The earth’s radius is 6.37×106m; it rotates once every 24 hours. What is the earth’s angular speed? What is the speed of a point on the equa

Question

The earth’s radius is 6.37×106m; it rotates once every 24 hours. What is the earth’s angular speed? What is the speed of a point on the equator? What is the speed of a point on the earth’s surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator? (Hint: what is the radius of the circle in which the point moves?)

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Sapo 5 years 2021-08-10T17:10:49+00:00 1 Answers 761 views 0

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  1. Xavia
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    2021-08-10T17:11:59+00:00
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    Answer:

    a) w = 7.27 * 10^-5 rad/s

    b) v1 = 463.1 m/s

    c) v1 = 440.433 m/s

    Explanation:

    Given:-

    – The radius of the earth,  R = 6.37 * 10 ^6 m

    – The time period for 1 revolution T = 24 hrs

    Find:

    What is the earth’s angular speed?

    What is the speed of a point on the equator?

    What is the speed of a point on the earth’s surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator?

    Solution:

    – The angular speed w of the earth can be related with the Time period T of the earth revolution by:

                                      w = 2π / T

                                      w = 2π / 24*3600

                                      w = 7.27 * 10^-5 rad/s

    – The speed of the point on the equator v1 can be determined from the linear and rotational motion kinematic relation.

                                     v1 = R*w

                                     v1 = (6.37 * 10 ^6)*(7.27 * 10^-5)

                                     v1 = 463.1 m/s

    – The angle θ subtended by a point on earth’s surface 1/5 th between the equator and the pole wrt equator is.

                                     π/2  ……….. s

                                     x     ………… 1/5 s

                                     x = π/2*5 = 18°    

    – The radius of the earth R’ at point where θ = 18° from the equator is:

                                    R’ = R*cos(18)

                                    R’ = (6.37 * 10 ^6)*cos(18)

                                    R’ = 6058230.0088 m

    – The speed of the point where θ = 18° from the equator v2 can be determined from the linear and rotational motion kinematic relation.

                                  v2 = R’*w

                                  v2 = (6058230.0088)*(7.27 * 10^-5)

                                  v2 = 440.433 m/s

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