Answer to Question #245770 in Other for Mohammed

Question

(Acceleration due to rotation and orbit of the Earth) A body is at rest at a location on the Earth's equator. Find its acceleration due to the Earth's rotation.[Take the Earth's radius at the equator to be 6400 km.] Find also the acceleration of the Earth in its orbit around the sun.[Take the Sun to be fixed and regard the Earth as a particle following a circular path with centre the Sun and radius 15*10¹0m.]

Answer 1

$$a=\omega^2R=(\dfrac{2\pi}{T})^2R$$

a)

$$R=6400\ km=6.4\times10^6\ m$$

$$T=24\ h=24\times3600\ s=8.64\times10^4\ s$$

$$a=(\dfrac{2\pi}{8.64\times10^4\ s})^2(6.4\times10^6\ m)$$

$$=0.033846\ m/s^2$$

$a=0.033846\ m/s^2$ directed towards the centre of the Earth.

b)

$$R=15\times10^{10}\ m$$

$$T=365\times24\ h=365\times24\times3600\ s=31536000\ s$$

$$a=(\dfrac{2\pi}{31536000\ s})^2(15\times10^{10}\ m)$$

$$=0.005954\ m/s^2$$

$a=0.005954\ m/s^2$ directed towards the centre of the Sun.