Question

a. Compute the centripetal acceleration of a point on the surface of the Earth at the equator caused by the rotation of the Earth about its axis. Enter the magnitude. The radius of the Earth is 6,371 km. The period is one day. m/s2 b. Suppose the Earth were spinning much faster. Find the period that results in the centipetal acceleration being equal to 9.8 m/s?. minutes When the centripetal acceleration equals the acceleration due to gravity, objects are in orbit and we would feel weightless. Luckily, since the actual centripetal acceleration is much less than 9.8 m/s, we don't need to worry about being flung off the spinning Earth.

Answer 1

Centripetal force is given as ,

${\displaystyle F_{c}=mr\omega ^{2}\,.}$

and relation between angular velocity and time period is ;

$\omega ={{2\pi } \over T}={2\pi f},$

We will utilize these relations to calculate what is desired.

I will attach the solution.

If any doubt, feel free to ask.

P.S. : A feedback is much appreciated.

Centripetal force on om obiect moning with angular relocity "w" in a circle of reading "se" is given as = m* - Centripetal acceleration - ac=1w² 21= 6371 km = 6371*103 m Ty period of earth's rotation = 24 hours = 864008 (1) Ac=r (25)² = 6371x108x un 2 86400 x 86400 0.034 ms² b) let's say desired period is to from relation in ean (2) aé = 4 (2) 9.8 = (6371* 189) **) so 1976)* Canne ses environ 6371103 9.8 637 LX3 = 27 6371 103 9.8 T = 5063.498 = 11:406 hours