Question

The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the plane of the equator. Assuming the earth is a sphere with a radius of 6.38 x 106 m, determine the speed and centripetal acceleration of a person situated (a) at the equator and (b) at a latitude of 61.0 ° north of the equator.

Answer 1

given,

Radius of sphere = 6.38 × 10⁶ m

time = 1 day = 86400 s

`\omega = (2\pi)/(T)`

`\omega = (2\pi)/( 86400 )`

`\omega = 7.272 * 10^(-5)\ rad/s`

a) at equator

`v = R_E \omega`

`v = 6.38 * 10^6* 7.272 * 10^(-5)`

v = 464 m/s

acceleration of the person

`a = \omega^2R_E`

`a = (7.272 * 10^(-5))^2(6.38 * 10^6)`

`a = 0.03374 m/s^2`

b) at a latitude of 61.0 ° north of the equator.

`R = R_E cos \theta`

`R = 6.38 * 10^6* cos 61^0`

`R = 3.093 * 10^6 m`

`v = R \omega`

`v = 3.093 * 10^6 * 7.272 * 10^(-5)`

`v = 225 m/s`

acceleration of the person

`a = \omega^2R_E`

`a = (7.272 * 10^(-5))^2(3.093 * 10^6 )`

`a = 0.01635 m/s^2`