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 65.0 ° north of the equator.

Answer 1

(a) v = 463.97 m/s,
`a_(c) = 0.0337\ m/s^(2)`

(b) v' = 196.02 m/s,
`a'_(c) = 0.0143\ m/s^(2)`

Solution:

As per the question:

Time period of the rotation of earth, T = 24 h =
`24* 3600 = 86400\ s`

Radius of the earth, R =
`6.38* 10^(6)\ m`

Angle,
`\theta = 65.0^(\circ)`

Now,

Angular velocity is given by:

`\omega = (2\pi)/(T) = (2\pi)/(86400) = 7.27* 10^(- 5) \rad`

(a) To calculate the speed and the centripetal acceleration at the equator:

Linear velocity or the speed, v =
`R\omega`

`v = 6.38* 10^(6)* 7.27* 10^(- 5) = 463.967\ m/s`

Centripetal Acceleration,
`a_(c) = (v^(2))/(R)`

`a_(c) = (463.967^(2))/(6.38* 10^(6)) = 0.0337\ m/s^(2)`

(b) To calculate the speed and acceleration at an altitude of
`65.0^(\circ)` N:

The horizontal component of the radius, R' =
`Rcos\theta`

`R' = 6.38* 10^(6)cos65.0^(\circ) = 2.69* 10^(6)\ m`

Now,

For the speed, v' =
`R'\omega = 2.69* 10^(6)* 7.27* 10^(- 5) = 196.02\ m/s`

For the centripetal acceleration,

`a'_(c) = (v'^(2))/(R')`

`a'_(c) = (196.02^(2))/(2.69* 10^(6)) = 0.01428\ m/s^(2)`