# Homework Help: Centripetal Acceleration of Earth

1. Sep 16, 2007

### shell4987

1. The problem statement, all variables and given/known data
(a) What is the magnitude of the centripetal acceleration of an object on Earth's equator owing to the rotation of Earth? (b) What would the period of rotation of Earth (in minutes) have to be for objects on the equator to have a centripetal acceleration with a magnitude of 9.80 m/s2?

2. Relevant equations
a=v squared/r and T=(2(pi)(r))/v

3. The attempt at a solution
How do I find out the radius of the Earth? Let alone the velocity of the Earth? I think if I knew those two that maybe I could solve this problem.

2. Sep 16, 2007

### Staff: Mentor

Look it up!

Once you have the radius, you should be able to figure out the speed of a point on the equator.

3. Sep 16, 2007

### nrqed

They assume that you will look up the radius of the Earth in your book. You have to calculate the speed using the fact that you know the period of rotation!

4. Sep 16, 2007

### shell4987

Okay, I looked up all of my information in the book and got the radius to be 6.37e6 and the velocity to be 11.2km/s, I attempted part (a) and got the answer to be 2.0e13m/s squared and that came out to be wrong, what did i do wrong for that, i used the a=v squared/r formula with 11200 m/s as velocity and 6.37e6m as the radius? Am I doing something wrong here?

And for part (b) I solved it and got it correct. Thank you.

5. Sep 16, 2007

### Staff: Mentor

Show how you calculated the velocity. Even if you assume that speed, how did you calculate an acceleration of 2.0e13!

6. Sep 16, 2007

### shell4987

I converted 11.2km/s (the velocity of the Earth) into meters by multiplying it by 1000, therefore getting 11200 m/s, then i put that into the a= v squared/r formula and used r as 6.37e6m... ahh i don't know what i'm doing wrong!

7. Sep 16, 2007

### Staff: Mentor

You want the rotational speed of the earth's surface. Figure it out using period (one day) and the earth's radius.