# Centripetal Acceleration at Equator

1. May 24, 2010

### physics(L)10

1. The problem statement, all variables and given/known data
An individual standing at the equator:
a)What is this individuals centripetal acceleration caused by the earths rotation?
b)How fast would the earth need to spin in order for the centripetal acceleration to be equal to g?

T=1 day
radius of earth = 6.4x10^6 m

2. Relevant equations
a=v^2/r, v=2pir/T

3. The attempt at a solution
We need to convert the period from day to seconds and then plug in the information.

a=4pi^2r/T^2 = 4pi^2(6.4x10^6)/(1x60x60x60)

2. May 24, 2010

### ehild

How do you convert one day to seconds?

ehild

3. May 24, 2010

### physics(L)10

1 day x 24hrs/day x 60min/hr x 60 sec/min .... My bad, I put 60 instead of 24

4. May 24, 2010

### ehild

Well, and do not forget to square T.

ehild

5. May 24, 2010

### physics(L)10

So I'm right?

6. May 24, 2010

### physics(L)10

Yeah I'm pretty sure I'm right, Thanks for your help :D :D And I figured out the last question thank you :D

7. May 24, 2010

### ehild

You are welcome :)

ehild