What is the Radial Acceleration of a Painting at 38.9° North?

[trainumc]

Homework Statement

Objects that are at rest relative to Earth's surface are in circular motion due to Earth's rotation. What is the radial acceleration of a painting hanging in a museum at a latitude of ϕ = 38.9° North? (Note that the object's radial acceleration is not directed toward the center of the Earth.)

Homework Equations

ac=v^2/r v=(4*Pi^2*r)/t^2

The Attempt at a Solution

ac=4Pi*2.3E13/576
im not sure what I'm doing wrong, any help? look at the attachment for more information

 

[Hootenanny]

Could you perhaps provide a little more detail in your solution. I'm not quite sure where your pulling some of those numbers from.

 

[nasu]

You forgot the square for pi (it's pi^2) and the units are mixed up.
You use the period in hours. I cannot figure out what unit you use for the radius...
Earth's radius is about 6400km (6.4 10^6 m). To get 10^13 we should measure it in fractions of microns or something...

 

[Hootenanny]

You forgot the square for pi (it's pi^2) and the units are mixed up.
You use the period in hours. I cannot figure out what unit you use for the radius...
Earth's radius is about 6400km (6.4 10^6 m). To get 10^13 we should measure it in fractions of microns or something...

The radius of the Earth is not necessarily the radius of the circular path, in this case the radius of the path is significantly less than the radius of the earth.

 

[nasu]

The radius of the Earth is not necessarily the radius of the circular path, in this case the radius of the path is significantly less than the radius of the earth.

No doubt. But at this stage I was just trying to do some estimate to figure out where that big number comes from.
Maybe the author will tell us himself.