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

Answer

given,

Radius of sphere = 6.38 × 10⁶ m

time  = 1 day = 86400 s

$\omega = \dfrac{2\pi}{T}$

$\omega = \dfrac{2\pi}{ 86400 }$

$\omega = 7.272 \times 10^{-5}\ rad/s$

a) at equator

$v = R_E \omega$

$v = 6.38 \times 10^6\times 7.272 \times 10^{-5}$

v = 464 m/s

acceleration of the person

$a = \omega^2R_E$

$a = (7.272 \times 10^{-5})^2(6.38 \times 10^6)$

$a = 0.03374 m/s^2$

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

$R = R_E cos \theta$

$R = 6.38 \times 10^6\times cos 61^0$

$R = 3.093 \times 10^6 m$

$v = R \omega$

$v = 3.093 \times 10^6 \times 7.272 \times 10^{-5}$

$v = 225 m/s$

acceleration of the person

$a = \omega^2R_E$

$a = (7.272 \times 10^{-5})^2(3.093 \times 10^6 )$

$a = 0.01635 m/s^2$