66.5^\circ } \right)\left( {\frac{{2\pi \;{\rm{rad}}}}{{1\;{\rm{day}}}}} \right)\left( {\frac{{1\;{\rm{day}}}}{{86400\;{\rm{s}}}}} \right)\\{\left( {{v_{\rm{a}}}} \right)_{\theta = 66.5^\circ }} &= 185\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)

Thus, the linear speed of the point on the Arctic Circle at latitude 66.5° N is \(185\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\).

The linear speed of the point on theArctic Circle at latitude 42.0° N can be calculated as follows:

\(\begin{aligned}{l}{\left( {{v_{\rm{a}}}} \right)_{\theta = 42.0^\circ }} &= r\omega \\{\left( {{v_{\rm{a}}}} \right)_{\theta = 42.0^\circ }} &= R\cos \theta \left( {\frac{{2\pi \;{\rm{rad}}}}{T}} \right)\\{\left( {{v_{\rm{a}}}} \right)_{\theta = 42.0^\circ }} &= \left( {6.38 \times {{10}^6}\;{\rm{m}}} \right)\cos \left( {42.0^\circ } \right)\left( {\frac{{2\pi\;{\rm{rad}}}}{{1\;{\rm{day}}}}}\right)\left({\frac{{1\;{\rm{day}}}}{{86400\;{\rm{s}}}}} \right)\\{\left( {{v_{\rm{a}}}} \right)_{\theta = 42.0^\circ }} &= 345\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)

Thus, the linear speed of the point on the Arctic Circle at latitude 42.0° N is \(345\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\).