# Math help!

1. Aug 28, 2007

### Confused_much

I'm stuck on this math problem for my physics class, I'm sure it's not that difficult, but it confuses me:

The Average distance between the Earth and the sun is 1.50 X 10^8 km.
a. Calculate the average speed in km/h of earth assuming a circular path about the sun. Use the equation v= 2TTr/T.

Last edited: Aug 28, 2007
2. Aug 28, 2007

### cristo

Staff Emeritus
Well, what have you tried? Have you tried using the equation for the first part? Do you know what each symbol represents? Please note that you must show working before we can help you with homework/coursework questions-- forum rules.

3. Aug 28, 2007

### Confused_much

Oh, ok.

Um...let's see...I know what the symbols stand for, it's just that I don't know if by r in this problem means the radius of the earth, and I don't even know what the time it takes.

4. Aug 28, 2007

### robert Ihnot

What figure do you have? Isn't it 1.5x10^8. Where does that fit in? The equation ought to have been copied better, its $$2\pi r$$, where you seem to have copied 2TTr.

How is time measured? What time is important here?

5. Aug 28, 2007

### Confused_much

Well, time is measured in seconds, but in this case, they want it in hours, right? The important time in this case is how long it will take for the Earth to go around the sun if their distance is 1.5x10^8.

6. Aug 29, 2007

### robert Ihnot

You are making progress, keep going!

7. Aug 29, 2007

### HallsofIvy

Here's a hint! The time it takes the earth to go around the sun is one year!

"Use the equation v= 2TTr/T" Frankly, this doesn't make much sense. You say that you know what the symbols mean but you haven't told us. I would think that
TTr/T= Tr. ??

8. Aug 29, 2007

### Nesk

I'm pretty sure that by TT, the OP means $$\pi$$.

In order to use the equation provided to gain the correct result, is is necessary to assume that Earth is in a circular orbit around the Sun. This is not exactly true, but it is a viable approximation for basic purposes.

With this in mind, consider how the numerator in the equation relates to Earth's orbit around the Sun.
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}