Question

We found the centripetal acceleration of the Earth as it revolves around the Sun. 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.)

Answer 1

Answer:

The centripetal acceleration by the rotation of the Earth is 3.31 10⁻² m/s²

Explanation:

The equation for centripetal acceleration is

     $a_{c}$ = v² / r

Where magnitude v is the distance it travels in a complete rotation; distance is the distance of a circle

       V = d / t = 2π r /T

With R the radius of the earth and T the period (1 day)

       T = 1 day (24h / 1day) 3600s / 1h)

       T = 86400 s

      V = 2 π 6.371 10⁶/86400

      V = 4.6 10² m / s

       $a_{c}$ = (4.6 10²)² / 6.371 10⁶

       $a_{c}$ = 3.31 10⁻² m/s²

The centripetal acceleration by the rotation of the Earth is 3.31 10⁻² m/s²

Answer 2

This question involves the concepts of centripetal acceleration and linear speed.

The centripetal acceleration of a point on the surface of the Earth at the equator caused by rotation of the Earth about its axis is "0.033 m/s²".

The centripetal acceleration is given by the following formula:

$a=\frac{v^2}{r}$

where,

a = centripetal acceleration = ?

r = radius = 6371 km = 6371000 m

T = time for a single rotation = (24 hrs)(3600 s/1 hr) = 86400 s

s = distance travelled in single rotation = 2πr = 2π(6371000 m)

s = 40,030,173.6 m

v = linear speed = $\frac{s}{T}=\frac{40,030,173.6\ m}{86400\ s}=463.31\ m/s$

Therefore,

$a=\frac{(463.31\ m/s)^2}{6371000\ m}$

a = 0.033 m/s²

Learn more about centripetal acceleration here:

https://brainly.com/question/17689540?referrer=searchResults