What is the centripetal acceleration of the Earth in its orbit around the Sun?

[curiousgeorge99]

Homework Statement

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.

Homework Equations

v= 2 pi r/ T
Ac = V^2/ r

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?

 

[Chi Meson]

I solve for V and get 29.48 m/s .

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

 

[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

 

[D H]

Taking 148022 KM

Where did you get that number? The moon is 385000 km away from the Earth.

The Google calculator is quite amazing:
http://www.google.com/search?hl=en&client=safari&rls=en&q=(1AU*2*pi/yr)^2/1AU&btnG=Search"

 

[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?

 

[D H]

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.

 

[curiousgeorge99]

Thank you so much!

I see my mistake.