Amanda567
Knowing that the diameter of the Earth at the equator is 12 740 km, compute the radial acceleration of a point on the surface of the Earth at the equator, due to the rotation of the Earth about its axis.

## Answers and Replies

RTW69
Please make an attempt at this problem as part of your request. Hint: You are looking for the acceleration directed to the center of the rotation.

What equations do you think might be useful for circular motion?

Amanda567
Sorry!
I know that centripetal acceleration is v^2/r
The diameter of the earth is 12740 km, and half of that is the radius of 6370km.
How would I find v?

RTW69
You are getting closer. V is the tangential velocity, what is the equation for that?

Amanda567
Would that be the period T, where that is the time to complete one revolution?
T=2pi(r)/v

RTW69
OK, Vt or tangential velocity = radius* angular velocity

What is the definition of angular velocity? You will need the 2*pi for sure, what else?

Amanda567
Angular velocity is what represents speed?
by: F=d/dt (mv)

RTW69
Angular velocity or w= the change in the angle of rotation/ change in time

Think about our earth making one complete rotation (how many radians is that?) divided by how long it takes to make that rotation in one day. Put that time in seconds.

Since you know the angular velocity multiply it by the radius to get Vt, the tangential velocity. Plug that value into V^2/r to get centripetal acceleration.

I get 3.33 E-2 m/s^2

• muhammed_oli
Amanda567
What is E-2?

RTW69
That means 3.33x10^-2 m/s^2 or 0.0333 m/s^2

Amanda567
w=2pi/86400sec
w=0.00007
0.00007(6370km)=0.46324km/sec
vt=0.46324km^2/6370
vt=0.0000336

Thats what i get, what am i doing wrong?

RTW69
Change KM to meters which means you multiply your final result by 1000

Amanda567
thanks so much!!