Question

What is the linear speed of a point (a) on the equator, (b) on theArctic Circle (latitude 66.5o N), and (c) at a latitudeof 45.0 degrees N, due to the Earth's rotation?
For part (a), I used the equation w = A/t to represent that theEarth rotates once a day on its axis. A point on the equatorthus rotates through 2πradians in 24 hours, so
w = A/t = (2π) / (24 hr) = (6.28 rad) / (24hr x 3600 s/hr) = 7.27 x 10-5 rad/s
Since the radian is a unitless measure it is not necessary toinclude it in the units for a quantity.
The radius of the Earth is re = 6.4 x 106 m,so:
v = wr = (7.27 x 10-5 rad/s)(6.4 x106 m) = 465 m/s = 1674 km/hr

Answer 1

Answer:

a)   v = 465.3 m / s , b)     v = 196.3 m / s , c)  v =327.2 m/s

Explanation:

Linear and angular magnitudes are related.

               v = w r

Where v and w are the linear and angular velocities, respectively and r is the radius of the circle.

Let's look for angular velocity

In rotated angle is a rotation of the Earth is 2π rad and the Time spent Period is24 h

Let's reduce the hours to seconds

       T = 24 h (3600 s / 1 h) = 86,400  s

       w = 2π / 68400

       w = 7.27 10⁻⁵ rad / s

Let's look for linear speeds

a) in Ecuador the radius of the circle is the radius of the planet

      r =  $R_{e}$ = 6.4 10⁶ m

      v = 7.27 10⁻⁵ 6.4 10⁶

      v = 465.3 m / s

b) As the sphere reduces the size of the circle, let's use trigonometry to find the triangle leg, the angle is measured from the x-axis (East

       cos 65 = r /  $R_{e}$

       r = $R_{e}$ cos 65

       r = 6.4 10⁶ cos 65

       r = 2.7 10⁶ m

  We calculate the linear speed

       v = 7.27 10⁻⁵ 2.7 10⁶

       v = 196.3 m / s

c) at an angle of 45

        r = 6.4 10⁶ cos45

        r = 4.5 10⁶ m

        v = 7.27 10⁻⁵ 4.5 10⁶

        v =327.2 m/s