# Speed of Earth at 41 degrees North.

1. Aug 23, 2007

### hawk320

1. The problem statement, all variables and given/known data
The earth's radius is 6370 km, if you are sitting in a classroom on the equator how quickly would you be traveling? If you were in NY which has a latitude of 41 degrees N, how quickly would you be traveling?

2. Relevant equations
2(pi)r=circumference

3. The attempt at a solution
I understand that you need to find the circumference and divide it by 24(hours). This gives you km/hr. So at the equator you would be going t about 1.67 * 10^3 km/hr. I don't understand how to find the circumference at 41 degrees N.

2. Aug 23, 2007

### G01

Think of the Earth as being made up of a whole lot of circles stacked on top of one another. The largest circle is the one at the equator, whose circumference you already found. The circle at 41 degrees north is smaller, and has a smaller radius correct? Now all you need is this radius. In order to find it, you are going to needs that angle.

HINT: Think triangles. Draw a circle representing the Earth. Now, draw in the radius at the equator and a smaller line above it representing the radius at 41 degrees north of the equator. Can you relate the lengths of these two radii using that angle of 41 degrees?

3. Aug 23, 2007

### Dick

It would be the x-component of a vector with length the radius of the earth and inclined 41 degrees to the x-axis. Can you find that?

4. Aug 23, 2007

### hawk320

I tried doing it as a right triangle which gave me the equation x(the radius at 41 degrees)= 6370/cos(41 degrees). The problem with that is that I get the speed of it to be 2.21 * 10^3km/hr, which is faster than at the equator. Which I believe is not possible because the equator should be the fastest (unless I am wrong about that).

5. Aug 23, 2007

### Dick

The earth's radius is the hyptenuse of your triangle. cos(theta)=adjacent/hypotenuse. The cos function shouldn't be in the denominator.

6. Aug 23, 2007

### hawk320

I got that from cos41 = 6730/x which is equal to cos41x = 6730 which is then equal to x=6730/cos41.

7. Aug 23, 2007

### G01

$$\cos 41 =\frac{x}{6730}$$

I think you may have your angle in the wrong place. The angle of 41 degrees is between the horizontal (equator line on that circle representing the Earth) and the hypotenuse. Then, using the properties of parallel lines, the angle in the triangle adjacent to the "x" side would also be 41 degrees.

Last edited: Aug 23, 2007
8. Aug 23, 2007

### mgb_phys

No, remember the 41deg is the angle between the equator(horizontal) and the line to the latitude of your classroom.

9. Aug 23, 2007

### hawk320

Thank you, the problem I was having was that I was solving for the wrong radius. Also what program do you use to type up the equations?

Last edited: Aug 23, 2007
10. Aug 23, 2007

### G01

Anytime. Glad you figured it out.

11. Aug 23, 2007

LaTex. There is a template in the menu above. It appears as a Summation symbol (sigma).

You can also click on any equation in any thread and a new window will open with the codes. That is how I am learning Latex. I think there is a thread where you can test your code using skills somewhere. I'll look around for it.

Casey

12. Aug 23, 2007

### hawk320

Thank you, that will make writing equations a lot easier.