# Homework Help: Centripetal Acceleration

1. Jun 4, 2007

### curiousgeorge99

1. The problem statement, all variables and given/known data
Calculate the centripetal acceleration, in units of m/s2, of the Earth in its orbit around the Sun. Assume that the Earth's orbit is a circle of radius 148,022 thousand km.

2. Relevant equations
v= 2 pi r/ T
Ac = V^2/ r

3. The attempt at a solution
I solve for V and get 29.48 m/s . For T, I use 3.15 x 10^7 seconds, and for r I use 1.48022 x 10^9

When I solve for Ac, I get 5.87 X 10-6 m/s2
I square 29.48 m/s and divide by 1.48022 x 10^9

The answer is 5.85 x 10-3 m/s2. I have no idea where I'm going wrong! I'm more concerned about why my answer is 10-6 when it should be 10-3 . Any ideas?

2. Jun 4, 2007

### Chi Meson

Do you really think the earth is travelling 30 m/s? 66 mph? That's not even speeding on a US highway. Might want to check your v again.

Last edited: Jun 4, 2007
3. Jun 4, 2007

### curiousgeorge99

That's true, it doesn't sound right does it?

However, if the answer is supposed to be m/s2 in the end, wouldn't my calculation for V still be correct? Taking 148022 KM and converting to meters becomes 148022000 M. When multiplied by 6.28, I then get 9.3 X10^8. The number of seconds in a year is 3.156 x 10^7.

So: 9.3 X10^8 / 3.156 x 10^7 = 29.5 m/s2

4. Jun 4, 2007

### D H

Staff Emeritus
Last edited by a moderator: Apr 22, 2017
5. Jun 4, 2007

### curiousgeorge99

I just copied the question from the book. Didn't see anything about the moon. Do I need to consider that for this question?

6. Jun 5, 2007

### D H

Staff Emeritus
No. I highlighted 148022 km because that number is the source your error. It is wrong. I simply used the Earth-moon distance to illustrate that you have the wrong value.

7. Jun 5, 2007

### curiousgeorge99

Thank you so much!!!

I see my mistake.